Dtjytyk

How to write a column outside the braces in a matrix?

"The cow" OR "a cow" OR "cows" in this context

Map of water taps to fill bottles

Can SQL Server create collisions in system generated constraint names?

Can we say “you can pay when the order gets ready”?

can anyone help me with this awful query plan?

Is the claim "Employers won't employ people with no 'social media presence'" realistic?

How to denote matrix elements succinctly?

Extension of 2-adic valuation to the real numbers

Multiple options vs single option UI

What does the integral of a function times a function of a random variable represent, conceptually?

How much cash can I safely carry into the USA and avoid civil forfeiture?

What makes accurate emulation of old systems a difficult task?

What happens to Mjolnir (Thor's hammer) at the end of Endgame?

Pre-plastic human skin alternative

How to stop co-workers from teasing me because I know Russian?

Re-entry to Germany after vacation using blue card

How did Captain America manage to do this?

Minor Revision with suggestion of an alternative proof by reviewer

What is the philosophical significance of speech acts/implicature?

Implications of cigar-shaped bodies having rings?

Can I grease a crank spindle/bracket without disassembling the crank set?

A ​Note ​on ​N!

"Whatever a Russian does, they end up making the Kalashnikov gun"? Are there any similar proverbs in English?

Finding a pattern, I'm stuck

Finding pattern (2,7,8,3,5)Finding pattern- $6$ $11$ $25$What is a permutation pattern?Pattern finding algorithmsNumber patterns - finding the patternPalindromic Numbers - Pattern “inside” Prime Numbers?Collatz conjecture pattern (3n + 1 problem).Completing sequence patternWhat is the general term of this series (obtained from an iterated mean)?Find the number pattern

3

$begingroup$

I have a friend who knows how much I love math, (I imagine new problems just to do the math behind them and to see if I can expand my understanding) and so he brings me the stuff that stumps him. Usually I end up finding the answer for him and then explaining how to him, but he gave me one 2 weeks ago and I cant figure it out. It's part of some math for calculating the area (or maybe perimeter?) of an elipse, the part he wanted me to figure out is: there is a part that goes π * (a + b) * (1 + 1/4 * h + 1/64 * h^2 + 1/256 * h^3 + 1/16384 * h^4 + 1/ 65536 * h^5 + 1/1048576 * h^6 ...) we only have those six, and he wanted me to find the pattern to calculate the rest of the denominators to extend it. But everything I have tried adds extra values within the first six or skips values within the first six. Does anyone know what the pattern is and what formula can be used to calculate the rest?

binomial-coefficients pattern-recognition pattern-matching

share|cite|improve this question

asked 8 hours ago

KaræthonKaræthon

183

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think it follows the increment pattern of 1,2,1,3,1,2,1,3... in the powers of 4 (starting from 0- 0,1,3,4,7,8,10,11...)
  $endgroup$
  – Ak19
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, i noticed a pattern like that, but can't figure out the logic behind it.
  $endgroup$
  – Karæthon
  8 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

I have a friend who knows how much I love math, (I imagine new problems just to do the math behind them and to see if I can expand my understanding) and so he brings me the stuff that stumps him. Usually I end up finding the answer for him and then explaining how to him, but he gave me one 2 weeks ago and I cant figure it out. It's part of some math for calculating the area (or maybe perimeter?) of an elipse, the part he wanted me to figure out is: there is a part that goes π * (a + b) * (1 + 1/4 * h + 1/64 * h^2 + 1/256 * h^3 + 1/16384 * h^4 + 1/ 65536 * h^5 + 1/1048576 * h^6 ...) we only have those six, and he wanted me to find the pattern to calculate the rest of the denominators to extend it. But everything I have tried adds extra values within the first six or skips values within the first six. Does anyone know what the pattern is and what formula can be used to calculate the rest?

binomial-coefficients pattern-recognition pattern-matching

share|cite|improve this question

asked 8 hours ago

KaræthonKaræthon

183

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think it follows the increment pattern of 1,2,1,3,1,2,1,3... in the powers of 4 (starting from 0- 0,1,3,4,7,8,10,11...)
  $endgroup$
  – Ak19
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, i noticed a pattern like that, but can't figure out the logic behind it.
  $endgroup$
  – Karæthon
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I have a friend who knows how much I love math, (I imagine new problems just to do the math behind them and to see if I can expand my understanding) and so he brings me the stuff that stumps him. Usually I end up finding the answer for him and then explaining how to him, but he gave me one 2 weeks ago and I cant figure it out. It's part of some math for calculating the area (or maybe perimeter?) of an elipse, the part he wanted me to figure out is: there is a part that goes π * (a + b) * (1 + 1/4 * h + 1/64 * h^2 + 1/256 * h^3 + 1/16384 * h^4 + 1/ 65536 * h^5 + 1/1048576 * h^6 ...) we only have those six, and he wanted me to find the pattern to calculate the rest of the denominators to extend it. But everything I have tried adds extra values within the first six or skips values within the first six. Does anyone know what the pattern is and what formula can be used to calculate the rest?

binomial-coefficients pattern-recognition pattern-matching

share|cite|improve this question

asked 8 hours ago

KaræthonKaræthon

183

New contributor

Karæthon 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 8 hours ago

KaræthonKaræthon

183

New contributor

Karæthon 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 8 hours ago

KaræthonKaræthon

183

New contributor

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

asked 8 hours ago

KaræthonKaræthon

183

New contributor

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

asked 8 hours ago

KaræthonKaræthon

183

2 Answers
2

active

oldest

votes

6

$begingroup$

If I could give some advice, these kinds of questions generally benefit from a look at the Online Encyclopedia of Integer Sequences.

