Cuban PrimesIs this number a prime?List Prime NumbersLeyland NumbersExcessive IntegersClosest Woodall PrimePalindromic Primes without 11Is it a weak prime?Primes other than OptimusPrimus-Orderus PrimesRepeated Digit PrimesBertrand's PrimesSome Lonely Primes

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Cuban Primes

Is this number a prime?List Prime NumbersLeyland NumbersExcessive IntegersClosest Woodall PrimePalindromic Primes without 11Is it a weak prime?Primes other than OptimusPrimus-Orderus PrimesRepeated Digit PrimesBertrand's PrimesSome Lonely Primes

Given a natural number $n$, return the $n$-th cuban prime.

Cuban Primes

A cuban prime is a prime number of the form

$$p = fracx^3-y^3x-y$$

where $y>0$ and $x = 1+y$ or $x = 2+y$

Details

You may use 0 or 1 based indexing, whatever suits you best.

You can return the $n$-th prime given the index $n$ or the first $n$ primes in increasing order, or alternatively you can return an infinite list/generator that produces the primes in increasing order.

Test cases

The first few terms are following:

(#1-13) 7, 13, 19, 37, 61, 109, 127, 193, 271, 331, 397, 433, 547,
(#14-24) 631, 769, 919, 1201, 1453, 1657, 1801, 1951, 2029, 2269, 2437,
(#25-34) 2791, 3169, 3469, 3571, 3889, 4219, 4447, 4801, 5167, 5419,
(#35-43) 6211, 7057, 7351, 8269, 9241, 10093, 10267, 11719, 12097,
(#44-52) 12289, 13267, 13669, 13873, 16651, 18253, 19441, 19927, 20173

More terms can be found on OEIS: They are split up in two sequences, depending on wheter $x = 1+y $ or $x = 2+y$: A002407 and A002648

edited May 14 at 18:28

asked May 14 at 13:10

flawr

27.8k670198

2

$begingroup$
Can we return the first n primes not sorted?
$endgroup$
– J42161217
May 14 at 13:29

$begingroup$
@J42161217 No, the primes should be in increasing order.
$endgroup$
– flawr
May 14 at 14:30

add a comment |

Given a natural number $n$, return the $n$-th cuban prime.

Cuban Primes

A cuban prime is a prime number of the form

$$p = fracx^3-y^3x-y$$

where $y>0$ and $x = 1+y$ or $x = 2+y$

Details

You may use 0 or 1 based indexing, whatever suits you best.

You can return the $n$-th prime given the index $n$ or the first $n$ primes in increasing order, or alternatively you can return an infinite list/generator that produces the primes in increasing order.

Test cases

The first few terms are following:

(#1-13) 7, 13, 19, 37, 61, 109, 127, 193, 271, 331, 397, 433, 547,
(#14-24) 631, 769, 919, 1201, 1453, 1657, 1801, 1951, 2029, 2269, 2437,
(#25-34) 2791, 3169, 3469, 3571, 3889, 4219, 4447, 4801, 5167, 5419,
(#35-43) 6211, 7057, 7351, 8269, 9241, 10093, 10267, 11719, 12097,
(#44-52) 12289, 13267, 13669, 13873, 16651, 18253, 19441, 19927, 20173

More terms can be found on OEIS: They are split up in two sequences, depending on wheter $x = 1+y $ or $x = 2+y$: A002407 and A002648

edited May 14 at 18:28

asked May 14 at 13:10

flawr

27.8k670198

2

$begingroup$
Can we return the first n primes not sorted?
$endgroup$
– J42161217
May 14 at 13:29

$begingroup$
@J42161217 No, the primes should be in increasing order.
$endgroup$
– flawr
May 14 at 14:30

add a comment |

Given a natural number $n$, return the $n$-th cuban prime.

Cuban Primes

A cuban prime is a prime number of the form

$$p = fracx^3-y^3x-y$$

where $y>0$ and $x = 1+y$ or $x = 2+y$

Details

You may use 0 or 1 based indexing, whatever suits you best.

You can return the $n$-th prime given the index $n$ or the first $n$ primes in increasing order, or alternatively you can return an infinite list/generator that produces the primes in increasing order.

Test cases

The first few terms are following:

(#1-13) 7, 13, 19, 37, 61, 109, 127, 193, 271, 331, 397, 433, 547,
(#14-24) 631, 769, 919, 1201, 1453, 1657, 1801, 1951, 2029, 2269, 2437,
(#25-34) 2791, 3169, 3469, 3571, 3889, 4219, 4447, 4801, 5167, 5419,
(#35-43) 6211, 7057, 7351, 8269, 9241, 10093, 10267, 11719, 12097,
(#44-52) 12289, 13267, 13669, 13873, 16651, 18253, 19441, 19927, 20173

More terms can be found on OEIS: They are split up in two sequences, depending on wheter $x = 1+y $ or $x = 2+y$: A002407 and A002648

edited May 14 at 18:28

asked May 14 at 13:10

flawr

27.8k670198

Given a natural number $n$, return the $n$-th cuban prime.

Cuban Primes

A cuban prime is a prime number of the form

$$p = fracx^3-y^3x-y$$

where $y>0$ and $x = 1+y$ or $x = 2+y$

Details

You may use 0 or 1 based indexing, whatever suits you best.

You can return the $n$-th prime given the index $n$ or the first $n$ primes in increasing order, or alternatively you can return an infinite list/generator that produces the primes in increasing order.

Test cases

The first few terms are following:

(#1-13) 7, 13, 19, 37, 61, 109, 127, 193, 271, 331, 397, 433, 547,
(#14-24) 631, 769, 919, 1201, 1453, 1657, 1801, 1951, 2029, 2269, 2437,
(#25-34) 2791, 3169, 3469, 3571, 3889, 4219, 4447, 4801, 5167, 5419,
(#35-43) 6211, 7057, 7351, 8269, 9241, 10093, 10267, 11719, 12097,
(#44-52) 12289, 13267, 13669, 13873, 16651, 18253, 19441, 19927, 20173

More terms can be found on OEIS: They are split up in two sequences, depending on wheter $x = 1+y $ or $x = 2+y$: A002407 and A002648

code-golf math number sequence primes

edited May 14 at 18:28

asked May 14 at 13:10

flawr

27.8k670198

edited May 14 at 18:28

asked May 14 at 13:10

flawr

27.8k670198

edited May 14 at 18:28

asked May 14 at 13:10

flawr

27.8k670198

asked May 14 at 13:10

flawr

27.8k670198

asked May 14 at 13:10

flawr

27.8k670198

2

$begingroup$
Can we return the first n primes not sorted?
$endgroup$
– J42161217
May 14 at 13:29

$begingroup$
@J42161217 No, the primes should be in increasing order.
$endgroup$
– flawr
May 14 at 14:30

add a comment |

2

$begingroup$
Can we return the first n primes not sorted?
$endgroup$
– J42161217
May 14 at 13:29

$begingroup$
@J42161217 No, the primes should be in increasing order.
$endgroup$
– flawr
May 14 at 14:30

Can we return the first n primes not sorted?

– J42161217
May 14 at 13:29

@J42161217 No, the primes should be in increasing order.

– flawr
May 14 at 14:30

add a comment |

16 Answers
16

active

oldest

votes

JavaScript (V8), 54 bytes

A full program that prints cuban primes forever.

for(x=0;;)for(k=N=~(3/4*++x*x);N%++k;);~k

Try it online!

^{NB: Unless you have infinite paper in your printer, do not attempt to run this in your browser console, where print() may have a different meaning.}

JavaScript (ES6), 63 61 60 59 bytes

Returns the $n$-th cuban prime, 1-indexed.

f=(n,x)=>(p=k=>N%++k?p(k):n-=!~k)(N=~(3/4*x*x))?f(n,-~x):-N

Try it online!

How?

This is based on the fact that cuban primes are primes of the form:

$$p_n=leftlfloorfrac3n^24rightrfloor+1,;nge3$$

The above formula can be written as:

$$p_n=begincases
dfrac3n^2+14;text if ntext is odd\
dfrac3n^2+44;text if ntext is even
endcases
$$

or for any $y>0$:

$$p_2y+1=dfrac3(2y+1)^2+14=3y^2+3y+1$$
$$p_2y+2=dfrac3(2y+2)^2+44=3y^2+6y+4$$

which is $dfracx^3-y^3x-y$ for $x=y+1$ and $x=y+2$ respectively.

edited May 15 at 11:41

answered May 14 at 13:30

Arnauld

85.1k7100349

add a comment |

05AB1E, 16 12 9 bytes

Generates an infinite list.

Saved 4 bytes with Kevin Cruijssen's port of Arnaulds formula.

Saved another 3 bytes thanks to Grimy

∞n3*4÷>ʒp

Try it online!

Explanation

∞ # on the list of infinite positive integers
 n3*4÷> # calculate (3*N^2)//4+1 for each
 ʒp # and filter to only keep primes

edited May 14 at 15:54

answered May 14 at 13:33

Emigna

49.5k534150

$begingroup$
You've made a typo in your explanation: "put a copy of N^2+3 on the stack" should be 3*N^2. Also, why the ) instead of ¯? Because it's easier to type? And for some reason I have the feeling the NnN‚3*¬sO‚ can be 1 byte shorter, but I'm not seeing it. A slight equal-byte alternative is Nn3*DN3*+‚. But I'm probably just seeing things that aren't there.. ;) Nice answer, so +1 from me.
$endgroup$
– Kevin Cruijssen
May 14 at 14:30

1

$begingroup$
I actually tried to port my answer to 05AB1E, but failed miserably. :D
$endgroup$
– Arnauld
May 14 at 15:09

1

$begingroup$
Actually, generating an infinite list is more convenient: 9 bytes with ∞n3*4÷>ʒp
$endgroup$
– Grimy
May 14 at 15:45

1

$begingroup$
OK, I'm not used to specs that contradict themselves. :-)
$endgroup$
– WGroleau
May 14 at 21:11

6

$begingroup$
@WGroleau I assume you've never developed software professionally then. I'm more concerned when I get specs that don't contradict themselves.
$endgroup$
– MikeTheLiar
May 14 at 21:29

|
show 7 more comments

R, 75 73 bytes

n=scan()
while(F<n)F=F+any(!(((T<-T+1)*1:4-1)/3)^.5%%1)*all(T%%(3:T-1))
T

Try it online!

-2 bytes by noticing that I can remove brackets if I use * instead of & (different precedence).

Outputs the nth Cuban prime (1-indexed).

It uses the fact (given in OEIS) that Cuban primes are of the form $p=1+3n^2$ or $4p=1+3n^2$ for some $n$, i.e. $n=sqrtfracacdot p-13$ is an integer for $a=1$ or $a=4$.

The trick is that no prime can be of the form $2p=1+3n^2$ or $3p=1+3n^2$ (*), so we can save 2 bytes by checking the formula for $ain1, 2, 3, 4$ (1:4) instead of $ain1, 4$ (c(1,4)).

Slightly ungolfed version of the code:

# F and T are implicitly initialized at 0 and 1
# F is number of Cuban primes found so far
# T is number currently being tested for being a Cuban prime
n = scan() # input
while(F<n)
 T = T+1 # increment T 
 F = F + # increment F if
 (!all(((T*1:4-1)/3)^.5 %% 1) # there is an integer of the form sqrt(((T*a)-1)/3)
 & all(T%%(3:T-1))) # and T is prime (not divisible by any number between 2 and T-1)
 
T # output T

(*) No prime can be of the form $3p=1+3n^2$, else $1=3(p-n^2)$ would be divisible by $3$.

No prime other than $p=2$ (which isn't a Cuban prime) can of the form $2p=1+3n^2$: $n$ would need to be odd, i.e. $n=2k+1$. Expanding gives $2p=4+12k(k+1)$, hence $p=2+6k(k+1)$ and $p$ would be even.

edited May 14 at 15:55

answered May 14 at 14:32

Robin Ryder

1,288215

$begingroup$
what about avoiding a loop by using an upper bound on the nth Cuban prime?
$endgroup$
– Xi'an
May 15 at 7:31

$begingroup$
@Xi'an I thought about that, but couldn't come up with such a bound. Do you have one?
$endgroup$
– Robin Ryder
May 15 at 13:32

add a comment |

Wolfram Language (Mathematica), 66 65 56 bytes

(f=1+⌊3#/4#⌋&;For[n=i=0,i<#,PrimeQ@f@++n&&i++];f@n)&

Try it online!

J42161217 -1 by using ⌊ ⌋ instead of Floor[ ]

attinat
- -1 by using ⌊3#/4#⌋ instead of ⌊3#^2/4⌋
- -8 for For[n=i=0,i<#,PrimeQ@f@++n&&i++] instead of n=2;i=#;While[i>0,i-=Boole@PrimeQ@f@++n]

edited May 14 at 22:56

answered May 14 at 18:27

speedstyle

695

1

$begingroup$
65 bytes. Welcome to ppcg. Nice first answer! +1
$endgroup$
– J42161217
May 14 at 18:39

$begingroup$
Thanks! (Long time lurker.) I couldn't quite parse your existing answer so I wrote my own and it came out a little shorter. I might do a Python one too.
$endgroup$
– speedstyle
May 14 at 18:57

2

$begingroup$
56 bytes
$endgroup$
– attinat
May 14 at 21:54

$begingroup$
@attinat I thought Arnauld's formula only worked for n>2 so I didn't start with 0 - although as in your example it works for all n (because it starts 1 1 4 7 13 ... so the primes are 7 13 ...)
$endgroup$
– speedstyle
May 14 at 22:33

add a comment |

Java 8, 94 88 86 84 bytes

v->for(int i=3,n,x;;System.out.print(x<1?++n+" ":""))for(x=n=i*i++*3/4;~n%x--<0;);

-6 bytes by using the Java prime-checker of @SaraJ, so make sure to upvote her!

