%I #21 Apr 15 2021 01:12:32
%S 8,8,8,5,7,6,6,0,9,2,4,7,5,0,6,7,9,9,6,3,7,3,5,2,6,7,4,5,2,4,0,2,5,3,
%T 8,5,3,1,2,7,8,3,3,1,8,2,2,3,0,2,8,1,0,5,1,6,7,7,2,4,3,8,8,0,3,2,6,3,
%U 8,2,5,9,2,9,2,8,3,7,7,1,5,3,2,1,8,4,9,5,0,1,3,5,9,9
%N Decimal expansion of 99^2/1103, an approximation to 2*Pi*sqrt(2) from Ramanujan.
%C Srinivasa Ramanujan produced this curious approximation to 2*Pi*sqrt(2) (A343392) with dividing 99^2 by prime 1103 (see link Prime Curios!). This approximation comes from the 1st term of the series (44) page 47 at the Ramanujan link.
%C This formula is correct to 5 places exactly with 2*Pi*sqrt(2) = 8.885765... while 99^2/1103 = 8.885766...
%C Indeed, in the Ramanujan paper, there is 1/(2*Pi*sqrt(2)) = 1103/99^2 + ..., and in the case of these two numbers, the approximation becomes correct to 8 places exactly with 1/(2*Pi*sqrt(2)) = 0.112539539... while 1103/99^2 = 0.112539536... (see David Wells).
%D David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Revised Edition, Penguin Books, London, England, 1997, entry 3.14159..., page 36.
%H Chris K. Caldwell and G. L. Honaker, Jr., <a href="https://primes.utm.edu/curios/page.php?short=1103">1103, 1st comment</a>, Prime Curios!
%H S. Ramanujan, <a href="http://ramanujan.sirinudi.org/Volumes/published/ram06.pdf">Modular equations and approximations to Pi</a>, Quarterly Journal of Mathematics, XLV, 1914, p. 47.
%F Equals 99^2/1103.
%e 8.88576609247506799637352674524025385312783318223...
%p evalf(99^2/1103,120);
%t RealDigits[99^2/1103, 10, 100][[1]] (* _Amiram Eldar_, Apr 13 2021 *)
%Y Cf. A343392.
%K nonn,cons
%O 1,1
%A _Bernard Schott_, Apr 13 2021