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A177783 Wolstenholme quotient of prime p=A000040(n), i.e., such integer m<p that harmonic number H(p-1) == m*p^2 (mod p^3). 3

%I #29 Feb 17 2016 09:53:50

%S 3,6,6,7,10,14,18,20,16,24,17,38,39,19,29,28,12,53,31,19,53,58,48,42,

%T 1,33,53,37,5,81,4,17,29,13,13,72,75,70,173,159,111,150,39,178,106,

%U 163,196,163,172,30,98,24,177,261,212,223,122,147,276,17,92,111,27,209,241

%N Wolstenholme quotient of prime p=A000040(n), i.e., such integer m<p that harmonic number H(p-1) == m*p^2 (mod p^3).

%C a(n) = 0 iff A000040(n) is a Wolstenholme prime (given by A088164).

%C For n>2 and p=A000040(n), H(p^2-p) == H(p^2-1) == a(n)*p (mod p^2).

%H David W. Boyd, <a href="http://www.emis.de/journals/EM/expmath/volumes/3/3.html">A p-adic study of the partial sums of the harmonic series</a>, Experimental Math., Vol. 3 (1994), No. 4, 287-302.

%H R. Mestrovic, <a href="http://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011)</a>, arXiv:1111.3057 [math.NT], 2011.

%H R. Mestrovic, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Mestrovic/mes4.html">On a Congruence Modulo n^3 Involving Two Consecutive Sums of Powers</a>, Journal of Integer Sequences, Vol. 17 (2014), 14.8.4.

%H Jianqiang Zhao, <a href="http://dx.doi.org/10.1016/j.jnt.2006.05.005">Bernoulli numbers, Wolstenholme's theorem, and p^5 variations of Lucas' theorem</a>, Journal of Number Theory, Volume 123, Issue 1, March 2007, Pages 18-26.

%F a(n) = H(p-1)/p^2 mod p = A001008(p-1)/A002805(p-1)/p^2 mod p = A034602(n)/2 mod p = (binomial(2*p-1,p)-1)/(2*p^3) mod p, where p = A000040(n).

%F a(n) = (-1/3)*B(p-3) mod p, with p=prime(n) and B(n) is the n-th Bernoulli number. - _Michel Marcus_, Feb 05 2016

%F a(n) = A087754(n)/4 mod A000040(n).

%o (PARI) { a(n) = my(p); p=prime(n); ((binomial(2*p-1,p)-1)/2/p^3)%p }

%Y Cf. A034602, A072984, A087754, A092101, A092103, A092193, A128673.

%K nonn

%O 3,1

%A _Max Alekseyev_, May 13 2010

%E Edited by _Max Alekseyev_, May 16 2010

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