-2 bytes thanks to @OlivierGrégoire. Since the first number we check is 7, we can drop the trailing %n from Sara's prime-checker, which is to terminate the loop for n=1.

-2 bytes thanks to @OlivierGrégoire by porting @Arnauld's answer.

Outputs space-delimited indefinitely.

Try it online.

Explanation (of the old 86 bytes version): TODO: Update explanation

Uses the formula of @Arnauld's JavaScript answer: $p_n=leftlfloorfrac3n^24rightrfloor+1,;nge3$.

v-> // Method with empty unused parameter and no return-type
 for(int i=3, // Loop-integer, starting at 3
 n,x // Temp integers
 ; // Loop indefinitely:
 ; // After every iteration:
 System.out.print( // Print:
 n==x? // If `n` equals `x`, which means `n` is a prime:
 n+" " // Print `n` with a space delimiter
 : // Else:
 "")) // Print nothing
 for(n=i*i++*3/4+1, // Set `n` to `(3*i^2)//4+1
 // (and increase `i` by 1 afterwards with `i++`)
 x=1; // Set `x` to 1
 n%++x // Loop as long as `n` modulo `x+1`
 // (after we've first increased `x` by 1 with `++x`)
 >0;); // is not 0 yet
 // (if `n` is equal to `x`, it means it's a prime)

edited May 16 at 15:43

answered May 15 at 6:48

Kevin Cruijssen

44.9k576225

$begingroup$
I don't really think it's feasible, but another way of finding the cuban primes uses this formula: v->for(int n=7,i=3,p,x,d,r=0;;i+=++r%2*3,n+=i,System.out.print(x>1?x+" ":""))for(x=n,d=1;++d<n;x=x%d<1?0:n);, maybe someone can use this to golf? I couldn't.
$endgroup$
– Olivier Grégoire
May 16 at 9:03

1

$begingroup$
@OlivierGrégoire You can golf yours a bit more by removing the unused ,p and changing i+=++r%2*3,n+=i to n+=i+=++r%2*3, but then I'll still end up at 106 bytes. Using Java 11's String#repeat with prime-regex is 105 bytes: v->(..+?)\1+"))System.out.println(n);.
$endgroup$
– Kevin Cruijssen
May 16 at 9:40

$begingroup$
Yeah, I guessed it wasn't much golfable despite my (now obvious) mistakes. Thanks for giving it a ride ;)
$endgroup$
– Olivier Grégoire
May 16 at 14:02

$begingroup$
@OlivierGrégoire Maybe also good to know for you, but there is apparently a shorter prime-check loop in Java. See my edit and SaraJ's prime-check answer.
$endgroup$
– Kevin Cruijssen
May 16 at 14:30

$begingroup$
I might be wrong, but the last %n isn't required, is it?
$endgroup$
– Olivier Grégoire
May 16 at 14:44

|
show 2 more comments

Wolfram Language (Mathematica), 83 bytes

(t=1;While[Length[l=Select[Join@@Array[(v=3#^2+1)+3#,v&,t++],PrimeQ]]<#];Sort@l)&

Try it online!

edited May 14 at 14:40

answered May 14 at 13:31

J42161217

15.1k21457

add a comment |

Jelly, 12 bytes

²×3:4‘
ÇẒ$#Ç

Try it online!

Based on @Arnauld’s method. Takes n on stdin and returns that many Cuban primes.

answered May 14 at 19:11

Nick Kennedy

2,56469

add a comment |

Wolfram Language (Mathematica), 83 bytes

This solution will output the n-th Cuban prime with the added benefits of being fast and remembering all previous results in the symbol f.

(d:=1+3y(c=1+y)+3b c;e:=If[PrimeQ@d,n++;f@n=d];For[n=y=b=0,n<#,e;b=1-b;e,y++];f@#)&

Try it online!

answered May 14 at 18:21

Kelly Lowder

3,050417

add a comment |

Whitespace, 180 bytes

[S S S T S N
_Push_2][S N
S _Duplicate][N
S S N
_Create_Label_OUTER_LOOP][S N
N
_Discard_top_stack][S S S T N
_Push_1][T S S S _Add][S N
S _Duplicate][S N
S _Duplicate][T S S N
_Multiply][S S S T T N
_Push_3][T S S N
_Multiply][S S S T S S N
_Push_4][T S T S _Integer_divide][S S S T N
_Push_1][T S S S _Add][S S S T N
_Push_1][S N
S _Duplicate_1][N
S S S N
_Create_Label_INNER_LOOP][S N
N
_Discard_top_stack][S S S T N
_Push_1][T S S S _Add][S N
S _Duplicate][S N
S _Duplicate][S T S S T T N
_Copy_0-based_3rd][T S S T _Subtract][N
T S T N
_Jump_to_Label_PRINT_if_0][S T S S T S N
_Copy_0-based_2nd][S N
T _Swap_top_two][T S T T _Modulo][S N
S _Duplicate][N
T S S S N
_Jump_to_Label_FALSE_if_0][N
S N
S N
_Jump_to_Label_INNER_LOOP][N
S S T N
_Create_Label_PRINT][T N
S T _Print_as_integer][S S S T S T S N
_Push_10_(newline)][T N
S S _Print_as_character][S N
S _Duplicate][N
S S S S N
_Create_Label_FALSE][S N
N
_Discard_top_stack][S N
N
_Discard_top_stack][N
S N
N
_Jump_to_Label_OUTER_LOOP]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Outputs newline-delimited indefinitely.

Try it online (with raw spaces, tabs, and new-lines only).

Explanation in pseudo-code:

Port of my Java 8 answer, which also uses the formula from @Arnauld's JavaScript answer: $p_n=leftlfloorfrac3n^24rightrfloor+1,;nge3$.

Integer i = 2
Start OUTER_LOOP:
 i = i + 1
 Integer n = i*i*3//4+1
 Integer x = 1
 Start INNER_LOOP:
 x = x + 1
 If(x == n):
 Call function PRINT
 If(n % x == 0):
 Go to next iteration of OUTER_LOOP
 Go to next iteration of INNER_LOOP

function PRINT:
 Print integer n
 Print character 'n'
 Go to next iteration of OUTER_LOOP

answered May 15 at 8:01

Kevin Cruijssen

44.9k576225

add a comment |

Python 3, 110 108 102 bytes

Similar method to my Mathematica answer (i.e. isPrime(1+⌊¾n²⌋) else n++)
using this golfed prime checker and returning an anonymous infinite generator

from itertools import*
(x for x in map(lambda n:1+3*n**2//4,count(2)) if all(x%j for j in range(2,x)))

Try it online!

mypetlion -2 because arguably anonymous generators are more allowed than named ones

-6 by starting count at 2 ₊₁ so that the and x>1 in the prime checker I borrowed is unnecessary _-7

edited May 15 at 9:28

answered May 14 at 20:44

speedstyle

695

$begingroup$
The answer going into a variable is usually not considered a valid form of "output". Could you rework your answer so that the result is either output to stdout or returned by a function?
$endgroup$
– mypetlion
May 14 at 23:01

1

$begingroup$
since anonymous functions are allowed, and the challenge explicitly allows an infinite generator, I've removed g=. I had only included it in the first place because it allowed a quick visual on TIO with print(next(g) for i in range(52)).
$endgroup$
– speedstyle
May 15 at 1:15

add a comment |

Japt, 14 13 bytes

Adapted from Arnauld's formula. 1-indexed.

@µXj}f@Ò(X²*¾

Try it

1 byte saved thanks to EmbodimentOfIgnorance.

edited May 15 at 9:46

answered May 14 at 21:25

Shaggy

19.5k31768

$begingroup$
13 bytes? Not tested thoroughly though.
$endgroup$
– Embodiment of Ignorance
May 15 at 2:15

$begingroup$
Thanks, @EmbodimentofIgnorance. I'd tried that but it didn't work; turns out I'd forgotten the (.
$endgroup$
– Shaggy
May 15 at 9:47

add a comment |

Racket, 124 bytes

(require math)(define(f n[i 3])(let([t(+(exact-floor(* 3/4 i i))1)][k(+ 1 i)])(if(prime? t)(if(= 0 n)t(f(- n 1)k))(f n k))))

Try it online!

Returns the n-th cuban prime, 0-indexed.

Uses the formula of @Arnauld's JavaScript answer

edited May 15 at 10:13

answered May 15 at 7:46

Galen Ivanov

8,25711137

add a comment |

Python 3, 83 bytes

prints the cuban primes forever.

P=k=1
while 1:P*=k*k;x=k;k+=1;P%k>0==((x/3)**.5%1)*((x/3+.25)**.5%1-.5)and print(k)

Try it online!

Based on this prime generator. For every prime it checks whether an integer y exists that fulfills the equation for either $x = 1+y$ or $x=2+y$.

$$ p=frac(1+y)^3-y^3(1+y)-y = 1 + 3y +3y^2 Leftrightarrow y = -frac12pmsqrtfrac14+fracp-13$$

$$ p=frac(2+y)^3-y^3(1+y)-y = 4 + 6y +3y^2 Leftrightarrow y = -1 pmsqrtfracp-13$$
As we only care whether $y$ has an integer solution, we can ignore the $pm$ and $-1$.

edited May 15 at 14:46

answered May 15 at 14:32

ovs

19.7k21161

add a comment |

Perl 6, 33 31 bytes

-2 bytes thanks to Grimy

¾*$++²xx*

Try it online!

Anonymous code block that returns a lazy infinite list of Cuban primes. This uses Arnauld's formula to generate possible cuban primes, then &is-prime to filter them.

Explanation:

 # Anonymous code block
 grep &is-prime, # Filter the primes from
 xx* # The infinite list
 ¾* # Of three quarters
 $++² # Of an increasing number squared
 1+| # Add one by ORing with 1

edited May 16 at 11:14

answered May 16 at 5:05

Jo King

28.4k366134

1

$begingroup$
1+0+| can be just 1+|
$endgroup$
– Grimy
May 16 at 10:49

add a comment |

Pari/GP, 51 bytes

Using Arnauld's formula.

n->a=0;for(i=1,n,until(isprime(p=3*a^24+1),a++));p

Try it online!

answered May 16 at 4:46

alephalpha

22.4k33096

add a comment |

APL(NARS), 98 chars, 196 bytes

r←h w;y;c;v
r←c←y←0⋄→4
→3×⍳∼0πv←1+3×y×1+y+←1⋄r←v⋄→0×⍳w≤c+←1
→2×⍳∼0πv+←3×y+1⋄c+←1⋄r←v
→2×⍳w>c

indented :

r←h w;y;c;v
r←c←y←0⋄→4
 →3×⍳∼0πv←1+3×y×1+y+←1⋄r←v⋄→0×⍳w≤c+←1
 →2×⍳∼0πv+←3×y+1⋄c+←1⋄r←v
 →2×⍳w>c

test:

 h ¨1..20
7 13 19 37 61 109 127 193 271 331 397 433 547 631 769 919 1201 1453 1657 1801 
 h 1000
25789873
 h 10000
4765143511

it is based on: if y in N, one possible Cuban Prime is

S1=1+3y(y+1)

the the next possible Cuban Prime will be

S2=3(y+1)+S1

edited May 23 at 10:04

answered May 16 at 8:33

RosLuP

2,346514

add a comment |

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JavaScript (V8), 54 bytes

A full program that prints cuban primes forever.

for(x=0;;)for(k=N=~(3/4*++x*x);N%++k;);~k

Try it online!

^{NB: Unless you have infinite paper in your printer, do not attempt to run this in your browser console, where print() may have a different meaning.}

JavaScript (ES6), 63 61 60 59 bytes

Returns the $n$-th cuban prime, 1-indexed.

f=(n,x)=>(p=k=>N%++k?p(k):n-=!~k)(N=~(3/4*x*x))?f(n,-~x):-N

Try it online!

How?

This is based on the fact that cuban primes are primes of the form:

$$p_n=leftlfloorfrac3n^24rightrfloor+1,;nge3$$

The above formula can be written as:

$$p_n=begincases
dfrac3n^2+14;text if ntext is odd\
dfrac3n^2+44;text if ntext is even
endcases
$$

or for any $y>0$:

$$p_2y+1=dfrac3(2y+1)^2+14=3y^2+3y+1$$
$$p_2y+2=dfrac3(2y+2)^2+44=3y^2+6y+4$$

which is $dfracx^3-y^3x-y$ for $x=y+1$ and $x=y+2$ respectively.

edited May 15 at 11:41

answered May 14 at 13:30

Arnauld

85.1k7100349

add a comment |

JavaScript (V8), 54 bytes

A full program that prints cuban primes forever.

for(x=0;;)for(k=N=~(3/4*++x*x);N%++k;);~k

Try it online!

