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A203168
Positions of 1 in the continued fraction expansion of Pi.
1
4, 6, 7, 8, 10, 12, 15, 16, 21, 24, 25, 29, 35, 41, 42, 45, 47, 51, 53, 54, 56, 57, 58, 60, 61, 63, 64, 66, 68, 69, 74, 79, 82, 84, 87, 89, 92, 94, 96, 98, 99, 104, 108, 113, 115, 116, 121, 125, 126, 134, 136, 138, 141, 144, 148, 149, 150, 154, 157, 158, 160
OFFSET
1,1
COMMENTS
In the Gauss-Kuzmin distribution, 1 appears with probability log_2(4/3) = 41.5037...%. Thus the n-th appearance of 1 in the continued fraction of a real number chosen uniformly from [0, 1) will be, with probability 1, n / (log_2(4/3)) + O(sqrt(n)). Does this sequence have the same asymptotic? - Charles R Greathouse IV, Dec 30 2011
FORMULA
A001203(a(n)) = 1.
MATHEMATICA
Flatten[Position[ContinuedFraction[Pi, 160], 1]]
PROG
(PARI) v=contfrac(Pi); for(i=1, #v, if(v[i]==1, print1(i", "))) \\ Charles R Greathouse IV, Dec 30 2011
CROSSREFS
KEYWORD
nonn,nice
AUTHOR
Ben Branman, Dec 29 2011
STATUS
approved