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

2

$begingroup$

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 non even numbers.

Any help would be highly appreciated.

sequences-and-series

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asked 23 mins ago

AnonymousAnonymous

131

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
  19 mins ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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 non even numbers.

Any help would be highly appreciated.

sequences-and-series

share|cite|improve this question

asked 23 mins ago

AnonymousAnonymous

131

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
  19 mins ago
  
  

  
  
  
  

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 non even numbers.

Any help would be highly appreciated.

sequences-and-series

share|cite|improve this question

asked 23 mins ago

AnonymousAnonymous

131

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

asked 23 mins ago

AnonymousAnonymous

131

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

asked 23 mins ago

AnonymousAnonymous

131

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 23 mins ago

AnonymousAnonymous

131

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 23 mins ago

AnonymousAnonymous

131

3 Answers
3

active

oldest

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4

$begingroup$

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

share|cite|improve this answer

answered 19 mins ago

FlowersFlowers

653410

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  3 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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

answered 3 mins ago

TravisTravis

63.8k769151

$endgroup$

add a comment | 

0

$begingroup$

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 17 mins ago

TravisTravis

63.8k769151

$endgroup$

add a comment | 

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4

$begingroup$

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

share|cite|improve this answer

answered 19 mins ago

FlowersFlowers

653410

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  3 mins ago
  

  
  
  
  

add a comment | 

4

$begingroup$

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

share|cite|improve this answer

answered 19 mins ago

FlowersFlowers

653410

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  3 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

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

share|cite|improve this answer

answered 19 mins ago

FlowersFlowers

653410

$endgroup$

share|cite|improve this answer

answered 19 mins ago

FlowersFlowers

653410

answered 19 mins ago

FlowersFlowers

653410

answered 19 mins ago

FlowersFlowers

653410

1

$begingroup$

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

answered 3 mins ago

TravisTravis

63.8k769151

$endgroup$

add a comment | 

1

$begingroup$

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

answered 3 mins ago

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

answered 3 mins ago

TravisTravis

63.8k769151

$endgroup$

share|cite

answered 3 mins ago

TravisTravis

63.8k769151

answered 3 mins ago

TravisTravis

63.8k769151

answered 3 mins ago

TravisTravis

63.8k769151

0

$begingroup$

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 17 mins ago

TravisTravis

63.8k769151

$endgroup$

add a comment | 

0

$begingroup$

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 17 mins ago

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 17 mins ago

TravisTravis

63.8k769151

$endgroup$

share|cite|improve this answer

answered 17 mins ago

TravisTravis

63.8k769151

answered 17 mins ago

TravisTravis

63.8k769151

answered 17 mins ago

TravisTravis

63.8k769151