^{NB: Unless you have infinite paper in your printer, do not attempt to run this in your browser console, where print() may have a different meaning.}

JavaScript (ES6), 63 61 60 59 bytes

Returns the $n$-th cuban prime, 1-indexed.

f=(n,x)=>(p=k=>N%++k?p(k):n-=!~k)(N=~(3/4*x*x))?f(n,-~x):-N

Try it online!

How?

This is based on the fact that cuban primes are primes of the form:

$$p_n=leftlfloorfrac3n^24rightrfloor+1,;nge3$$

The above formula can be written as:

$$p_n=begincases
dfrac3n^2+14;text if ntext is odd\
dfrac3n^2+44;text if ntext is even
endcases
$$

or for any $y>0$:

$$p_2y+1=dfrac3(2y+1)^2+14=3y^2+3y+1$$
$$p_2y+2=dfrac3(2y+2)^2+44=3y^2+6y+4$$

which is $dfracx^3-y^3x-y$ for $x=y+1$ and $x=y+2$ respectively.

edited May 15 at 11:41

answered May 14 at 13:30

Arnauld

85.1k7100349

add a comment |

JavaScript (V8), 54 bytes

A full program that prints cuban primes forever.

for(x=0;;)for(k=N=~(3/4*++x*x);N%++k;);~k

Try it online!

^{NB: Unless you have infinite paper in your printer, do not attempt to run this in your browser console, where print() may have a different meaning.}

JavaScript (ES6), 63 61 60 59 bytes

Returns the $n$-th cuban prime, 1-indexed.

f=(n,x)=>(p=k=>N%++k?p(k):n-=!~k)(N=~(3/4*x*x))?f(n,-~x):-N

Try it online!

How?

This is based on the fact that cuban primes are primes of the form:

$$p_n=leftlfloorfrac3n^24rightrfloor+1,;nge3$$

The above formula can be written as:

$$p_n=begincases
dfrac3n^2+14;text if ntext is odd\
dfrac3n^2+44;text if ntext is even
endcases
$$

or for any $y>0$:

$$p_2y+1=dfrac3(2y+1)^2+14=3y^2+3y+1$$
$$p_2y+2=dfrac3(2y+2)^2+44=3y^2+6y+4$$

which is $dfracx^3-y^3x-y$ for $x=y+1$ and $x=y+2$ respectively.

edited May 15 at 11:41

answered May 14 at 13:30

Arnauld

85.1k7100349

JavaScript (V8), 54 bytes

A full program that prints cuban primes forever.

for(x=0;;)for(k=N=~(3/4*++x*x);N%++k;);~k

Try it online!

^{NB: Unless you have infinite paper in your printer, do not attempt to run this in your browser console, where print() may have a different meaning.}

JavaScript (ES6), 63 61 60 59 bytes

Returns the $n$-th cuban prime, 1-indexed.

f=(n,x)=>(p=k=>N%++k?p(k):n-=!~k)(N=~(3/4*x*x))?f(n,-~x):-N

Try it online!

How?

This is based on the fact that cuban primes are primes of the form:

$$p_n=leftlfloorfrac3n^24rightrfloor+1,;nge3$$

The above formula can be written as:

$$p_n=begincases
dfrac3n^2+14;text if ntext is odd\
dfrac3n^2+44;text if ntext is even
endcases
$$

or for any $y>0$:

$$p_2y+1=dfrac3(2y+1)^2+14=3y^2+3y+1$$
$$p_2y+2=dfrac3(2y+2)^2+44=3y^2+6y+4$$

which is $dfracx^3-y^3x-y$ for $x=y+1$ and $x=y+2$ respectively.

edited May 15 at 11:41

answered May 14 at 13:30

Arnauld

85.1k7100349

edited May 15 at 11:41

answered May 14 at 13:30

Arnauld

85.1k7100349

answered May 14 at 13:30

Arnauld

85.1k7100349

answered May 14 at 13:30

Arnauld

85.1k7100349

add a comment |

05AB1E, 16 12 9 bytes

Generates an infinite list.

Saved 4 bytes with Kevin Cruijssen's port of Arnaulds formula.

Saved another 3 bytes thanks to Grimy

∞n3*4÷>ʒp

Try it online!

Explanation

∞ # on the list of infinite positive integers
 n3*4÷> # calculate (3*N^2)//4+1 for each
 ʒp # and filter to only keep primes

edited May 14 at 15:54

answered May 14 at 13:33

Emigna

49.5k534150

$begingroup$
You've made a typo in your explanation: "put a copy of N^2+3 on the stack" should be 3*N^2. Also, why the ) instead of ¯? Because it's easier to type? And for some reason I have the feeling the NnN‚3*¬sO‚ can be 1 byte shorter, but I'm not seeing it. A slight equal-byte alternative is Nn3*DN3*+‚. But I'm probably just seeing things that aren't there.. ;) Nice answer, so +1 from me.
$endgroup$
– Kevin Cruijssen
May 14 at 14:30

1

$begingroup$
I actually tried to port my answer to 05AB1E, but failed miserably. :D
$endgroup$
– Arnauld
May 14 at 15:09

1

$begingroup$
Actually, generating an infinite list is more convenient: 9 bytes with ∞n3*4÷>ʒp
$endgroup$
– Grimy
May 14 at 15:45

1

$begingroup$
OK, I'm not used to specs that contradict themselves. :-)
$endgroup$
– WGroleau
May 14 at 21:11

6

$begingroup$
@WGroleau I assume you've never developed software professionally then. I'm more concerned when I get specs that don't contradict themselves.
$endgroup$
– MikeTheLiar
May 14 at 21:29

|
show 7 more comments

05AB1E, 16 12 9 bytes

Generates an infinite list.

Saved 4 bytes with Kevin Cruijssen's port of Arnaulds formula.

Saved another 3 bytes thanks to Grimy

∞n3*4÷>ʒp

Try it online!

Explanation

∞ # on the list of infinite positive integers
 n3*4÷> # calculate (3*N^2)//4+1 for each
 ʒp # and filter to only keep primes

edited May 14 at 15:54

answered May 14 at 13:33

Emigna

49.5k534150

$begingroup$
You've made a typo in your explanation: "put a copy of N^2+3 on the stack" should be 3*N^2. Also, why the ) instead of ¯? Because it's easier to type? And for some reason I have the feeling the NnN‚3*¬sO‚ can be 1 byte shorter, but I'm not seeing it. A slight equal-byte alternative is Nn3*DN3*+‚. But I'm probably just seeing things that aren't there.. ;) Nice answer, so +1 from me.
$endgroup$
– Kevin Cruijssen
May 14 at 14:30

1

$begingroup$
I actually tried to port my answer to 05AB1E, but failed miserably. :D
$endgroup$
– Arnauld
May 14 at 15:09

1

$begingroup$
Actually, generating an infinite list is more convenient: 9 bytes with ∞n3*4÷>ʒp
$endgroup$
– Grimy
May 14 at 15:45

1

$begingroup$
OK, I'm not used to specs that contradict themselves. :-)
$endgroup$
– WGroleau
May 14 at 21:11

6

$begingroup$
@WGroleau I assume you've never developed software professionally then. I'm more concerned when I get specs that don't contradict themselves.
$endgroup$
– MikeTheLiar
May 14 at 21:29

|
show 7 more comments

05AB1E, 16 12 9 bytes

Generates an infinite list.

Saved 4 bytes with Kevin Cruijssen's port of Arnaulds formula.

Saved another 3 bytes thanks to Grimy

∞n3*4÷>ʒp

Try it online!

Explanation

∞ # on the list of infinite positive integers
 n3*4÷> # calculate (3*N^2)//4+1 for each
 ʒp # and filter to only keep primes

edited May 14 at 15:54

answered May 14 at 13:33

Emigna

49.5k534150

05AB1E, 16 12 9 bytes

Generates an infinite list.

Saved 4 bytes with Kevin Cruijssen's port of Arnaulds formula.

Saved another 3 bytes thanks to Grimy

∞n3*4÷>ʒp

Try it online!

Explanation

∞ # on the list of infinite positive integers
 n3*4÷> # calculate (3*N^2)//4+1 for each
 ʒp # and filter to only keep primes

edited May 14 at 15:54

answered May 14 at 13:33

Emigna

49.5k534150

edited May 14 at 15:54

answered May 14 at 13:33

Emigna

49.5k534150

answered May 14 at 13:33

Emigna

49.5k534150

answered May 14 at 13:33

Emigna

49.5k534150

$begingroup$
You've made a typo in your explanation: "put a copy of N^2+3 on the stack" should be 3*N^2. Also, why the ) instead of ¯? Because it's easier to type? And for some reason I have the feeling the NnN‚3*¬sO‚ can be 1 byte shorter, but I'm not seeing it. A slight equal-byte alternative is Nn3*DN3*+‚. But I'm probably just seeing things that aren't there.. ;) Nice answer, so +1 from me.
$endgroup$
– Kevin Cruijssen
May 14 at 14:30

1

$begingroup$
I actually tried to port my answer to 05AB1E, but failed miserably. :D
$endgroup$
– Arnauld
May 14 at 15:09

1

$begingroup$
Actually, generating an infinite list is more convenient: 9 bytes with ∞n3*4÷>ʒp
$endgroup$
– Grimy
May 14 at 15:45

1

$begingroup$
OK, I'm not used to specs that contradict themselves. :-)
$endgroup$
– WGroleau
May 14 at 21:11

6

$begingroup$
@WGroleau I assume you've never developed software professionally then. I'm more concerned when I get specs that don't contradict themselves.
$endgroup$
– MikeTheLiar
May 14 at 21:29

|
show 7 more comments

$begingroup$
You've made a typo in your explanation: "put a copy of N^2+3 on the stack" should be 3*N^2. Also, why the ) instead of ¯? Because it's easier to type? And for some reason I have the feeling the NnN‚3*¬sO‚ can be 1 byte shorter, but I'm not seeing it. A slight equal-byte alternative is Nn3*DN3*+‚. But I'm probably just seeing things that aren't there.. ;) Nice answer, so +1 from me.
$endgroup$
– Kevin Cruijssen
May 14 at 14:30

1

$begingroup$
I actually tried to port my answer to 05AB1E, but failed miserably. :D
$endgroup$
– Arnauld
May 14 at 15:09

1

$begingroup$
Actually, generating an infinite list is more convenient: 9 bytes with ∞n3*4÷>ʒp
$endgroup$
– Grimy
May 14 at 15:45

1

$begingroup$
OK, I'm not used to specs that contradict themselves. :-)
$endgroup$
– WGroleau
May 14 at 21:11

6

$begingroup$
@WGroleau I assume you've never developed software professionally then. I'm more concerned when I get specs that don't contradict themselves.
$endgroup$
– MikeTheLiar
May 14 at 21:29

You've made a typo in your explanation: "put a copy of N^2+3 on the stack" should be 3*N^2. Also, why the ) instead of ¯? Because it's easier to type? And for some reason I have the feeling the NnN‚3*¬sO‚ can be 1 byte shorter, but I'm not seeing it. A slight equal-byte alternative is Nn3*DN3*+‚. But I'm probably just seeing things that aren't there.. ;) Nice answer, so +1 from me.

– Kevin Cruijssen
May 14 at 14:30

I actually tried to port my answer to 05AB1E, but failed miserably. :D

– Arnauld
May 14 at 15:09

Actually, generating an infinite list is more convenient: 9 bytes with ∞n3*4÷>ʒp

– Grimy
May 14 at 15:45

OK, I'm not used to specs that contradict themselves. :-)

– WGroleau
May 14 at 21:11

@WGroleau I assume you've never developed software professionally then. I'm more concerned when I get specs that don't contradict themselves.

– MikeTheLiar
May 14 at 21:29

|
show 7 more comments

R, 75 73 bytes

n=scan()
while(F<n)F=F+any(!(((T<-T+1)*1:4-1)/3)^.5%%1)*all(T%%(3:T-1))
T

Try it online!

-2 bytes by noticing that I can remove brackets if I use * instead of & (different precedence).

Outputs the nth Cuban prime (1-indexed).

It uses the fact (given in OEIS) that Cuban primes are of the form $p=1+3n^2$ or $4p=1+3n^2$ for some $n$, i.e. $n=sqrtfracacdot p-13$ is an integer for $a=1$ or $a=4$.

The trick is that no prime can be of the form $2p=1+3n^2$ or $3p=1+3n^2$ (*), so we can save 2 bytes by checking the formula for $ain1, 2, 3, 4$ (1:4) instead of $ain1, 4$ (c(1,4)).

Slightly ungolfed version of the code:

# F and T are implicitly initialized at 0 and 1
# F is number of Cuban primes found so far
# T is number currently being tested for being a Cuban prime
n = scan() # input
while(F<n)
 T = T+1 # increment T 
 F = F + # increment F if
 (!all(((T*1:4-1)/3)^.5 %% 1) # there is an integer of the form sqrt(((T*a)-1)/3)
 & all(T%%(3:T-1))) # and T is prime (not divisible by any number between 2 and T-1)
 
T # output T

(*) No prime can be of the form $3p=1+3n^2$, else $1=3(p-n^2)$ would be divisible by $3$.

