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a(n) is the number of decimal places to which the n-th convergent of the continued fraction expansion of the golden section matches the correct value.
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%I #24 Nov 04 2024 18:13:20

%S 0,-1,0,1,1,1,2,2,2,3,2,4,4,5,5,5,6,5,7,7,8,7,9,9,9,10,10,10,11,10,12,

%T 12,13,12,13,14,15,14,15,16,16,16,17,17,17,18,18,19,18,20,20,21,21,21,

%U 22,22,23,23,24,23,24,25,25,26,26,27,27,27,28,28,29,29,29,30,30

%N a(n) is the number of decimal places to which the n-th convergent of the continued fraction expansion of the golden section matches the correct value.

%C The correct decimal value of the golden section is given in A001622; the continued fraction terms of the golden section is given in A000012.

%C For the number of correct decimal digits of the golden section see A318057.

%C The denominator of the k-th convergent obtained from a continued fraction tend to k*A001622; the error between the k-th convergent and the constant itself tends to 1/(2*k*A001622), or in binary digits 2*k*log(A001622)/log(2) bits after the binary point.

%F Limit_{n -> oo} a(n)/n = 2*log(A001622)/log(10) = 2*A002390/log(10) = A202543/log(10) = 2*A097348.

%e n convergent decimal expansion a(n)

%e == ============= ========================= ====

%e 1 1 / 1 1.0 0

%e 2 2 / 1 2.0 -1

%e 3 3 / 2 1.5 0

%e 4 5 / 3 1.66 1

%e 5 8 / 5 1.60 1

%e 6 13 / 8 1.62 1

%e 7 21 / 13 1.615 2

%e 8 34 / 21 1.619 2

%e 9 55 / 34 1.617 2

%e 10 89 / 55 1.6181 3

%e oo lim = A001622 1.6180339887498948482 --

%o (Python)

%o p, q, i, base = 1, 1, 0, 10

%o while i < 20200:

%o p, q, i = p+q, p, i+1

%o a0, p, q = p//q, q, p

%o i, p, dd = 0, p*base, [0]

%o while i < 30000:

%o d, p, i = p//q, (p%q)*base, i+1

%o dd = dd+[d]

%o n, pn, qn = 0, 1, 0

%o while n < 20000:

%o n, pn, qn = n+1, pn+qn, pn

%o if pn//qn != a0:

%o print(n, "- manual!")

%o else:

%o i, p, q, di = 0, (pn%qn)*base, qn, 0

%o while di == dd[i]:

%o i, di, p = i+1, p//q, (p%q)*base

%o print(n, i-1)

%Y Cf. A000012, A001622, A002390, A097348, A202543, A318057.

%K sign,base,changed

%O 1,7

%A _A.H.M. Smeets_, Aug 14 2018