%I #10 Apr 11 2018 12:05:11
%S 49,34,42,42,39,3,62,57,19,62,10,36,14,53,57,16,60,22,41,21,25,0,56,
%T 21,24,52,33,28,51,34,60,8,47,39,42,33,14,66,50,48,45,28,61,50,27,8,
%U 30,59,32,15,3,1,54,12,30,20,14,12,10,49,33,49,54,13,26,42,8,58,12,63,19,16,48,15,2,13,1,23,2,44,64,25,40,0,16,58,44,31,62,47,61,46,9,2,50,1,62,34,31
%N p-adic digits of the unique simple zero of the p-adic zeta-function zeta_{(p,l)} with (p,l) = (67,58).
%C The p-adic digits are used to compute A299467(n) = least even integer k such that numerator(B_k) == 0 (mod 67^n) (see 2nd formula below and the program in A299467).
%C The algorithm used in the Mathematica program below is from Kellner 2007, Prop. 5.3, p. 428.
%C The corresponding sequences for (p,l) = (37,32) and (p,l) = (59,44) are A299468 and A299469, respectively.
%H Bernd C. Kellner and Jonathan Sondow, <a href="/A299470/b299470.txt">Table of n, a(n) for n = 0..98</a>
%H Bernd C. Kellner, <a href="http://bernoulli.org/">The Bernoulli Number Page</a>
%H Bernd C. Kellner, <a href="http://dx.doi.org/10.1090/S0025-5718-06-01887-4">On irregular prime power divisors of the Bernoulli numbers</a>, Math. Comp. 76 (2007) 405-441.
%F 0 <= a(n) <= 66.
%F l + (p - 1)*Sum_{i=0..n-2} a(i)*p^i = A299467(n) with (p,l) = (67,58).
%e The zero is given by a(0) + a(1)*p + a(2)*p^2 + ... with p = 67.
%t n = 99; p = 67; l = 58;
%t ModR[x_, m_] := Mod[Mod[Numerator[x], m] PowerMod[Denominator[x], -1, m], m];
%t B[n_] := -(1 - p^(n - 1)) BernoulliB[n]/n;
%t T[r_, k_, x_] := Sum[(-1)^(j + k) Binomial[j, k] Binomial[x, j], {j, k, r}];
%t zt = Table[ModR[B[l + (p - 1) k]/p, p^n], {k, 0, n}];
%t Z[n_] := zt[[n + 1]]; d = Mod[Z[0] - Z[1], p]; t = 0; L = {};
%t For[r = 1, r <= n, r++, x = Mod[Sum[Z[k] T[r, k, t], {k, 0, r}], p^r];
%t s = ModR[x/(d*p^(r - 1)), p]; AppendTo[L, s]; t += s*p^(r - 1)];
%t Print[L]
%Y Cf. A251782, A299466, A299467, A299468, A299469.
%K nonn
%O 0,1
%A _Bernd C. Kellner_ and _Jonathan Sondow_, Apr 08 2018