No prime other than $p=2$ (which isn't a Cuban prime) can of the form $2p=1+3n^2$: $n$ would need to be odd, i.e. $n=2k+1$. Expanding gives $2p=4+12k(k+1)$, hence $p=2+6k(k+1)$ and $p$ would be even.

edited May 14 at 15:55

answered May 14 at 14:32

Robin Ryder

1,288215

$begingroup$
what about avoiding a loop by using an upper bound on the nth Cuban prime?
$endgroup$
– Xi'an
May 15 at 7:31

$begingroup$
@Xi'an I thought about that, but couldn't come up with such a bound. Do you have one?
$endgroup$
– Robin Ryder
May 15 at 13:32

add a comment |

R, 75 73 bytes

n=scan()
while(F<n)F=F+any(!(((T<-T+1)*1:4-1)/3)^.5%%1)*all(T%%(3:T-1))
T

Try it online!

-2 bytes by noticing that I can remove brackets if I use * instead of & (different precedence).

Outputs the nth Cuban prime (1-indexed).

It uses the fact (given in OEIS) that Cuban primes are of the form $p=1+3n^2$ or $4p=1+3n^2$ for some $n$, i.e. $n=sqrtfracacdot p-13$ is an integer for $a=1$ or $a=4$.

The trick is that no prime can be of the form $2p=1+3n^2$ or $3p=1+3n^2$ (*), so we can save 2 bytes by checking the formula for $ain1, 2, 3, 4$ (1:4) instead of $ain1, 4$ (c(1,4)).

Slightly ungolfed version of the code:

# F and T are implicitly initialized at 0 and 1
# F is number of Cuban primes found so far
# T is number currently being tested for being a Cuban prime
n = scan() # input
while(F<n)
 T = T+1 # increment T 
 F = F + # increment F if
 (!all(((T*1:4-1)/3)^.5 %% 1) # there is an integer of the form sqrt(((T*a)-1)/3)
 & all(T%%(3:T-1))) # and T is prime (not divisible by any number between 2 and T-1)
 
T # output T

(*) No prime can be of the form $3p=1+3n^2$, else $1=3(p-n^2)$ would be divisible by $3$.

No prime other than $p=2$ (which isn't a Cuban prime) can of the form $2p=1+3n^2$: $n$ would need to be odd, i.e. $n=2k+1$. Expanding gives $2p=4+12k(k+1)$, hence $p=2+6k(k+1)$ and $p$ would be even.

edited May 14 at 15:55

answered May 14 at 14:32

Robin Ryder

1,288215

$begingroup$
what about avoiding a loop by using an upper bound on the nth Cuban prime?
$endgroup$
– Xi'an
May 15 at 7:31

$begingroup$
@Xi'an I thought about that, but couldn't come up with such a bound. Do you have one?
$endgroup$
– Robin Ryder
May 15 at 13:32

add a comment |

R, 75 73 bytes

n=scan()
while(F<n)F=F+any(!(((T<-T+1)*1:4-1)/3)^.5%%1)*all(T%%(3:T-1))
T

Try it online!

-2 bytes by noticing that I can remove brackets if I use * instead of & (different precedence).

Outputs the nth Cuban prime (1-indexed).

It uses the fact (given in OEIS) that Cuban primes are of the form $p=1+3n^2$ or $4p=1+3n^2$ for some $n$, i.e. $n=sqrtfracacdot p-13$ is an integer for $a=1$ or $a=4$.

The trick is that no prime can be of the form $2p=1+3n^2$ or $3p=1+3n^2$ (*), so we can save 2 bytes by checking the formula for $ain1, 2, 3, 4$ (1:4) instead of $ain1, 4$ (c(1,4)).

Slightly ungolfed version of the code:

# F and T are implicitly initialized at 0 and 1
# F is number of Cuban primes found so far
# T is number currently being tested for being a Cuban prime
n = scan() # input
while(F<n)
 T = T+1 # increment T 
 F = F + # increment F if
 (!all(((T*1:4-1)/3)^.5 %% 1) # there is an integer of the form sqrt(((T*a)-1)/3)
 & all(T%%(3:T-1))) # and T is prime (not divisible by any number between 2 and T-1)
 
T # output T

(*) No prime can be of the form $3p=1+3n^2$, else $1=3(p-n^2)$ would be divisible by $3$.

No prime other than $p=2$ (which isn't a Cuban prime) can of the form $2p=1+3n^2$: $n$ would need to be odd, i.e. $n=2k+1$. Expanding gives $2p=4+12k(k+1)$, hence $p=2+6k(k+1)$ and $p$ would be even.

edited May 14 at 15:55

answered May 14 at 14:32

Robin Ryder

1,288215

R, 75 73 bytes

n=scan()
while(F<n)F=F+any(!(((T<-T+1)*1:4-1)/3)^.5%%1)*all(T%%(3:T-1))
T

Try it online!

-2 bytes by noticing that I can remove brackets if I use * instead of & (different precedence).

Outputs the nth Cuban prime (1-indexed).

It uses the fact (given in OEIS) that Cuban primes are of the form $p=1+3n^2$ or $4p=1+3n^2$ for some $n$, i.e. $n=sqrtfracacdot p-13$ is an integer for $a=1$ or $a=4$.

The trick is that no prime can be of the form $2p=1+3n^2$ or $3p=1+3n^2$ (*), so we can save 2 bytes by checking the formula for $ain1, 2, 3, 4$ (1:4) instead of $ain1, 4$ (c(1,4)).

Slightly ungolfed version of the code:

# F and T are implicitly initialized at 0 and 1
# F is number of Cuban primes found so far
# T is number currently being tested for being a Cuban prime
n = scan() # input
while(F<n)
 T = T+1 # increment T 
 F = F + # increment F if
 (!all(((T*1:4-1)/3)^.5 %% 1) # there is an integer of the form sqrt(((T*a)-1)/3)
 & all(T%%(3:T-1))) # and T is prime (not divisible by any number between 2 and T-1)
 
T # output T

(*) No prime can be of the form $3p=1+3n^2$, else $1=3(p-n^2)$ would be divisible by $3$.

No prime other than $p=2$ (which isn't a Cuban prime) can of the form $2p=1+3n^2$: $n$ would need to be odd, i.e. $n=2k+1$. Expanding gives $2p=4+12k(k+1)$, hence $p=2+6k(k+1)$ and $p$ would be even.

edited May 14 at 15:55

answered May 14 at 14:32

Robin Ryder

1,288215

edited May 14 at 15:55

answered May 14 at 14:32

Robin Ryder

1,288215

answered May 14 at 14:32

Robin Ryder

1,288215

answered May 14 at 14:32

Robin Ryder

1,288215

$begingroup$
what about avoiding a loop by using an upper bound on the nth Cuban prime?
$endgroup$
– Xi'an
May 15 at 7:31

$begingroup$
@Xi'an I thought about that, but couldn't come up with such a bound. Do you have one?
$endgroup$
– Robin Ryder
May 15 at 13:32

add a comment |

$begingroup$
what about avoiding a loop by using an upper bound on the nth Cuban prime?
$endgroup$
– Xi'an
May 15 at 7:31

$begingroup$
@Xi'an I thought about that, but couldn't come up with such a bound. Do you have one?
$endgroup$
– Robin Ryder
May 15 at 13:32

what about avoiding a loop by using an upper bound on the nth Cuban prime?

– Xi'an
May 15 at 7:31

@Xi'an I thought about that, but couldn't come up with such a bound. Do you have one?

– Robin Ryder
May 15 at 13:32

add a comment |

Wolfram Language (Mathematica), 66 65 56 bytes

(f=1+⌊3#/4#⌋&;For[n=i=0,i<#,PrimeQ@f@++n&&i++];f@n)&

Try it online!

J42161217 -1 by using ⌊ ⌋ instead of Floor[ ]

attinat
- -1 by using ⌊3#/4#⌋ instead of ⌊3#^2/4⌋
- -8 for For[n=i=0,i<#,PrimeQ@f@++n&&i++] instead of n=2;i=#;While[i>0,i-=Boole@PrimeQ@f@++n]

edited May 14 at 22:56

answered May 14 at 18:27

speedstyle

695

1

$begingroup$
65 bytes. Welcome to ppcg. Nice first answer! +1
$endgroup$
– J42161217
May 14 at 18:39

$begingroup$
Thanks! (Long time lurker.) I couldn't quite parse your existing answer so I wrote my own and it came out a little shorter. I might do a Python one too.
$endgroup$
– speedstyle
May 14 at 18:57

2

$begingroup$
56 bytes
$endgroup$
– attinat
May 14 at 21:54

$begingroup$
@attinat I thought Arnauld's formula only worked for n>2 so I didn't start with 0 - although as in your example it works for all n (because it starts 1 1 4 7 13 ... so the primes are 7 13 ...)
$endgroup$
– speedstyle
May 14 at 22:33

add a comment |

Wolfram Language (Mathematica), 66 65 56 bytes

(f=1+⌊3#/4#⌋&;For[n=i=0,i<#,PrimeQ@f@++n&&i++];f@n)&

Try it online!

J42161217 -1 by using ⌊ ⌋ instead of Floor[ ]

attinat
- -1 by using ⌊3#/4#⌋ instead of ⌊3#^2/4⌋
- -8 for For[n=i=0,i<#,PrimeQ@f@++n&&i++] instead of n=2;i=#;While[i>0,i-=Boole@PrimeQ@f@++n]

edited May 14 at 22:56

answered May 14 at 18:27

speedstyle

695

1

$begingroup$
65 bytes. Welcome to ppcg. Nice first answer! +1
$endgroup$
– J42161217
May 14 at 18:39

$begingroup$
Thanks! (Long time lurker.) I couldn't quite parse your existing answer so I wrote my own and it came out a little shorter. I might do a Python one too.
$endgroup$
– speedstyle
May 14 at 18:57

2

$begingroup$
56 bytes
$endgroup$
– attinat
May 14 at 21:54

$begingroup$
@attinat I thought Arnauld's formula only worked for n>2 so I didn't start with 0 - although as in your example it works for all n (because it starts 1 1 4 7 13 ... so the primes are 7 13 ...)
$endgroup$
– speedstyle
May 14 at 22:33

add a comment |

Wolfram Language (Mathematica), 66 65 56 bytes

(f=1+⌊3#/4#⌋&;For[n=i=0,i<#,PrimeQ@f@++n&&i++];f@n)&

Try it online!

J42161217 -1 by using ⌊ ⌋ instead of Floor[ ]

attinat
- -1 by using ⌊3#/4#⌋ instead of ⌊3#^2/4⌋
- -8 for For[n=i=0,i<#,PrimeQ@f@++n&&i++] instead of n=2;i=#;While[i>0,i-=Boole@PrimeQ@f@++n]

edited May 14 at 22:56

answered May 14 at 18:27

speedstyle

695

Wolfram Language (Mathematica), 66 65 56 bytes

(f=1+⌊3#/4#⌋&;For[n=i=0,i<#,PrimeQ@f@++n&&i++];f@n)&

Try it online!

J42161217 -1 by using ⌊ ⌋ instead of Floor[ ]

attinat
- -1 by using ⌊3#/4#⌋ instead of ⌊3#^2/4⌋
- -8 for For[n=i=0,i<#,PrimeQ@f@++n&&i++] instead of n=2;i=#;While[i>0,i-=Boole@PrimeQ@f@++n]

edited May 14 at 22:56

answered May 14 at 18:27

speedstyle

695

edited May 14 at 22:56

answered May 14 at 18:27

speedstyle

695

answered May 14 at 18:27

speedstyle

695

answered May 14 at 18:27

speedstyle

695

1

$begingroup$
65 bytes. Welcome to ppcg. Nice first answer! +1
$endgroup$
– J42161217
May 14 at 18:39

$begingroup$
Thanks! (Long time lurker.) I couldn't quite parse your existing answer so I wrote my own and it came out a little shorter. I might do a Python one too.
$endgroup$
– speedstyle
May 14 at 18:57

2

$begingroup$
56 bytes
$endgroup$
– attinat
May 14 at 21:54

$begingroup$
@attinat I thought Arnauld's formula only worked for n>2 so I didn't start with 0 - although as in your example it works for all n (because it starts 1 1 4 7 13 ... so the primes are 7 13 ...)
$endgroup$
– speedstyle
May 14 at 22:33

add a comment |

1

$begingroup$
65 bytes. Welcome to ppcg. Nice first answer! +1
$endgroup$
– J42161217
May 14 at 18:39

$begingroup$
Thanks! (Long time lurker.) I couldn't quite parse your existing answer so I wrote my own and it came out a little shorter. I might do a Python one too.
$endgroup$
– speedstyle
May 14 at 18:57

2

$begingroup$
56 bytes
$endgroup$
– attinat
May 14 at 21:54

$begingroup$
@attinat I thought Arnauld's formula only worked for n>2 so I didn't start with 0 - although as in your example it works for all n (because it starts 1 1 4 7 13 ... so the primes are 7 13 ...)
$endgroup$
– speedstyle
May 14 at 22:33

65 bytes. Welcome to ppcg. Nice first answer! +1

– J42161217
May 14 at 18:39

Thanks! (Long time lurker.) I couldn't quite parse your existing answer so I wrote my own and it came out a little shorter. I might do a Python one too.

