Mryntu

Which one is the true statement?

Is a distribution that is normal, but highly skewed considered Gaussian?

Is there a difference between "Fahrstuhl" and "Aufzug"

Domestic-to-international connection at Orlando (MCO)

What is meant by "large scale tonal organization?"

Legal workarounds for testamentary trust perceived as unfair

Make solar eclipses exceedingly rare, but still have new moons

TikZ: How to reverse arrow direction without switching start/end point?

The past simple of "gaslight" – "gaslighted" or "gaslit"?

Bartok - Syncopation (1): Meaning of notes in between Grand Staff

Using Rolle's theorem to show an equation has only one real root

Why doesn't UK go for the same deal Japan has with EU to resolve Brexit?

Reference request: Grassmannian and Plucker coordinates in type B, C, D

A Man With a Stainless Steel Endoskeleton (like The Terminator) Fighting Cloaked Aliens Only He Can See

Is it ever safe to open a suspicious HTML file (e.g. email attachment)?

Writing differences on a blackboard

Why the difference in type-inference over the as-pattern in two similar function definitions?

Is there a way to save my career from absolute disaster?

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

If the heap is zero-initialized for security, then why is the stack merely uninitialized?

Find non-case sensitive string in a mixed list of elements?

The exact meaning of 'Mom made me a sandwich'

Is it possible to use a NPN BJT as switch, from single power source?

How to count occurrences of text in a file?

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

share|cite|improve this question

asked 20 mins ago

AnonymousAnonymous

111

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
  15 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 20 mins ago

AnonymousAnonymous

111

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
  15 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 20 mins ago

AnonymousAnonymous

111

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 20 mins ago

AnonymousAnonymous

111

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 20 mins ago

AnonymousAnonymous

111

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 20 mins ago

AnonymousAnonymous

111

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 20 mins ago

AnonymousAnonymous

111

3 Answers
3

active

oldest

votes

4

$begingroup$

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

share|cite|improve this answer

answered 16 mins ago

FlowersFlowers

638410

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

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 13 mins ago

TravisTravis

63.8k769151

$endgroup$

add a comment | 

0

$begingroup$

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 24 secs ago

TravisTravis

63.8k769151

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

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
);



);

Anonymous is a new contributor. Be nice, and check out our Code of Conduct.

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

4

$begingroup$

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

share|cite|improve this answer

answered 16 mins ago

FlowersFlowers

638410

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

4

$begingroup$

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

share|cite|improve this answer

answered 16 mins ago

FlowersFlowers

638410

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

share|cite|improve this answer

answered 16 mins ago

FlowersFlowers

638410

$endgroup$

share|cite|improve this answer

answered 16 mins ago

FlowersFlowers

638410

answered 16 mins ago

FlowersFlowers

638410

answered 16 mins ago

FlowersFlowers

638410

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 13 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 13 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 13 mins ago

TravisTravis

63.8k769151

$endgroup$

share|cite|improve this answer

answered 13 mins ago

TravisTravis

63.8k769151

answered 13 mins ago

TravisTravis

63.8k769151

answered 13 mins ago

TravisTravis

63.8k769151

0

$begingroup$

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 24 secs ago

TravisTravis

63.8k769151

$endgroup$

add a comment | 

0

$begingroup$

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 24 secs 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 24 secs ago

TravisTravis

63.8k769151

$endgroup$

share|cite

answered 24 secs ago

TravisTravis

63.8k769151

answered 24 secs ago

TravisTravis

63.8k769151

answered 24 secs ago

TravisTravis

63.8k769151