cp's OEIS Frontend

A126809 Minimum number of terms required in the Gregory-Leibniz series, i.e., 4(1 - 1/3 + 1/5 - 1/7 + 1/9 - ...), to obtain a value of Pi correct to n decimal digits.

%I A126809 #25 Jan 26 2025 17:27:15
%S A126809 3,19,119,1688,10794,136121,1530012,18660270,156001209,1695509436,
%T A126809 11136696006,102111268282,1260654956982,10725187563686,
%U A126809 147895359776637,1313133218365935,16250291773636035,118166387818704586,1860961545617561679,15963377896404315146
%N A126809 Minimum number of terms required in the Gregory-Leibniz series, i.e., 4(1 - 1/3 + 1/5 - 1/7 + 1/9 - ...), to obtain a value of Pi correct to n decimal digits.
%C A126809 Calculations by _Jud McCranie_.
%C A126809 The m-th partial sum of the Gregory-Leibniz series approaches Pi - (-1)^m * (1/m - 1/(4*m^3) + 5/(16*m^5) - 61/(64*m^7) + 1385/(256*m^9) - ...), where the numerators 1, 1, 5, 61, 1385, ... are the Euler (or secant or "Zig") numbers, A000364. I.e., 4*Sum_{i=1..m} (-1)^(i+1)/(2*i-1) approaches Pi - 2*(-1)^m*Sum_{j>=0} (-1)^j*A000364(j)/(2*m)^(2*j+1); this latter alternating sum eventually diverges, but for any number of digits n > 1, we can take just its first term (i.e., the j=0 term), Pi - (-1)^m/m, set it equal to the lower or upper limit of the desired interval around Pi (whichever limit is farther from Pi), and round m up as necessary to get either a(n) or (possibly) a number that differs from a(n) by only 2 (see Example section). - _Jon E. Schoenfield_, Nov 11 2016
%H A126809 Jon E. Schoenfield, <a href="/A126809/b126809.txt">Table of n, a(n) for n = 1..100</a>
%F A126809 From _Jon E. Schoenfield_, Nov 11 2016: (Start)
%F A126809 The following has been verified to give the correct value of a(n) for all n from a(2)=19 through at least a(10000) = 11673...08624 (a 10000-digit number), and very likely for all n beyond 10000 as well (since the discarded terms 1/(4m^3), 5/(16m^5), etc. rapidly become small, so the simple approximation S(m) = Pi - (-1)^m/m becomes increasingly accurate):
%F A126809 Let x = Pi * 10^(n-1). If x - floor(x) < 1/2, then a(n) = 1 + 2*floor((1/(ceiling(x)/10^(n-1) - Pi) + 1)/2); otherwise, a(n) = 2*ceiling((1/2)/(Pi - floor(x)/10^(n-1))). (End)
%e A126809 E.g., a(2)=19 because if 4 is multiplied by the sum of the first 19 terms of the alternating series, then the result begins with 3.1 (the first two decimal digits of Pi) for the first time.
%e A126809 At n=3, we want the smallest m such that the partial sum S(m) = 4(1 - 1/3 + 1/5 - 1/7 + ... - (-1)^m/(2m-1)) is in the half-open interval [3.14,3.15). S(m) < Pi iff m is even, so for even m, setting Pi - (-1)^m/m = Pi - 1/m equal to 3.14 gives m=627.882..., and rounding up to the next even number gives 628. The other end of the interval, however, being farther from Pi, will be reached at a smaller value of m; for odd m, setting Pi - (-1)^m/m = Pi + 1/m equal to 3.15 gives m=118.943..., and rounding up to the next odd number gives 119. As it turns out, m=117 is the last odd number to fail (giving a sum of 3.150139...); m=119 succeeds (sum=3.149995...). No even number less than 628 yields a sum in the interval, so a(3)=119. - _Jon E. Schoenfield_, Nov 11 2016
%Y A126809 Cf. A000364.
%Y A126809 For a similar problem involving the convergence of a non-alternating series, see A274982. - _Jon E. Schoenfield_, Nov 11 2016
%K A126809 nonn,base
%O A126809 1,1
%A A126809 _G. L. Honaker, Jr._, Mar 14 2007
%E A126809 a(6)-a(8) from _Mike Keith_, Mar 18 2007
%E A126809 Edited by _Jon E. Schoenfield_, Nov 11 2016