– speedstyle
May 14 at 18:57

56 bytes

– attinat
May 14 at 21:54

@attinat I thought Arnauld's formula only worked for n>2 so I didn't start with 0 - although as in your example it works for all n (because it starts 1 1 4 7 13 ... so the primes are 7 13 ...)

– speedstyle
May 14 at 22:33

add a comment |

Java 8, 94 88 86 84 bytes

v->for(int i=3,n,x;;System.out.print(x<1?++n+" ":""))for(x=n=i*i++*3/4;~n%x--<0;);

Outputs space-delimited indefinitely.

Try it online.

Explanation (of the old 86 bytes version): TODO: Update explanation

Uses the formula of @Arnauld's JavaScript answer: $p_n=leftlfloorfrac3n^24rightrfloor+1,;nge3$.

v-> // Method with empty unused parameter and no return-type
 for(int i=3, // Loop-integer, starting at 3
 n,x // Temp integers
 ; // Loop indefinitely:
 ; // After every iteration:
 System.out.print( // Print:
 n==x? // If `n` equals `x`, which means `n` is a prime:
 n+" " // Print `n` with a space delimiter
 : // Else:
 "")) // Print nothing
 for(n=i*i++*3/4+1, // Set `n` to `(3*i^2)//4+1
 // (and increase `i` by 1 afterwards with `i++`)
 x=1; // Set `x` to 1
 n%++x // Loop as long as `n` modulo `x+1`
 // (after we've first increased `x` by 1 with `++x`)
 >0;); // is not 0 yet
 // (if `n` is equal to `x`, it means it's a prime)

edited May 16 at 15:43

answered May 15 at 6:48

Kevin Cruijssen

44.9k576225

$begingroup$
I don't really think it's feasible, but another way of finding the cuban primes uses this formula: v->for(int n=7,i=3,p,x,d,r=0;;i+=++r%2*3,n+=i,System.out.print(x>1?x+" ":""))for(x=n,d=1;++d<n;x=x%d<1?0:n);, maybe someone can use this to golf? I couldn't.
$endgroup$
– Olivier Grégoire
May 16 at 9:03

1

$begingroup$
@OlivierGrégoire You can golf yours a bit more by removing the unused ,p and changing i+=++r%2*3,n+=i to n+=i+=++r%2*3, but then I'll still end up at 106 bytes. Using Java 11's String#repeat with prime-regex is 105 bytes: v->(..+?)\1+"))System.out.println(n);.
$endgroup$
– Kevin Cruijssen
May 16 at 9:40

$begingroup$
Yeah, I guessed it wasn't much golfable despite my (now obvious) mistakes. Thanks for giving it a ride ;)
$endgroup$
– Olivier Grégoire
May 16 at 14:02

$begingroup$
@OlivierGrégoire Maybe also good to know for you, but there is apparently a shorter prime-check loop in Java. See my edit and SaraJ's prime-check answer.
$endgroup$
– Kevin Cruijssen
May 16 at 14:30

$begingroup$
I might be wrong, but the last %n isn't required, is it?
$endgroup$
– Olivier Grégoire
May 16 at 14:44

|
show 2 more comments

Java 8, 94 88 86 84 bytes

v->for(int i=3,n,x;;System.out.print(x<1?++n+" ":""))for(x=n=i*i++*3/4;~n%x--<0;);

Outputs space-delimited indefinitely.

Try it online.

Explanation (of the old 86 bytes version): TODO: Update explanation

Uses the formula of @Arnauld's JavaScript answer: $p_n=leftlfloorfrac3n^24rightrfloor+1,;nge3$.

v-> // Method with empty unused parameter and no return-type
 for(int i=3, // Loop-integer, starting at 3
 n,x // Temp integers
 ; // Loop indefinitely:
 ; // After every iteration:
 System.out.print( // Print:
 n==x? // If `n` equals `x`, which means `n` is a prime:
 n+" " // Print `n` with a space delimiter
 : // Else:
 "")) // Print nothing
 for(n=i*i++*3/4+1, // Set `n` to `(3*i^2)//4+1
 // (and increase `i` by 1 afterwards with `i++`)
 x=1; // Set `x` to 1
 n%++x // Loop as long as `n` modulo `x+1`
 // (after we've first increased `x` by 1 with `++x`)
 >0;); // is not 0 yet
 // (if `n` is equal to `x`, it means it's a prime)

edited May 16 at 15:43

answered May 15 at 6:48

Kevin Cruijssen

44.9k576225

$begingroup$
I don't really think it's feasible, but another way of finding the cuban primes uses this formula: v->for(int n=7,i=3,p,x,d,r=0;;i+=++r%2*3,n+=i,System.out.print(x>1?x+" ":""))for(x=n,d=1;++d<n;x=x%d<1?0:n);, maybe someone can use this to golf? I couldn't.
$endgroup$
– Olivier Grégoire
May 16 at 9:03

1

$begingroup$
@OlivierGrégoire You can golf yours a bit more by removing the unused ,p and changing i+=++r%2*3,n+=i to n+=i+=++r%2*3, but then I'll still end up at 106 bytes. Using Java 11's String#repeat with prime-regex is 105 bytes: v->(..+?)\1+"))System.out.println(n);.
$endgroup$
– Kevin Cruijssen
May 16 at 9:40

$begingroup$
Yeah, I guessed it wasn't much golfable despite my (now obvious) mistakes. Thanks for giving it a ride ;)
$endgroup$
– Olivier Grégoire
May 16 at 14:02

$begingroup$
@OlivierGrégoire Maybe also good to know for you, but there is apparently a shorter prime-check loop in Java. See my edit and SaraJ's prime-check answer.
$endgroup$
– Kevin Cruijssen
May 16 at 14:30

$begingroup$
I might be wrong, but the last %n isn't required, is it?
$endgroup$
– Olivier Grégoire
May 16 at 14:44

|
show 2 more comments

Java 8, 94 88 86 84 bytes

v->for(int i=3,n,x;;System.out.print(x<1?++n+" ":""))for(x=n=i*i++*3/4;~n%x--<0;);

Outputs space-delimited indefinitely.

Try it online.

Explanation (of the old 86 bytes version): TODO: Update explanation

Uses the formula of @Arnauld's JavaScript answer: $p_n=leftlfloorfrac3n^24rightrfloor+1,;nge3$.

v-> // Method with empty unused parameter and no return-type
 for(int i=3, // Loop-integer, starting at 3
 n,x // Temp integers
 ; // Loop indefinitely:
 ; // After every iteration:
 System.out.print( // Print:
 n==x? // If `n` equals `x`, which means `n` is a prime:
 n+" " // Print `n` with a space delimiter
 : // Else:
 "")) // Print nothing
 for(n=i*i++*3/4+1, // Set `n` to `(3*i^2)//4+1
 // (and increase `i` by 1 afterwards with `i++`)
 x=1; // Set `x` to 1
 n%++x // Loop as long as `n` modulo `x+1`
 // (after we've first increased `x` by 1 with `++x`)
 >0;); // is not 0 yet
 // (if `n` is equal to `x`, it means it's a prime)

edited May 16 at 15:43

answered May 15 at 6:48

Kevin Cruijssen

44.9k576225

Java 8, 94 88 86 84 bytes

v->for(int i=3,n,x;;System.out.print(x<1?++n+" ":""))for(x=n=i*i++*3/4;~n%x--<0;);

Outputs space-delimited indefinitely.

Try it online.

Explanation (of the old 86 bytes version): TODO: Update explanation

Uses the formula of @Arnauld's JavaScript answer: $p_n=leftlfloorfrac3n^24rightrfloor+1,;nge3$.

v-> // Method with empty unused parameter and no return-type
 for(int i=3, // Loop-integer, starting at 3
 n,x // Temp integers
 ; // Loop indefinitely:
 ; // After every iteration:
 System.out.print( // Print:
 n==x? // If `n` equals `x`, which means `n` is a prime:
 n+" " // Print `n` with a space delimiter
 : // Else:
 "")) // Print nothing
 for(n=i*i++*3/4+1, // Set `n` to `(3*i^2)//4+1
 // (and increase `i` by 1 afterwards with `i++`)
 x=1; // Set `x` to 1
 n%++x // Loop as long as `n` modulo `x+1`
 // (after we've first increased `x` by 1 with `++x`)
 >0;); // is not 0 yet
 // (if `n` is equal to `x`, it means it's a prime)

edited May 16 at 15:43

answered May 15 at 6:48

Kevin Cruijssen

44.9k576225

edited May 16 at 15:43

answered May 15 at 6:48

Kevin Cruijssen

44.9k576225

answered May 15 at 6:48

Kevin Cruijssen

44.9k576225

answered May 15 at 6:48

Kevin Cruijssen

44.9k576225

$begingroup$
I don't really think it's feasible, but another way of finding the cuban primes uses this formula: v->for(int n=7,i=3,p,x,d,r=0;;i+=++r%2*3,n+=i,System.out.print(x>1?x+" ":""))for(x=n,d=1;++d<n;x=x%d<1?0:n);, maybe someone can use this to golf? I couldn't.
$endgroup$
– Olivier Grégoire
May 16 at 9:03

1

$begingroup$
@OlivierGrégoire You can golf yours a bit more by removing the unused ,p and changing i+=++r%2*3,n+=i to n+=i+=++r%2*3, but then I'll still end up at 106 bytes. Using Java 11's String#repeat with prime-regex is 105 bytes: v->(..+?)\1+"))System.out.println(n);.
$endgroup$
– Kevin Cruijssen
May 16 at 9:40

$begingroup$
Yeah, I guessed it wasn't much golfable despite my (now obvious) mistakes. Thanks for giving it a ride ;)
$endgroup$
– Olivier Grégoire
May 16 at 14:02

$begingroup$
@OlivierGrégoire Maybe also good to know for you, but there is apparently a shorter prime-check loop in Java. See my edit and SaraJ's prime-check answer.
$endgroup$
– Kevin Cruijssen
May 16 at 14:30

$begingroup$
I might be wrong, but the last %n isn't required, is it?
$endgroup$
– Olivier Grégoire
May 16 at 14:44

|
show 2 more comments

$begingroup$
I don't really think it's feasible, but another way of finding the cuban primes uses this formula: v->for(int n=7,i=3,p,x,d,r=0;;i+=++r%2*3,n+=i,System.out.print(x>1?x+" ":""))for(x=n,d=1;++d<n;x=x%d<1?0:n);, maybe someone can use this to golf? I couldn't.
$endgroup$
– Olivier Grégoire
May 16 at 9:03

1

$begingroup$
@OlivierGrégoire You can golf yours a bit more by removing the unused ,p and changing i+=++r%2*3,n+=i to n+=i+=++r%2*3, but then I'll still end up at 106 bytes. Using Java 11's String#repeat with prime-regex is 105 bytes: v->(..+?)\1+"))System.out.println(n);.
$endgroup$
– Kevin Cruijssen
May 16 at 9:40

$begingroup$
Yeah, I guessed it wasn't much golfable despite my (now obvious) mistakes. Thanks for giving it a ride ;)
$endgroup$
– Olivier Grégoire
May 16 at 14:02

$begingroup$
@OlivierGrégoire Maybe also good to know for you, but there is apparently a shorter prime-check loop in Java. See my edit and SaraJ's prime-check answer.
$endgroup$
– Kevin Cruijssen
May 16 at 14:30

$begingroup$
I might be wrong, but the last %n isn't required, is it?
$endgroup$
– Olivier Grégoire
May 16 at 14:44

I don't really think it's feasible, but another way of finding the cuban primes uses this formula: v->for(int n=7,i=3,p,x,d,r=0;;i+=++r%2*3,n+=i,System.out.print(x>1?x+" ":""))for(x=n,d=1;++d<n;x=x%d<1?0:n);, maybe someone can use this to golf? I couldn't.

– Olivier Grégoire
May 16 at 9:03

@OlivierGrégoire You can golf yours a bit more by removing the unused ,p and changing i+=++r%2*3,n+=i to n+=i+=++r%2*3, but then I'll still end up at 106 bytes. Using Java 11's String#repeat with prime-regex is 105 bytes: v->(..+?)\1+"))System.out.println(n);.

– Kevin Cruijssen
May 16 at 9:40

Yeah, I guessed it wasn't much golfable despite my (now obvious) mistakes. Thanks for giving it a ride ;)

– Olivier Grégoire
May 16 at 14:02

@OlivierGrégoire Maybe also good to know for you, but there is apparently a shorter prime-check loop in Java. See my edit and SaraJ's prime-check answer.

– Kevin Cruijssen
May 16 at 14:30

I might be wrong, but the last %n isn't required, is it?