Considering it has to do with ellipses, the sequence of denominators should be A056982.

share|cite|improve this answer

answered 8 hours ago

FimpellizieriFimpellizieri

17.3k11836

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oooo, never seen that site before.... Me likey, and yes that does appear to be the sequence, but I'm not understanding what it is telling me to do to calculate those numbers...
  $endgroup$
  – Karæthon
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The sequence is given as $4$ to the power of A005187, which in turn is given as a recurrence relation (or other non-simple descriptions). It does not appear to have a simple formula involving only sums, products and exponentiation, which is often the case in maths. There are other ways to describe it though (with words, via generating functions, or with formulas using other more complex operations).
  $endgroup$
  – Fimpellizieri
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, I think I understand, it works on my calculator so...
  $endgroup$
  – Karæthon
  8 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

The $n^textth$ denominator appears to be

$$4^large2n-text(the number of 1s in the base-2 representation of n)$$

starting with $n=0.$

share|cite|improve this answer

answered 8 hours ago

Mitchell SpectorMitchell Spector

7,1572824

$endgroup$

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
);



);

Karæthon 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%2f3203539%2ffinding-a-pattern-im-stuck%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

6

$begingroup$

If I could give some advice, these kinds of questions generally benefit from a look at the Online Encyclopedia of Integer Sequences.

Considering it has to do with ellipses, the sequence of denominators should be A056982.

share|cite|improve this answer

answered 8 hours ago

FimpellizieriFimpellizieri

17.3k11836

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oooo, never seen that site before.... Me likey, and yes that does appear to be the sequence, but I'm not understanding what it is telling me to do to calculate those numbers...
  $endgroup$
  – Karæthon
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The sequence is given as $4$ to the power of A005187, which in turn is given as a recurrence relation (or other non-simple descriptions). It does not appear to have a simple formula involving only sums, products and exponentiation, which is often the case in maths. There are other ways to describe it though (with words, via generating functions, or with formulas using other more complex operations).
  $endgroup$
  – Fimpellizieri
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, I think I understand, it works on my calculator so...
  $endgroup$
  – Karæthon
  8 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

If I could give some advice, these kinds of questions generally benefit from a look at the Online Encyclopedia of Integer Sequences.

Considering it has to do with ellipses, the sequence of denominators should be A056982.

share|cite|improve this answer

answered 8 hours ago

FimpellizieriFimpellizieri

17.3k11836

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oooo, never seen that site before.... Me likey, and yes that does appear to be the sequence, but I'm not understanding what it is telling me to do to calculate those numbers...
  $endgroup$
  – Karæthon
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The sequence is given as $4$ to the power of A005187, which in turn is given as a recurrence relation (or other non-simple descriptions). It does not appear to have a simple formula involving only sums, products and exponentiation, which is often the case in maths. There are other ways to describe it though (with words, via generating functions, or with formulas using other more complex operations).
  $endgroup$
  – Fimpellizieri
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, I think I understand, it works on my calculator so...
  $endgroup$
  – Karæthon
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

If I could give some advice, these kinds of questions generally benefit from a look at the Online Encyclopedia of Integer Sequences.

Considering it has to do with ellipses, the sequence of denominators should be A056982.

share|cite|improve this answer

answered 8 hours ago

FimpellizieriFimpellizieri

17.3k11836

$endgroup$

share|cite|improve this answer

answered 8 hours ago

FimpellizieriFimpellizieri

17.3k11836

answered 8 hours ago

FimpellizieriFimpellizieri

17.3k11836

answered 8 hours ago

FimpellizieriFimpellizieri

17.3k11836

0

$begingroup$

The $n^textth$ denominator appears to be

$$4^large2n-text(the number of 1s in the base-2 representation of n)$$

starting with $n=0.$

share|cite|improve this answer

answered 8 hours ago

Mitchell SpectorMitchell Spector

7,1572824

$endgroup$

add a comment | 

0

$begingroup$

The $n^textth$ denominator appears to be

$$4^large2n-text(the number of 1s in the base-2 representation of n)$$

starting with $n=0.$

share|cite|improve this answer

answered 8 hours ago

Mitchell SpectorMitchell Spector

7,1572824

$endgroup$

add a comment | 

$begingroup$

The $n^textth$ denominator appears to be

$$4^large2n-text(the number of 1s in the base-2 representation of n)$$

starting with $n=0.$

share|cite|improve this answer

answered 8 hours ago

Mitchell SpectorMitchell Spector

7,1572824

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Mitchell SpectorMitchell Spector

7,1572824

answered 8 hours ago

Mitchell SpectorMitchell Spector

7,1572824

answered 8 hours ago

Mitchell SpectorMitchell Spector

7,1572824