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A318057
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.
1
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, 32, 36, 32, 38, 32, 41, 42, 44, 45, 46, 47, 50, 48, 52, 54, 53, 56, 53, 58, 59, 60, 64, 62, 66, 62, 67, 69, 71, 73, 75, 74, 77, 78, 80, 82, 81, 84, 81, 87, 81, 88, 90
OFFSET
1,2
COMMENTS
The correct binary value of the golden section is given in A068432; the continued fraction terms of the golden section is given in A000012.
For the number of correct decimal digits of the golden section see A318058.
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.
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).
FORMULA
Lim {n -> oo} a(n)/n = 2*log(A001622)/log(2) = 2*A002390/log(2) = A202543/log(2) = 2*A242208.
EXAMPLE
n convergent binary expansion a(n)
== ============= ========================== ====
1 1 / 1 1.0 0
2 2 / 1 10.0 -2
3 3 / 2 1.1000 3
4 5 / 3 1.101 2
5 8 / 5 1.100110 5
6 13 / 8 1.101 2
7 21 / 13 1.1001110 6
8 34 / 21 1.1001111001 9
9 55 / 34 1.10011110000 10
10 89 / 55 1.1001111001 9
oo lim = A068432 1.1001111000110111011110 --
PROG
(Python)
p, q, i, base = 1, 1, 0, 2
while i < 20200:
p, q, i = p+q, p, i+1
a0, p, q = p//q, q, p
i, p, dd = 0, p*base, [0]
while i < 30000:
d, p, i = p//q, (p%q)*base, i+1
dd = dd+[d]
n, pn, qn = 0, 1, 0
while n < 20000:
n, pn, qn = n+1, pn+qn, pn
if pn//qn != a0:
print(n, "- manual!")
else:
i, p, q, di = 0, (pn%qn)*base, qn, 0
while di == dd[i]:
i, di, p = i+1, p//q, (p%q)*base
print(n, i-1)
KEYWORD
sign,base
AUTHOR
A.H.M. Smeets, Aug 14 2018
STATUS
approved