– Olivier Grégoire
May 16 at 14:44

|
show 2 more comments

Wolfram Language (Mathematica), 83 bytes

(t=1;While[Length[l=Select[Join@@Array[(v=3#^2+1)+3#,v&,t++],PrimeQ]]<#];Sort@l)&

Try it online!

edited May 14 at 14:40

answered May 14 at 13:31

J42161217

15.1k21457

add a comment |

Wolfram Language (Mathematica), 83 bytes

(t=1;While[Length[l=Select[Join@@Array[(v=3#^2+1)+3#,v&,t++],PrimeQ]]<#];Sort@l)&

Try it online!

edited May 14 at 14:40

answered May 14 at 13:31

J42161217

15.1k21457

add a comment |

Wolfram Language (Mathematica), 83 bytes

(t=1;While[Length[l=Select[Join@@Array[(v=3#^2+1)+3#,v&,t++],PrimeQ]]<#];Sort@l)&

Try it online!

edited May 14 at 14:40

answered May 14 at 13:31

J42161217

15.1k21457

Wolfram Language (Mathematica), 83 bytes

(t=1;While[Length[l=Select[Join@@Array[(v=3#^2+1)+3#,v&,t++],PrimeQ]]<#];Sort@l)&

Try it online!

edited May 14 at 14:40

answered May 14 at 13:31

J42161217

15.1k21457

edited May 14 at 14:40

answered May 14 at 13:31

J42161217

15.1k21457

answered May 14 at 13:31

J42161217

15.1k21457

answered May 14 at 13:31

J42161217

15.1k21457

add a comment |

Jelly, 12 bytes

²×3:4‘
ÇẒ$#Ç

Try it online!

Based on @Arnauld’s method. Takes n on stdin and returns that many Cuban primes.

answered May 14 at 19:11

Nick Kennedy

2,56469

add a comment |

Jelly, 12 bytes

²×3:4‘
ÇẒ$#Ç

Try it online!

Based on @Arnauld’s method. Takes n on stdin and returns that many Cuban primes.

answered May 14 at 19:11

Nick Kennedy

2,56469

add a comment |

Jelly, 12 bytes

²×3:4‘
ÇẒ$#Ç

Try it online!

Based on @Arnauld’s method. Takes n on stdin and returns that many Cuban primes.

answered May 14 at 19:11

Nick Kennedy

2,56469

Jelly, 12 bytes

²×3:4‘
ÇẒ$#Ç

Try it online!

Based on @Arnauld’s method. Takes n on stdin and returns that many Cuban primes.

answered May 14 at 19:11

Nick Kennedy

2,56469

answered May 14 at 19:11

Nick Kennedy

2,56469

answered May 14 at 19:11

Nick Kennedy

2,56469

answered May 14 at 19:11

Nick Kennedy

2,56469

add a comment |

Wolfram Language (Mathematica), 83 bytes

This solution will output the n-th Cuban prime with the added benefits of being fast and remembering all previous results in the symbol f.

(d:=1+3y(c=1+y)+3b c;e:=If[PrimeQ@d,n++;f@n=d];For[n=y=b=0,n<#,e;b=1-b;e,y++];f@#)&

Try it online!

answered May 14 at 18:21

Kelly Lowder

3,050417

add a comment |

Wolfram Language (Mathematica), 83 bytes

This solution will output the n-th Cuban prime with the added benefits of being fast and remembering all previous results in the symbol f.

(d:=1+3y(c=1+y)+3b c;e:=If[PrimeQ@d,n++;f@n=d];For[n=y=b=0,n<#,e;b=1-b;e,y++];f@#)&

Try it online!

answered May 14 at 18:21

Kelly Lowder

3,050417

add a comment |

Wolfram Language (Mathematica), 83 bytes

This solution will output the n-th Cuban prime with the added benefits of being fast and remembering all previous results in the symbol f.

(d:=1+3y(c=1+y)+3b c;e:=If[PrimeQ@d,n++;f@n=d];For[n=y=b=0,n<#,e;b=1-b;e,y++];f@#)&

Try it online!

answered May 14 at 18:21

Kelly Lowder

3,050417

Wolfram Language (Mathematica), 83 bytes

This solution will output the n-th Cuban prime with the added benefits of being fast and remembering all previous results in the symbol f.

(d:=1+3y(c=1+y)+3b c;e:=If[PrimeQ@d,n++;f@n=d];For[n=y=b=0,n<#,e;b=1-b;e,y++];f@#)&

Try it online!

answered May 14 at 18:21

Kelly Lowder

3,050417

answered May 14 at 18:21

Kelly Lowder

3,050417

answered May 14 at 18:21

Kelly Lowder

3,050417

answered May 14 at 18:21

Kelly Lowder

3,050417

add a comment |

Whitespace, 180 bytes

[S S S T S N
_Push_2][S N
S _Duplicate][N
S S N
_Create_Label_OUTER_LOOP][S N
N
_Discard_top_stack][S S S T N
_Push_1][T S S S _Add][S N
S _Duplicate][S N
S _Duplicate][T S S N
_Multiply][S S S T T N
_Push_3][T S S N
_Multiply][S S S T S S N
_Push_4][T S T S _Integer_divide][S S S T N
_Push_1][T S S S _Add][S S S T N
_Push_1][S N
S _Duplicate_1][N
S S S N
_Create_Label_INNER_LOOP][S N
N
_Discard_top_stack][S S S T N
_Push_1][T S S S _Add][S N
S _Duplicate][S N
S _Duplicate][S T S S T T N
_Copy_0-based_3rd][T S S T _Subtract][N
T S T N
_Jump_to_Label_PRINT_if_0][S T S S T S N
_Copy_0-based_2nd][S N
T _Swap_top_two][T S T T _Modulo][S N
S _Duplicate][N
T S S S N
_Jump_to_Label_FALSE_if_0][N
S N
S N
_Jump_to_Label_INNER_LOOP][N
S S T N
_Create_Label_PRINT][T N
S T _Print_as_integer][S S S T S T S N
_Push_10_(newline)][T N
S S _Print_as_character][S N
S _Duplicate][N
S S S S N
_Create_Label_FALSE][S N
N
_Discard_top_stack][S N
N
_Discard_top_stack][N
S N
N
_Jump_to_Label_OUTER_LOOP]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Outputs newline-delimited indefinitely.

Try it online (with raw spaces, tabs, and new-lines only).

Explanation in pseudo-code:

Port of my Java 8 answer, which also uses the formula from @Arnauld's JavaScript answer: $p_n=leftlfloorfrac3n^24rightrfloor+1,;nge3$.

Integer i = 2
Start OUTER_LOOP:
 i = i + 1
 Integer n = i*i*3//4+1
 Integer x = 1
 Start INNER_LOOP:
 x = x + 1
 If(x == n):
 Call function PRINT
 If(n % x == 0):
 Go to next iteration of OUTER_LOOP
 Go to next iteration of INNER_LOOP

function PRINT:
 Print integer n
 Print character 'n'
 Go to next iteration of OUTER_LOOP

answered May 15 at 8:01

Kevin Cruijssen

44.9k576225

add a comment |

Whitespace, 180 bytes

[S S S T S N
_Push_2][S N
S _Duplicate][N
S S N
_Create_Label_OUTER_LOOP][S N
N
_Discard_top_stack][S S S T N
_Push_1][T S S S _Add][S N
S _Duplicate][S N
S _Duplicate][T S S N
_Multiply][S S S T T N
_Push_3][T S S N
_Multiply][S S S T S S N
_Push_4][T S T S _Integer_divide][S S S T N
_Push_1][T S S S _Add][S S S T N
_Push_1][S N
S _Duplicate_1][N
S S S N
_Create_Label_INNER_LOOP][S N
N
_Discard_top_stack][S S S T N
_Push_1][T S S S _Add][S N
S _Duplicate][S N
S _Duplicate][S T S S T T N
_Copy_0-based_3rd][T S S T _Subtract][N
T S T N
_Jump_to_Label_PRINT_if_0][S T S S T S N
_Copy_0-based_2nd][S N
T _Swap_top_two][T S T T _Modulo][S N
S _Duplicate][N
T S S S N
_Jump_to_Label_FALSE_if_0][N
S N
S N
_Jump_to_Label_INNER_LOOP][N
S S T N
_Create_Label_PRINT][T N
S T _Print_as_integer][S S S T S T S N
_Push_10_(newline)][T N
S S _Print_as_character][S N
S _Duplicate][N
S S S S N
_Create_Label_FALSE][S N
N
_Discard_top_stack][S N
N
_Discard_top_stack][N
S N
N
_Jump_to_Label_OUTER_LOOP]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Outputs newline-delimited indefinitely.

Try it online (with raw spaces, tabs, and new-lines only).

Explanation in pseudo-code:

Port of my Java 8 answer, which also uses the formula from @Arnauld's JavaScript answer: $p_n=leftlfloorfrac3n^24rightrfloor+1,;nge3$.

Integer i = 2
Start OUTER_LOOP:
 i = i + 1
 Integer n = i*i*3//4+1
 Integer x = 1
 Start INNER_LOOP:
 x = x + 1
 If(x == n):
 Call function PRINT
 If(n % x == 0):
 Go to next iteration of OUTER_LOOP
 Go to next iteration of INNER_LOOP

function PRINT:
 Print integer n
 Print character 'n'
 Go to next iteration of OUTER_LOOP

answered May 15 at 8:01

Kevin Cruijssen

44.9k576225

add a comment |

Whitespace, 180 bytes

[S S S T S N
_Push_2][S N
S _Duplicate][N
S S N
_Create_Label_OUTER_LOOP][S N
N
_Discard_top_stack][S S S T N
_Push_1][T S S S _Add][S N
S _Duplicate][S N
S _Duplicate][T S S N
_Multiply][S S S T T N
_Push_3][T S S N
_Multiply][S S S T S S N
_Push_4][T S T S _Integer_divide][S S S T N
_Push_1][T S S S _Add][S S S T N
_Push_1][S N
S _Duplicate_1][N
S S S N
_Create_Label_INNER_LOOP][S N
N
_Discard_top_stack][S S S T N
_Push_1][T S S S _Add][S N
S _Duplicate][S N
S _Duplicate][S T S S T T N
_Copy_0-based_3rd][T S S T _Subtract][N
T S T N
_Jump_to_Label_PRINT_if_0][S T S S T S N
_Copy_0-based_2nd][S N
T _Swap_top_two][T S T T _Modulo][S N
S _Duplicate][N
T S S S N
_Jump_to_Label_FALSE_if_0][N
S N
S N
_Jump_to_Label_INNER_LOOP][N
S S T N
_Create_Label_PRINT][T N
S T _Print_as_integer][S S S T S T S N
_Push_10_(newline)][T N
S S _Print_as_character][S N
S _Duplicate][N
S S S S N
_Create_Label_FALSE][S N
N
_Discard_top_stack][S N
N
_Discard_top_stack][N
S N
N
_Jump_to_Label_OUTER_LOOP]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Outputs newline-delimited indefinitely.

Try it online (with raw spaces, tabs, and new-lines only).

Explanation in pseudo-code:

Port of my Java 8 answer, which also uses the formula from @Arnauld's JavaScript answer: $p_n=leftlfloorfrac3n^24rightrfloor+1,;nge3$.

Integer i = 2
Start OUTER_LOOP:
 i = i + 1
 Integer n = i*i*3//4+1
 Integer x = 1
 Start INNER_LOOP:
 x = x + 1
 If(x == n):
 Call function PRINT
 If(n % x == 0):
 Go to next iteration of OUTER_LOOP
 Go to next iteration of INNER_LOOP

function PRINT:
 Print integer n
 Print character 'n'
 Go to next iteration of OUTER_LOOP

answered May 15 at 8:01

Kevin Cruijssen

44.9k576225

Whitespace, 180 bytes

[S S S T S N
_Push_2][S N
S _Duplicate][N
S S N
_Create_Label_OUTER_LOOP][S N
N
_Discard_top_stack][S S S T N
_Push_1][T S S S _Add][S N
S _Duplicate][S N
S _Duplicate][T S S N
_Multiply][S S S T T N
_Push_3][T S S N
_Multiply][S S S T S S N
_Push_4][T S T S _Integer_divide][S S S T N
_Push_1][T S S S _Add][S S S T N
_Push_1][S N
S _Duplicate_1][N
S S S N
_Create_Label_INNER_LOOP][S N
N
_Discard_top_stack][S S S T N
_Push_1][T S S S _Add][S N
S _Duplicate][S N
S _Duplicate][S T S S T T N
_Copy_0-based_3rd][T S S T _Subtract][N
T S T N
_Jump_to_Label_PRINT_if_0][S T S S T S N
_Copy_0-based_2nd][S N
T _Swap_top_two][T S T T _Modulo][S N
S _Duplicate][N
T S S S N
_Jump_to_Label_FALSE_if_0][N
S N
S N
_Jump_to_Label_INNER_LOOP][N
S S T N
_Create_Label_PRINT][T N
S T _Print_as_integer][S S S T S T S N
_Push_10_(newline)][T N
S S _Print_as_character][S N
S _Duplicate][N
S S S S N
_Create_Label_FALSE][S N
N
_Discard_top_stack][S N
N
_Discard_top_stack][N
S N
N
_Jump_to_Label_OUTER_LOOP]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Outputs newline-delimited indefinitely.

Try it online (with raw spaces, tabs, and new-lines only).

Explanation in pseudo-code:

Port of my Java 8 answer, which also uses the formula from @Arnauld's JavaScript answer: $p_n=leftlfloorfrac3n^24rightrfloor+1,;nge3$.

