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How to find the nth term in the following sequence: $1,1,2,2,4,4,8,8,16,16$

4

1

$begingroup$

I'm having difficulty in finding the formula for the sequence above, when I put this in WolframAlpha it gave me a rather complex formula which I'm not convinced even works properly but I'm sure there's a simple way to achieve this. I've searched for many similar sequences but couldn't find anything that helped me.

I'm thinking I'll most likely need to have a condition for even numbers and another for noneven numbers.

Any help would be highly appreciated.

sequences-and-series

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edited yesterday

YuiTo Cheng

2,1863937

asked yesterday

AnonymousAnonymous

232

New contributor

Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How about using the floor function?
  $endgroup$
  – John. P
  yesterday
  
  

  
  
  
  

add a comment | 

4

1

$begingroup$

I'm having difficulty in finding the formula for the sequence above, when I put this in WolframAlpha it gave me a rather complex formula which I'm not convinced even works properly but I'm sure there's a simple way to achieve this. I've searched for many similar sequences but couldn't find anything that helped me.

I'm thinking I'll most likely need to have a condition for even numbers and another for noneven numbers.

Any help would be highly appreciated.

sequences-and-series

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,1863937

asked yesterday

AnonymousAnonymous

232

New contributor

Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How about using the floor function?
  $endgroup$
  – John. P
  yesterday
  
  

  
  
  
  

add a comment | 

$begingroup$

I'm having difficulty in finding the formula for the sequence above, when I put this in WolframAlpha it gave me a rather complex formula which I'm not convinced even works properly but I'm sure there's a simple way to achieve this. I've searched for many similar sequences but couldn't find anything that helped me.

I'm thinking I'll most likely need to have a condition for even numbers and another for noneven numbers.

Any help would be highly appreciated.

sequences-and-series

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,1863937

asked yesterday

AnonymousAnonymous

232

New contributor

Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,1863937

asked yesterday

AnonymousAnonymous

232

New contributor

Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,1863937

asked yesterday

AnonymousAnonymous

232

New contributor

Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited yesterday

YuiTo Cheng

2,1863937

edited yesterday

YuiTo Cheng

2,1863937

asked yesterday

AnonymousAnonymous

232

New contributor

Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

AnonymousAnonymous

232

3 Answers
3

active

oldest

votes

6

$begingroup$

These are just powers of two. So: $2^lfloor n / 2rfloor$

share|cite|improve this answer

answered yesterday

FlowersFlowers

673410

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
  $endgroup$
  – Anonymous
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is assuming zero-indexing. So the first element is $n=0$. If you want it to be one-indexed then just subtract 1 from n in the formula.
  $endgroup$
  – Flowers
  yesterday
  

  
  
  
  

add a comment | 

3

$begingroup$

Alternatively, this is an example of a sequence where the $n$th term is a fixed linear combination of the immediately previous terms: We can write it as
$$a_n = 2 a_n - 2, qquad a_0 = a_1 = 1.$$
Using the ansatz $a_n = C r^n$ and substituting in the recursion formula gives $C r^n = 2 C r^n - 2$. Rearranging and clearing gives the characteristic equation $r^2 - 2 = 0$, whose solutions are $pm sqrt2$. So, the general solution is
$$a_n = A (sqrt2)^n + B(-sqrt2)^n = (sqrt2)^n [A + B(-1)^n] .$$
Substituting the initial values $a_0 = a_1 = 1$ gives a linear system in the coefficients $A, B$.

share|cite|improve this answer

answered yesterday

TravisTravis

63.8k769151

$endgroup$

add a comment | 

1

$begingroup$

The sequence is the powers of two, each repeated twice. We can encode the latter feature using the quantity $lfloor fracn2 rfloor$, which has values $0, 0, 1, 1, 2, 2, ldots$.

So, the sequence is given (for appropriate indexing) by $$color#df0000boxeda_n := 2^lfloor n / 2 rfloor .$$

share|cite|improve this answer

answered yesterday

TravisTravis

63.8k769151

$endgroup$

add a comment | 

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6

$begingroup$

These are just powers of two. So: $2^lfloor n / 2rfloor$

share|cite|improve this answer

answered yesterday

FlowersFlowers

673410

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
  $endgroup$
  – Anonymous
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is assuming zero-indexing. So the first element is $n=0$. If you want it to be one-indexed then just subtract 1 from n in the formula.
  $endgroup$
  – Flowers
  yesterday
  

  
  
  
  

add a comment | 

6

$begingroup$

These are just powers of two. So: $2^lfloor n / 2rfloor$

share|cite|improve this answer

answered yesterday

FlowersFlowers

673410

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
  $endgroup$
  – Anonymous
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is assuming zero-indexing. So the first element is $n=0$. If you want it to be one-indexed then just subtract 1 from n in the formula.
  $endgroup$
  – Flowers
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

