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.

3, 19, 119, 1688, 10794, 136121, 1530012, 18660270, 156001209, 1695509436, 11136696006, 102111268282, 1260654956982, 10725187563686, 147895359776637, 1313133218365935, 16250291773636035, 118166387818704586, 1860961545617561679, 15963377896404315146

Offset: 1

Author

G. L. Honaker, Jr., Mar 14 2007

Keywords

• nonn
• base

Comments

Calculations by Jud McCranie.

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

Examples

			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.
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
		

Links

• Jon E. Schoenfield, Table of n, a(n) for n = 1..100

Crossrefs

Cf. A000364.

For a similar problem involving the convergence of a non-alternating series, see A274982. - Jon E. Schoenfield, Nov 11 2016

Formula

From Jon E. Schoenfield, Nov 11 2016: (Start)

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):

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)

Extensions

a(6)-a(8) from Mike Keith, Mar 18 2007

Edited by Jon E. Schoenfield, Nov 11 2016