Integer i = 2
Start OUTER_LOOP:
 i = i + 1
 Integer n = i*i*3//4+1
 Integer x = 1
 Start INNER_LOOP:
 x = x + 1
 If(x == n):
 Call function PRINT
 If(n % x == 0):
 Go to next iteration of OUTER_LOOP
 Go to next iteration of INNER_LOOP

function PRINT:
 Print integer n
 Print character 'n'
 Go to next iteration of OUTER_LOOP

answered May 15 at 8:01

Kevin Cruijssen

44.9k576225

answered May 15 at 8:01

Kevin Cruijssen

44.9k576225

answered May 15 at 8:01

Kevin Cruijssen

44.9k576225

answered May 15 at 8:01

Kevin Cruijssen

44.9k576225

add a comment |

Python 3, 110 108 102 bytes

Similar method to my Mathematica answer (i.e. isPrime(1+⌊¾n²⌋) else n++)
using this golfed prime checker and returning an anonymous infinite generator

from itertools import*
(x for x in map(lambda n:1+3*n**2//4,count(2)) if all(x%j for j in range(2,x)))

Try it online!

mypetlion -2 because arguably anonymous generators are more allowed than named ones

-6 by starting count at 2 ₊₁ so that the and x>1 in the prime checker I borrowed is unnecessary _-7

edited May 15 at 9:28

answered May 14 at 20:44

speedstyle

695

$begingroup$
The answer going into a variable is usually not considered a valid form of "output". Could you rework your answer so that the result is either output to stdout or returned by a function?
$endgroup$
– mypetlion
May 14 at 23:01

1

$begingroup$
since anonymous functions are allowed, and the challenge explicitly allows an infinite generator, I've removed g=. I had only included it in the first place because it allowed a quick visual on TIO with print(next(g) for i in range(52)).
$endgroup$
– speedstyle
May 15 at 1:15

add a comment |

Python 3, 110 108 102 bytes

Similar method to my Mathematica answer (i.e. isPrime(1+⌊¾n²⌋) else n++)
using this golfed prime checker and returning an anonymous infinite generator

from itertools import*
(x for x in map(lambda n:1+3*n**2//4,count(2)) if all(x%j for j in range(2,x)))

Try it online!

mypetlion -2 because arguably anonymous generators are more allowed than named ones

-6 by starting count at 2 ₊₁ so that the and x>1 in the prime checker I borrowed is unnecessary _-7

edited May 15 at 9:28

answered May 14 at 20:44

speedstyle

695

$begingroup$
The answer going into a variable is usually not considered a valid form of "output". Could you rework your answer so that the result is either output to stdout or returned by a function?
$endgroup$
– mypetlion
May 14 at 23:01

1

$begingroup$
since anonymous functions are allowed, and the challenge explicitly allows an infinite generator, I've removed g=. I had only included it in the first place because it allowed a quick visual on TIO with print(next(g) for i in range(52)).
$endgroup$
– speedstyle
May 15 at 1:15

add a comment |

Python 3, 110 108 102 bytes

Similar method to my Mathematica answer (i.e. isPrime(1+⌊¾n²⌋) else n++)
using this golfed prime checker and returning an anonymous infinite generator

from itertools import*
(x for x in map(lambda n:1+3*n**2//4,count(2)) if all(x%j for j in range(2,x)))

Try it online!

mypetlion -2 because arguably anonymous generators are more allowed than named ones

-6 by starting count at 2 ₊₁ so that the and x>1 in the prime checker I borrowed is unnecessary _-7

edited May 15 at 9:28

answered May 14 at 20:44

speedstyle

695

Python 3, 110 108 102 bytes

Similar method to my Mathematica answer (i.e. isPrime(1+⌊¾n²⌋) else n++)
using this golfed prime checker and returning an anonymous infinite generator

from itertools import*
(x for x in map(lambda n:1+3*n**2//4,count(2)) if all(x%j for j in range(2,x)))

Try it online!

mypetlion -2 because arguably anonymous generators are more allowed than named ones

-6 by starting count at 2 ₊₁ so that the and x>1 in the prime checker I borrowed is unnecessary _-7

edited May 15 at 9:28

answered May 14 at 20:44

speedstyle

695

edited May 15 at 9:28

answered May 14 at 20:44

speedstyle

695

answered May 14 at 20:44

speedstyle

695

answered May 14 at 20:44

speedstyle

695

$begingroup$
The answer going into a variable is usually not considered a valid form of "output". Could you rework your answer so that the result is either output to stdout or returned by a function?
$endgroup$
– mypetlion
May 14 at 23:01

1

$begingroup$
since anonymous functions are allowed, and the challenge explicitly allows an infinite generator, I've removed g=. I had only included it in the first place because it allowed a quick visual on TIO with print(next(g) for i in range(52)).
$endgroup$
– speedstyle
May 15 at 1:15

add a comment |

$begingroup$
The answer going into a variable is usually not considered a valid form of "output". Could you rework your answer so that the result is either output to stdout or returned by a function?
$endgroup$
– mypetlion
May 14 at 23:01

1

$begingroup$
since anonymous functions are allowed, and the challenge explicitly allows an infinite generator, I've removed g=. I had only included it in the first place because it allowed a quick visual on TIO with print(next(g) for i in range(52)).
$endgroup$
– speedstyle
May 15 at 1:15

The answer going into a variable is usually not considered a valid form of "output". Could you rework your answer so that the result is either output to stdout or returned by a function?

– mypetlion
May 14 at 23:01

since anonymous functions are allowed, and the challenge explicitly allows an infinite generator, I've removed g=. I had only included it in the first place because it allowed a quick visual on TIO with print(next(g) for i in range(52)).

– speedstyle
May 15 at 1:15

add a comment |

Japt, 14 13 bytes

Adapted from Arnauld's formula. 1-indexed.

@µXj}f@Ò(X²*¾

Try it

1 byte saved thanks to EmbodimentOfIgnorance.

edited May 15 at 9:46

answered May 14 at 21:25

Shaggy

19.5k31768

$begingroup$
13 bytes? Not tested thoroughly though.
$endgroup$
– Embodiment of Ignorance
May 15 at 2:15

$begingroup$
Thanks, @EmbodimentofIgnorance. I'd tried that but it didn't work; turns out I'd forgotten the (.
$endgroup$
– Shaggy
May 15 at 9:47

add a comment |

Japt, 14 13 bytes

Adapted from Arnauld's formula. 1-indexed.

@µXj}f@Ò(X²*¾

Try it

1 byte saved thanks to EmbodimentOfIgnorance.

edited May 15 at 9:46

answered May 14 at 21:25

Shaggy

19.5k31768

$begingroup$
13 bytes? Not tested thoroughly though.
$endgroup$
– Embodiment of Ignorance
May 15 at 2:15

$begingroup$
Thanks, @EmbodimentofIgnorance. I'd tried that but it didn't work; turns out I'd forgotten the (.
$endgroup$
– Shaggy
May 15 at 9:47

add a comment |

Japt, 14 13 bytes

Adapted from Arnauld's formula. 1-indexed.

@µXj}f@Ò(X²*¾

Try it

1 byte saved thanks to EmbodimentOfIgnorance.

edited May 15 at 9:46

answered May 14 at 21:25

Shaggy

19.5k31768

Japt, 14 13 bytes

Adapted from Arnauld's formula. 1-indexed.

@µXj}f@Ò(X²*¾

Try it

1 byte saved thanks to EmbodimentOfIgnorance.

edited May 15 at 9:46

answered May 14 at 21:25

Shaggy

19.5k31768

edited May 15 at 9:46

answered May 14 at 21:25

Shaggy

19.5k31768

answered May 14 at 21:25

Shaggy

19.5k31768

answered May 14 at 21:25

Shaggy

19.5k31768

$begingroup$
13 bytes? Not tested thoroughly though.
$endgroup$
– Embodiment of Ignorance
May 15 at 2:15

$begingroup$
Thanks, @EmbodimentofIgnorance. I'd tried that but it didn't work; turns out I'd forgotten the (.
$endgroup$
– Shaggy
May 15 at 9:47

add a comment |

$begingroup$
13 bytes? Not tested thoroughly though.
$endgroup$
– Embodiment of Ignorance
May 15 at 2:15

$begingroup$
Thanks, @EmbodimentofIgnorance. I'd tried that but it didn't work; turns out I'd forgotten the (.
$endgroup$
– Shaggy
May 15 at 9:47

13 bytes? Not tested thoroughly though.

– Embodiment of Ignorance
May 15 at 2:15

Thanks, @EmbodimentofIgnorance. I'd tried that but it didn't work; turns out I'd forgotten the (.

– Shaggy
May 15 at 9:47

add a comment |

Racket, 124 bytes

(require math)(define(f n[i 3])(let([t(+(exact-floor(* 3/4 i i))1)][k(+ 1 i)])(if(prime? t)(if(= 0 n)t(f(- n 1)k))(f n k))))

Try it online!

Returns the n-th cuban prime, 0-indexed.

Uses the formula of @Arnauld's JavaScript answer

edited May 15 at 10:13

answered May 15 at 7:46

Galen Ivanov

8,25711137

add a comment |

Racket, 124 bytes

(require math)(define(f n[i 3])(let([t(+(exact-floor(* 3/4 i i))1)][k(+ 1 i)])(if(prime? t)(if(= 0 n)t(f(- n 1)k))(f n k))))

Try it online!

Returns the n-th cuban prime, 0-indexed.

Uses the formula of @Arnauld's JavaScript answer

edited May 15 at 10:13

answered May 15 at 7:46

Galen Ivanov

8,25711137

add a comment |

Racket, 124 bytes

(require math)(define(f n[i 3])(let([t(+(exact-floor(* 3/4 i i))1)][k(+ 1 i)])(if(prime? t)(if(= 0 n)t(f(- n 1)k))(f n k))))

Try it online!

Returns the n-th cuban prime, 0-indexed.

Uses the formula of @Arnauld's JavaScript answer

edited May 15 at 10:13

answered May 15 at 7:46

Galen Ivanov

8,25711137

Racket, 124 bytes

(require math)(define(f n[i 3])(let([t(+(exact-floor(* 3/4 i i))1)][k(+ 1 i)])(if(prime? t)(if(= 0 n)t(f(- n 1)k))(f n k))))

Try it online!

Returns the n-th cuban prime, 0-indexed.

Uses the formula of @Arnauld's JavaScript answer

edited May 15 at 10:13

answered May 15 at 7:46

Galen Ivanov

8,25711137

edited May 15 at 10:13

answered May 15 at 7:46

Galen Ivanov

8,25711137

answered May 15 at 7:46

Galen Ivanov

8,25711137

answered May 15 at 7:46

Galen Ivanov

8,25711137

add a comment |

Python 3, 83 bytes

prints the cuban primes forever.

P=k=1
while 1:P*=k*k;x=k;k+=1;P%k>0==((x/3)**.5%1)*((x/3+.25)**.5%1-.5)and print(k)

Try it online!

Based on this prime generator. For every prime it checks whether an integer y exists that fulfills the equation for either $x = 1+y$ or $x=2+y$.

$$ p=frac(1+y)^3-y^3(1+y)-y = 1 + 3y +3y^2 Leftrightarrow y = -frac12pmsqrtfrac14+fracp-13$$

$$ p=frac(2+y)^3-y^3(1+y)-y = 4 + 6y +3y^2 Leftrightarrow y = -1 pmsqrtfracp-13$$
As we only care whether $y$ has an integer solution, we can ignore the $pm$ and $-1$.

edited May 15 at 14:46

answered May 15 at 14:32

ovs

19.7k21161

add a comment |

Python 3, 83 bytes

prints the cuban primes forever.

P=k=1
while 1:P*=k*k;x=k;k+=1;P%k>0==((x/3)**.5%1)*((x/3+.25)**.5%1-.5)and print(k)

Try it online!

Based on this prime generator. For every prime it checks whether an integer y exists that fulfills the equation for either $x = 1+y$ or $x=2+y$.

$$ p=frac(1+y)^3-y^3(1+y)-y = 1 + 3y +3y^2 Leftrightarrow y = -frac12pmsqrtfrac14+fracp-13$$

$$ p=frac(2+y)^3-y^3(1+y)-y = 4 + 6y +3y^2 Leftrightarrow y = -1 pmsqrtfracp-13$$
As we only care whether $y$ has an integer solution, we can ignore the $pm$ and $-1$.

edited May 15 at 14:46

answered May 15 at 14:32

ovs

19.7k21161

add a comment |

Python 3, 83 bytes

prints the cuban primes forever.

P=k=1
while 1:P*=k*k;x=k;k+=1;P%k>0==((x/3)**.5%1)*((x/3+.25)**.5%1-.5)and print(k)

Try it online!

Based on this prime generator. For every prime it checks whether an integer y exists that fulfills the equation for either $x = 1+y$ or $x=2+y$.

$$ p=frac(1+y)^3-y^3(1+y)-y = 1 + 3y +3y^2 Leftrightarrow y = -frac12pmsqrtfrac14+fracp-13$$

$$ p=frac(2+y)^3-y^3(1+y)-y = 4 + 6y +3y^2 Leftrightarrow y = -1 pmsqrtfracp-13$$
As we only care whether $y$ has an integer solution, we can ignore the $pm$ and $-1$.

edited May 15 at 14:46

answered May 15 at 14:32

ovs

19.7k21161

Python 3, 83 bytes

prints the cuban primes forever.

