%I #8 Apr 11 2018 12:05:02

%S 15,25,40,36,18,11,17,28,58,9,51,13,25,41,44,17,43,35,21,10,21,38,9,

%T 12,40,43,45,30,41,0,3,25,34,49,45,9,19,48,57,11,13,29,28,44,41,37,33,

%U 29,43,8,57,12,48,15,15,53,57,16,51,16,54,30,9,26,8,49,22,58,11,42,28,36,33,45,24,32,18,12,29,45,40,27,19,40,41,11,42,49,35,41,57,54,33,0,34,34,49,6,31

%N p-adic digits of the unique simple zero of the p-adic zeta-function zeta_{(p,l)} with (p,l) = (59,44).

%C The p-adic digits are used to compute A299466(n) = least even integer k such that numerator(B_k) == 0 (mod 59^n) (see 2nd formula below and the program in A299466).

%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) = (67,58) are A299468 and A299470, respectively.

%H Bernd C. Kellner and Jonathan Sondow, <a href="/A299469/b299469.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) <= 58.

%F l + (p - 1)*Sum_{i=0..n-2} a(i)*p^i = A299466(n) with (p,l) = (59,44).

%e The zero is given by a(0) + a(1)*p + a(2)*p^2 + ... with p = 59.

%t n = 99; p = 59; l = 44;

%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, A299470.

%K nonn

%O 0,1

%A _Bernd C. Kellner_ and _Jonathan Sondow_, Apr 08 2018