Dgdrxrt

Why can't wing-mounted spoilers be used to steepen approaches?

Single author papers against my advisor's will?

Would an alien lifeform be able to achieve space travel if lacking in vision?

Why doesn't a hydraulic lever violate conservation of energy?

How do you keep chess fun when your opponent constantly beats you?

Why are PDP-7-style microprogrammed instructions out of vogue?

Solving overdetermined system by QR decomposition

Was credit for the black hole image misappropriated?

How to read αἱμύλιος or when to aspirate

How to determine omitted units in a publication

What happens to a Warlock's expended Spell Slots when they gain a Level?

Do warforged have souls?

Working through the single responsibility principle (SRP) in Python when calls are expensive

Is this wall load bearing? Blueprints and photos attached

Is there a writing software that you can sort scenes like slides in PowerPoint?

Accepted by European university, rejected by all American ones I applied to? Possible reasons?

How to make Illustrator type tool selection automatically adapt with text length

Student Loan from years ago pops up and is taking my salary

What force causes entropy to increase?

Is an up-to-date browser secure on an out-of-date OS?

Do working physicists consider Newtonian mechanics to be "falsified"?

Fixing different display colors within string

Circular reasoning in L'Hopital's rule

For what reasons would an animal species NOT cross a *horizontal* land bridge?

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

share|cite|improve this question

edited Mar 31 at 7:04

YuiTo Cheng

2,4064937

asked Mar 31 at 6:30

AnonymousAnonymous

233

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How about using the floor function?
  $endgroup$
  – John. P
  Mar 31 at 6:35
  
  

  
  
  
  

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 Mar 31 at 7:04

YuiTo Cheng

2,4064937

asked Mar 31 at 6:30

AnonymousAnonymous

233

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How about using the floor function?
  $endgroup$
  – John. P
  Mar 31 at 6:35
  
  

  
  
  
  

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 Mar 31 at 7:04

YuiTo Cheng

2,4064937

asked Mar 31 at 6:30

AnonymousAnonymous

233

$endgroup$

share|cite|improve this question

edited Mar 31 at 7:04

YuiTo Cheng

2,4064937

asked Mar 31 at 6:30

AnonymousAnonymous

233

share|cite|improve this question

edited Mar 31 at 7:04

YuiTo Cheng

2,4064937

asked Mar 31 at 6:30

AnonymousAnonymous

233

edited Mar 31 at 7:04

YuiTo Cheng

2,4064937

edited Mar 31 at 7:04

YuiTo Cheng

2,4064937

asked Mar 31 at 6:30

AnonymousAnonymous

233

asked Mar 31 at 6:30

AnonymousAnonymous

233

3 Answers
3

active

oldest

votes

7

$begingroup$

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

share|cite|improve this answer

answered Mar 31 at 6:34

FlowersFlowers

683410

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
  $endgroup$
  – Anonymous
  Mar 31 at 6:47
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 31 at 6:50
  

  
  
  
  

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 Mar 31 at 6:50

TravisTravis

64.3k769151

$endgroup$

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 Mar 31 at 6:37

TravisTravis

64.3k769151

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3169109%2fhow-to-find-the-nth-term-in-the-following-sequence-1-1-2-2-4-4-8-8-16-16%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

7

$begingroup$

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

share|cite|improve this answer

answered Mar 31 at 6:34

FlowersFlowers

683410

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
  $endgroup$
  – Anonymous
  Mar 31 at 6:47
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 31 at 6:50
  

  
  
  
  

add a comment | 

7

$begingroup$

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

share|cite|improve this answer

answered Mar 31 at 6:34

FlowersFlowers

683410

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
  $endgroup$
  – Anonymous
  Mar 31 at 6:47
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 31 at 6:50
  

  
  
  
  

add a comment | 

$begingroup$

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

share|cite|improve this answer

answered Mar 31 at 6:34

FlowersFlowers

683410

$endgroup$

share|cite|improve this answer

answered Mar 31 at 6:34

FlowersFlowers

683410

answered Mar 31 at 6:34

FlowersFlowers

683410

answered Mar 31 at 6:34

FlowersFlowers

683410

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 Mar 31 at 6:50

TravisTravis

64.3k769151

$endgroup$

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 Mar 31 at 6:50

TravisTravis

64.3k769151

$endgroup$

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 Mar 31 at 6:50

TravisTravis

64.3k769151

$endgroup$

share|cite|improve this answer

answered Mar 31 at 6:50

TravisTravis

64.3k769151

answered Mar 31 at 6:50

TravisTravis

64.3k769151

answered Mar 31 at 6:50

TravisTravis

64.3k769151

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 Mar 31 at 6:37

TravisTravis

64.3k769151

$endgroup$

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 Mar 31 at 6:37

TravisTravis

64.3k769151

$endgroup$

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 Mar 31 at 6:37

TravisTravis

64.3k769151

$endgroup$

share|cite|improve this answer

answered Mar 31 at 6:37

TravisTravis

64.3k769151

answered Mar 31 at 6:37

TravisTravis

64.3k769151

answered Mar 31 at 6:37

TravisTravis

64.3k769151