These are just powers of two. So: $2^lfloor n / 2rfloor$

share|cite|improve this answer

answered yesterday

FlowersFlowers

673410

$endgroup$

share|cite|improve this answer

answered yesterday

FlowersFlowers

673410

answered yesterday

FlowersFlowers

673410

answered yesterday

FlowersFlowers

673410

3

$begingroup$

Alternatively, this is an example of a sequence where the $n$th term is a fixed linear combination of the immediately previous terms: We can write it as
$$a_n = 2 a_n - 2, qquad a_0 = a_1 = 1.$$
Using the ansatz $a_n = C r^n$ and substituting in the recursion formula gives $C r^n = 2 C r^n - 2$. Rearranging and clearing gives the characteristic equation $r^2 - 2 = 0$, whose solutions are $pm sqrt2$. So, the general solution is
$$a_n = A (sqrt2)^n + B(-sqrt2)^n = (sqrt2)^n [A + B(-1)^n] .$$
Substituting the initial values $a_0 = a_1 = 1$ gives a linear system in the coefficients $A, B$.

share|cite|improve this answer

answered yesterday

TravisTravis

63.8k769151

$endgroup$

add a comment | 

3

$begingroup$

Alternatively, this is an example of a sequence where the $n$th term is a fixed linear combination of the immediately previous terms: We can write it as
$$a_n = 2 a_n - 2, qquad a_0 = a_1 = 1.$$
Using the ansatz $a_n = C r^n$ and substituting in the recursion formula gives $C r^n = 2 C r^n - 2$. Rearranging and clearing gives the characteristic equation $r^2 - 2 = 0$, whose solutions are $pm sqrt2$. So, the general solution is
$$a_n = A (sqrt2)^n + B(-sqrt2)^n = (sqrt2)^n [A + B(-1)^n] .$$
Substituting the initial values $a_0 = a_1 = 1$ gives a linear system in the coefficients $A, B$.

share|cite|improve this answer

answered yesterday

TravisTravis

63.8k769151

$endgroup$

add a comment | 

$begingroup$

Alternatively, this is an example of a sequence where the $n$th term is a fixed linear combination of the immediately previous terms: We can write it as
$$a_n = 2 a_n - 2, qquad a_0 = a_1 = 1.$$
Using the ansatz $a_n = C r^n$ and substituting in the recursion formula gives $C r^n = 2 C r^n - 2$. Rearranging and clearing gives the characteristic equation $r^2 - 2 = 0$, whose solutions are $pm sqrt2$. So, the general solution is
$$a_n = A (sqrt2)^n + B(-sqrt2)^n = (sqrt2)^n [A + B(-1)^n] .$$
Substituting the initial values $a_0 = a_1 = 1$ gives a linear system in the coefficients $A, B$.

share|cite|improve this answer

answered yesterday

TravisTravis

63.8k769151

$endgroup$

share|cite|improve this answer

answered yesterday

TravisTravis

63.8k769151

answered yesterday

TravisTravis

63.8k769151

answered yesterday

TravisTravis

63.8k769151

1

$begingroup$

The sequence is the powers of two, each repeated twice. We can encode the latter feature using the quantity $lfloor fracn2 rfloor$, which has values $0, 0, 1, 1, 2, 2, ldots$.

So, the sequence is given (for appropriate indexing) by $$color#df0000boxeda_n := 2^lfloor n / 2 rfloor .$$

share|cite|improve this answer

answered yesterday

TravisTravis

63.8k769151

$endgroup$

add a comment | 

1

$begingroup$

The sequence is the powers of two, each repeated twice. We can encode the latter feature using the quantity $lfloor fracn2 rfloor$, which has values $0, 0, 1, 1, 2, 2, ldots$.

So, the sequence is given (for appropriate indexing) by $$color#df0000boxeda_n := 2^lfloor n / 2 rfloor .$$

share|cite|improve this answer

answered yesterday

TravisTravis

63.8k769151

$endgroup$

add a comment | 

$begingroup$

The sequence is the powers of two, each repeated twice. We can encode the latter feature using the quantity $lfloor fracn2 rfloor$, which has values $0, 0, 1, 1, 2, 2, ldots$.

So, the sequence is given (for appropriate indexing) by $$color#df0000boxeda_n := 2^lfloor n / 2 rfloor .$$

share|cite|improve this answer

answered yesterday

TravisTravis

63.8k769151

$endgroup$

share|cite|improve this answer

answered yesterday

TravisTravis

63.8k769151

answered yesterday

TravisTravis

63.8k769151

answered yesterday

TravisTravis

63.8k769151