Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%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