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A295282 a(n) > n is chosen to minimize the difference between ratios a(n):n and n:(a(n) - n), so that they are matching approximations to the golden ratio. 1

%I #12 Dec 19 2017 02:46:50

%S 2,3,5,7,8,10,11,13,15,16,18,19,21,23,24,26,28,29,31,32,34,36,37,39,

%T 40,42,44,45,47,49,50,52,53,55,57,58,60,61,63,65,66,68,70,71,73,74,76,

%U 78,79,81,83,84,86,87,89,91,92,94,95,97,99,100,102,104,105,107,108

%N a(n) > n is chosen to minimize the difference between ratios a(n):n and n:(a(n) - n), so that they are matching approximations to the golden ratio.

%C The difference between the matching ratios is evaluated by dividing the larger by the smaller.

%C Take a rectangle A with sides n and n+m; remove a square of side n from one end to form rectangle B with sides n and m; scale B in the ratio (n+m):n to form rectangle C with a side n+m. Place A and C alongside, with edges of length n+m coinciding, to form rectangle D. For n > 0, let m_n be the m that has the coincident edges dividing D in nearest to equal proportions, then a(n) = n + m_n.

%C Compared with other neighboring values of n, the resulting proportions can be made most nearly equal when n is a Fibonacci number F(k) = A000045(k), k > 1, in which case a(n) is F(k+1). In contrast, if 2n is a Fibonacci number F(k), then a relatively good choice for rectangle A's longer side would be F(k+1)/2, except that F(k+1) is odd when F(k) is even, so F(k+1)/2 is halfway between integers.

%C a(n) is usually the same as A007067(n), but when 2n is a Fibonacci number, they sometimes differ. The first differences are a(4) = 7 = A007067(4) + 1 and a(72) = 117 = A007067(72) + 1. The author expects a(n) to differ from A007067(n) if and only if n is in A060645. The terms of A060645 are half the value of alternate even Fibonacci numbers.

%C More specifically, for k > 0: F(3k) is an even Fibonacci number, F(3k+1) is odd and a(F(3k)/2) = F(3k+1)/2 + 1/2; whereas A007067(F(6k+3)/2) = F(6k+4)/2 + 1/2, but A007067(F(6k)/2) = F(6k+1)/2 - 1/2.

%F a(n) = (m+n) > n so as to minimize (max((m+n)/n, n/m) / min((m+n)/n, n/m)).

%F a(n+1) = a(n) + 2 - floor((a(n)+2) * (a(n)+1-n) * (a(n)+1) * (a(n)-n) / (n+1)^4), with a(0) = 0 for the purpose of this calculation.

%e The matching ratios and the differences between them begin:

%e 2:1 1:1 2.0

%e 3:2 2:1 1.3333...

%e 5:3 3:2 1.1111...

%e 7:4 4:3 1.3125

%e 8:5 5:3 1.0416...

%e 10:6 6:4 1.1111...

%e 11:7 7:4 1.1136...

%e 13:8 8:5 1.015625

%e 15:9 9:6 1.1111...

%e 16:10 10:6 1.0416...

%e 18:11 11:7 1.0413...

%e 19:12 12:7 1.0827...

%e 21:13 13:8 1.0059...

%e 23:14 14:9 1.0561...

%e 24:15 15:9 1.0416...

%e 26:16 16:10 1.015625

%e 28:17 17:11 1.0657...

%e 29:18 18:11 1.0156...

%e 31:19 19:12 1.0304...

%e 32:20 20:12 1.0416...

%e 34:21 21:13 1.0022...

%e ...

%e For n = 4:

%e if a(4) = 5, the matching ratios would be a(4):4 = 5:4 and 4:(a(4)-4) = 4:1, with the difference between them (larger divided by smaller) = (4/1) / (5/4) = 16/5 = 3.2;

%e if a(4) = 6, ratios would be 6:4 and 4:2, with difference = (4/2) / (6/4) = 16/12 = 1.333...;

%e if a(4) = 7, ratios would be 7:4 and 4:3, with difference = (7/4) / (4/3) = 21/16 = 1.3125;

%e if a(4) = 8, ratios would be 8:4 and 4:4, with difference = (8/4) / (4/4) = 32/16 = 2.0.

%e Any larger value for a(4) would give a difference between the ratios that exceeded 2.0, so a(4) = 7, as this achieves the minimum difference.

%e This example translates as follows into the geometry described early in the comments:

%e n 4 4

%e m 2 3

%e n+m 6 7

%e Rectangle A n X (n+m) 4 X 6 4 X 7

%e Rectangle B m X n 2 X 4 3 X 4

%e Scaling ratio n:(n+m) 4:6 4:7

%e m scaled up m*(n+m)/n 2*6/4 3*7/4

%e = side of C l 3 5.25

%e Rectangle C l X (n+m) 3 X 6 5.25 X 7

%e Rectangle D (n+l) X (n+m) 7 X 6 9.25 X 7

%e proportion C/D l/(n+l) 3/7 5.25/9.25

%e - as decimal 0.4285... 0.5675...

%e - its difference from 0.5 0.0714... 0.0675...

%Y A001622 gives the value of the golden ratio.

%Y Cf. A000045, A007067, A022342, A060645.

%K nonn

%O 1,1

%A _Peter Munn_, Nov 19 2017

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