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%I #23 Dec 06 2019 21:43:55
%S 4,6,7,8,10,12,15,16,21,24,25,29,35,41,42,45,47,51,53,54,56,57,58,60,
%T 61,63,64,66,68,69,74,79,82,84,87,89,92,94,96,98,99,104,108,113,115,
%U 116,121,125,126,134,136,138,141,144,148,149,150,154,157,158,160
%N Positions of 1 in the continued fraction expansion of Pi.
%C 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
%H T. D. Noe, <a href="/A203168/b203168.txt">Table of n, a(n) for n = 1..1000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PiContinuedFraction.html">Pi Continued Fraction</a>
%H <a href="/index/Con#confC">Index entries for continued fractions for constants</a>
%H <a href="/index/Ph#Pi314">Index entries for sequences related to the number Pi</a>
%F A001203(a(n)) = 1.
%t Flatten[Position[ContinuedFraction[Pi, 160], 1]]
%o (PARI) v=contfrac(Pi);for(i=1,#v,if(v[i]==1,print1(i", "))) \\ _Charles R Greathouse IV_, Dec 30 2011
%Y Cf. A001203, A033089, A000796.
%K nonn,nice
%O 1,1
%A _Ben Branman_, Dec 29 2011
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