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
LINKS
T. D. Noe, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Pi Continued Fraction
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