P=k=1
while 1:P*=k*k;x=k;k+=1;P%k>0==((x/3)**.5%1)*((x/3+.25)**.5%1-.5)and print(k)

Try it online!

Based on this prime generator. For every prime it checks whether an integer y exists that fulfills the equation for either $x = 1+y$ or $x=2+y$.

$$ p=frac(1+y)^3-y^3(1+y)-y = 1 + 3y +3y^2 Leftrightarrow y = -frac12pmsqrtfrac14+fracp-13$$

$$ p=frac(2+y)^3-y^3(1+y)-y = 4 + 6y +3y^2 Leftrightarrow y = -1 pmsqrtfracp-13$$
As we only care whether $y$ has an integer solution, we can ignore the $pm$ and $-1$.

edited May 15 at 14:46

answered May 15 at 14:32

ovs

19.7k21161

edited May 15 at 14:46

answered May 15 at 14:32

ovs

19.7k21161

answered May 15 at 14:32

ovs

19.7k21161

answered May 15 at 14:32

ovs

19.7k21161

add a comment |

Perl 6, 33 31 bytes

-2 bytes thanks to Grimy

¾*$++²xx*

Try it online!

Anonymous code block that returns a lazy infinite list of Cuban primes. This uses Arnauld's formula to generate possible cuban primes, then &is-prime to filter them.

Explanation:

 # Anonymous code block
 grep &is-prime, # Filter the primes from
 xx* # The infinite list
 ¾* # Of three quarters
 $++² # Of an increasing number squared
 1+| # Add one by ORing with 1

edited May 16 at 11:14

answered May 16 at 5:05

Jo King

28.4k366134

1

$begingroup$
1+0+| can be just 1+|
$endgroup$
– Grimy
May 16 at 10:49

add a comment |

Perl 6, 33 31 bytes

-2 bytes thanks to Grimy

¾*$++²xx*

Try it online!

Anonymous code block that returns a lazy infinite list of Cuban primes. This uses Arnauld's formula to generate possible cuban primes, then &is-prime to filter them.

Explanation:

 # Anonymous code block
 grep &is-prime, # Filter the primes from
 xx* # The infinite list
 ¾* # Of three quarters
 $++² # Of an increasing number squared
 1+| # Add one by ORing with 1

edited May 16 at 11:14

answered May 16 at 5:05

Jo King

28.4k366134

1

$begingroup$
1+0+| can be just 1+|
$endgroup$
– Grimy
May 16 at 10:49

add a comment |

Perl 6, 33 31 bytes

-2 bytes thanks to Grimy

¾*$++²xx*

Try it online!

Anonymous code block that returns a lazy infinite list of Cuban primes. This uses Arnauld's formula to generate possible cuban primes, then &is-prime to filter them.

Explanation:

 # Anonymous code block
 grep &is-prime, # Filter the primes from
 xx* # The infinite list
 ¾* # Of three quarters
 $++² # Of an increasing number squared
 1+| # Add one by ORing with 1

edited May 16 at 11:14

answered May 16 at 5:05

Jo King

28.4k366134

Perl 6, 33 31 bytes

-2 bytes thanks to Grimy

¾*$++²xx*

Try it online!

Anonymous code block that returns a lazy infinite list of Cuban primes. This uses Arnauld's formula to generate possible cuban primes, then &is-prime to filter them.

Explanation:

 # Anonymous code block
 grep &is-prime, # Filter the primes from
 xx* # The infinite list
 ¾* # Of three quarters
 $++² # Of an increasing number squared
 1+| # Add one by ORing with 1

edited May 16 at 11:14

answered May 16 at 5:05

Jo King

28.4k366134

edited May 16 at 11:14

answered May 16 at 5:05

Jo King

28.4k366134

answered May 16 at 5:05

Jo King

28.4k366134

answered May 16 at 5:05

Jo King

28.4k366134

1

$begingroup$
1+0+| can be just 1+|
$endgroup$
– Grimy
May 16 at 10:49

add a comment |

1

$begingroup$
1+0+| can be just 1+|
$endgroup$
– Grimy
May 16 at 10:49

1+0+| can be just 1+|

– Grimy
May 16 at 10:49

add a comment |

Pari/GP, 51 bytes

Using Arnauld's formula.

n->a=0;for(i=1,n,until(isprime(p=3*a^24+1),a++));p

Try it online!

answered May 16 at 4:46

alephalpha

22.4k33096

add a comment |

Pari/GP, 51 bytes

Using Arnauld's formula.

n->a=0;for(i=1,n,until(isprime(p=3*a^24+1),a++));p

Try it online!

answered May 16 at 4:46

alephalpha

22.4k33096

add a comment |

Pari/GP, 51 bytes

Using Arnauld's formula.

n->a=0;for(i=1,n,until(isprime(p=3*a^24+1),a++));p

Try it online!

answered May 16 at 4:46

alephalpha

22.4k33096

Pari/GP, 51 bytes

Using Arnauld's formula.

n->a=0;for(i=1,n,until(isprime(p=3*a^24+1),a++));p

Try it online!

answered May 16 at 4:46

alephalpha

22.4k33096

answered May 16 at 4:46

alephalpha

22.4k33096

answered May 16 at 4:46

alephalpha

22.4k33096

answered May 16 at 4:46

alephalpha

22.4k33096

add a comment |

APL(NARS), 98 chars, 196 bytes

r←h w;y;c;v
r←c←y←0⋄→4
→3×⍳∼0πv←1+3×y×1+y+←1⋄r←v⋄→0×⍳w≤c+←1
→2×⍳∼0πv+←3×y+1⋄c+←1⋄r←v
→2×⍳w>c

indented :

r←h w;y;c;v
r←c←y←0⋄→4
 →3×⍳∼0πv←1+3×y×1+y+←1⋄r←v⋄→0×⍳w≤c+←1
 →2×⍳∼0πv+←3×y+1⋄c+←1⋄r←v
 →2×⍳w>c

test:

 h ¨1..20
7 13 19 37 61 109 127 193 271 331 397 433 547 631 769 919 1201 1453 1657 1801 
 h 1000
25789873
 h 10000
4765143511

it is based on: if y in N, one possible Cuban Prime is

S1=1+3y(y+1)

the the next possible Cuban Prime will be

S2=3(y+1)+S1

edited May 23 at 10:04

answered May 16 at 8:33

RosLuP

2,346514

add a comment |

APL(NARS), 98 chars, 196 bytes

r←h w;y;c;v
r←c←y←0⋄→4
→3×⍳∼0πv←1+3×y×1+y+←1⋄r←v⋄→0×⍳w≤c+←1
→2×⍳∼0πv+←3×y+1⋄c+←1⋄r←v
→2×⍳w>c

indented :

r←h w;y;c;v
r←c←y←0⋄→4
 →3×⍳∼0πv←1+3×y×1+y+←1⋄r←v⋄→0×⍳w≤c+←1
 →2×⍳∼0πv+←3×y+1⋄c+←1⋄r←v
 →2×⍳w>c

test:

 h ¨1..20
7 13 19 37 61 109 127 193 271 331 397 433 547 631 769 919 1201 1453 1657 1801 
 h 1000
25789873
 h 10000
4765143511

it is based on: if y in N, one possible Cuban Prime is

S1=1+3y(y+1)

the the next possible Cuban Prime will be

S2=3(y+1)+S1

edited May 23 at 10:04

answered May 16 at 8:33

RosLuP

2,346514

add a comment |

APL(NARS), 98 chars, 196 bytes

r←h w;y;c;v
r←c←y←0⋄→4
→3×⍳∼0πv←1+3×y×1+y+←1⋄r←v⋄→0×⍳w≤c+←1
→2×⍳∼0πv+←3×y+1⋄c+←1⋄r←v
→2×⍳w>c

indented :

r←h w;y;c;v
r←c←y←0⋄→4
 →3×⍳∼0πv←1+3×y×1+y+←1⋄r←v⋄→0×⍳w≤c+←1
 →2×⍳∼0πv+←3×y+1⋄c+←1⋄r←v
 →2×⍳w>c

test:

 h ¨1..20
7 13 19 37 61 109 127 193 271 331 397 433 547 631 769 919 1201 1453 1657 1801 
 h 1000
25789873
 h 10000
4765143511

it is based on: if y in N, one possible Cuban Prime is

S1=1+3y(y+1)

the the next possible Cuban Prime will be

S2=3(y+1)+S1

edited May 23 at 10:04

answered May 16 at 8:33

RosLuP

2,346514

APL(NARS), 98 chars, 196 bytes

r←h w;y;c;v
r←c←y←0⋄→4
→3×⍳∼0πv←1+3×y×1+y+←1⋄r←v⋄→0×⍳w≤c+←1
→2×⍳∼0πv+←3×y+1⋄c+←1⋄r←v
→2×⍳w>c

indented :

r←h w;y;c;v
r←c←y←0⋄→4
 →3×⍳∼0πv←1+3×y×1+y+←1⋄r←v⋄→0×⍳w≤c+←1
 →2×⍳∼0πv+←3×y+1⋄c+←1⋄r←v
 →2×⍳w>c

test:

 h ¨1..20
7 13 19 37 61 109 127 193 271 331 397 433 547 631 769 919 1201 1453 1657 1801 
 h 1000
25789873
 h 10000
4765143511

it is based on: if y in N, one possible Cuban Prime is

S1=1+3y(y+1)

the the next possible Cuban Prime will be

S2=3(y+1)+S1

edited May 23 at 10:04

answered May 16 at 8:33

RosLuP

2,346514

edited May 23 at 10:04

answered May 16 at 8:33

RosLuP

2,346514

answered May 16 at 8:33

RosLuP

2,346514

answered May 16 at 8:33

RosLuP

2,346514

add a comment |

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.

…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

draft saved

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If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.

…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

draft saved

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​Cuban​ ​PrimesIs this number a prime?List Prime NumbersLeyland NumbersExcessive IntegersClosest Woodall PrimePalindromic Primes without 11Is it a weak prime?Primes other than OptimusPrimus-Orderus PrimesRepeated Digit PrimesBertrand's PrimesSome Lonely Primes

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Cuban Primes

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Cuban Primes

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Test cases

16 Answers 16

JavaScript (V8), 54 bytes

JavaScript (ES6), 63 61 60 59 bytes

How?

05AB1E, 16 12 9 bytes

R, 75 73 bytes

Wolfram Language (Mathematica), 66 65 56 bytes

Java 8, 94 88 86 84 bytes

Wolfram Language (Mathematica), 83 bytes

Jelly, 12 bytes

Wolfram Language (Mathematica), 83 bytes

Whitespace, 180 bytes

Python 3, 110 108 102 bytes

Japt, 14 13 bytes

Racket, 124 bytes

Python 3, 83 bytes

Perl 6, 33 31 bytes

Explanation:

Pari/GP, 51 bytes

APL(NARS), 98 chars, 196 bytes

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16 Answers 16

16 Answers 16

JavaScript (V8), 54 bytes

JavaScript (ES6), 63 61 60 59 bytes

How?

JavaScript (V8), 54 bytes

JavaScript (ES6), 63 61 60 59 bytes

How?

JavaScript (V8), 54 bytes

JavaScript (ES6), 63 61 60 59 bytes

How?

JavaScript (V8), 54 bytes

JavaScript (ES6), 63 61 60 59 bytes

How?

05AB1E, 16 12 9 bytes

05AB1E, 16 12 9 bytes

05AB1E, 16 12 9 bytes

05AB1E, 16 12 9 bytes

R, 75 73 bytes

R, 75 73 bytes

R, 75 73 bytes

R, 75 73 bytes

Wolfram Language (Mathematica), 66 65 56 bytes

Wolfram Language (Mathematica), 66 65 56 bytes

Wolfram Language (Mathematica), 66 65 56 bytes

Wolfram Language (Mathematica), 66 65 56 bytes

Java 8, 94 88 86 84 bytes

Java 8, 94 88 86 84 bytes

Java 8, 94 88 86 84 bytes

Java 8, 94 88 86 84 bytes

Wolfram Language (Mathematica), 83 bytes

Wolfram Language (Mathematica), 83 bytes

Wolfram Language (Mathematica), 83 bytes

Wolfram Language (Mathematica), 83 bytes

Jelly, 12 bytes

Jelly, 12 bytes

Jelly, 12 bytes

Jelly, 12 bytes

Wolfram Language (Mathematica), 83 bytes

Wolfram Language (Mathematica), 83 bytes

Wolfram Language (Mathematica), 83 bytes

Wolfram Language (Mathematica), 83 bytes

Whitespace, 180 bytes

Cuban PrimesIs this number a prime?List Prime NumbersLeyland NumbersExcessive IntegersClosest Woodall PrimePalindromic Primes without 11Is it a weak prime?Primes other than OptimusPrimus-Orderus PrimesRepeated Digit PrimesBertrand's PrimesSome Lonely Primes

16 Answers
16

16 Answers
16

16 Answers
16