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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?

Karæthon
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  • 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...) – 19aksh Apr 26 '19 at 17:29
  • Yeah, i noticed a pattern like that, but can't figure out the logic behind it. – Karæthon Apr 26 '19 at 17:31

2 Answers2

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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.

Fimpellizzeri
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  • 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... – Karæthon Apr 26 '19 at 17:29
  • 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). – Fimpellizzeri Apr 26 '19 at 17:49
  • Ok, I think I understand, it works on my calculator so... – Karæthon Apr 26 '19 at 17:51
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The $n^\text{th}$ denominator appears to be

$$4^{\large{2n-\text{(the number of 1s in the base-2 representation of }n)}}$$

starting with $n=0.$

Mitchell Spector
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