OFFSET
1,1
COMMENTS
Let s(k) be the sum of the first k terms of the Leibniz series, and let eps(k) = |Pi-floor(Pi*10^(k-1))/10^(k-1)|. Then a(n) is either the minimum k such that |Pi-s(k)| < Min(eps(n), 1/10^(n-1)-eps(n)), or one less than the minimum such k.
a(n) is the minimum k such that the k-th and k+1-th partial sums agree with Pi in the first n decimal digits.
It is easy to calculate a(n) for large n because the partial sums of the Leibniz series can be expressed in terms of the digamma function, which can be computed efficiently using an asymptotic series.
LINKS
Ben Whitmore, Table of n, a(n) for n = 1..500
EXAMPLE
a(1) = 7 because the sum of the first 6 terms is approximately 2.9706, while the sum of the first 7 terms is approximately 3.2837, and all partial sums of more than 7 terms also have 3 as the first digit.
a(20) = 26769019461318409710 because the sum of the first 26769019461318409709 terms is approximately 3.1415926535897932385000000000000000000007642, while the sum of the first 26769019461318409710 terms is approximately 3.14159265358979323842528, which agrees with Pi in the first 20 digits, as do all partial sums with more terms.
MATHEMATICA
a[n_]:=Module[{eps, num, check, pi},
Block[{$MaxExtraPrecision=Infinity},
pi[k_]:=If[EvenQ[k],
Pi+PolyGamma[0, N[1/4, n+30]+k/2]-PolyGamma[0, 3/4+k/2],
pi[k+1]+4/(2k+1)
];
eps=Min[#, 10^-(n-1)-#]&[Abs[Pi-Floor[Pi 10^(n-1)]/10^(n-1)]];
num=2Ceiling[k/2]/.FindRoot[PolyGamma[k/2+3/4]-PolyGamma[k/2+1/4]==eps,
{k, 10^(n-1)}, WorkingPrecision->2n+30];
check[a_]:=RealDigits[pi[a], 10, n][[1, -1]]==RealDigits[Pi, 10, n][[1, -1]];
num-3+Position[check/@Range[num-3, num], False][[-1, 1]]
]
];
a/@Range[20]
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Ben Whitmore, Mar 16 2021
STATUS
approved