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a(n) is the number of binary places to which n-th convergent of continued fraction expansion of the golden section matches the correct value.
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%I #19 Nov 04 2024 18:15:31

%S 0,-2,3,2,5,2,6,9,10,9,13,12,15,16,19,16,20,22,24,25,27,29,28,30,33,

%T 32,36,32,38,32,41,42,44,45,46,47,50,48,52,54,53,56,53,58,59,60,64,62,

%U 66,62,67,69,71,73,75,74,77,78,80,82,81,84,81,87,81,88,90

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

%C The correct binary value of the golden section is given in A068432; 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 A318058.

%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.

%C The sequence for quaternary digits is obtained by floor(a(n)/2), the sequence for octal digits is obtained by floor(a(n)/3), and the sequence for hexadecimal digits is obtained by floor(a(n)/4).

%F Lim {n -> oo} a(n)/n = 2*log(A001622)/log(2) = 2*A002390/log(2) = A202543/log(2) = 2*A242208.

%e n convergent binary expansion a(n)

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

%e 1 1 / 1 1.0 0

%e 2 2 / 1 10.0 -2

%e 3 3 / 2 1.1000 3

%e 4 5 / 3 1.101 2

%e 5 8 / 5 1.100110 5

%e 6 13 / 8 1.101 2

%e 7 21 / 13 1.1001110 6

%e 8 34 / 21 1.1001111001 9

%e 9 55 / 34 1.10011110000 10

%e 10 89 / 55 1.1001111001 9

%e oo lim = A068432 1.1001111000110111011110 --

%o (Python)

%o p, q, i, base = 1, 1, 0, 2

%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, A068432, A202543, A242208, A318058.

%K sign,base,changed

%O 1,2

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