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A244919 For odd prime p, largest k such that binomial(2p-1, p-1) is congruent to 1 modulo p^k. 3

%I #15 Aug 24 2014 10:30:13

%S 2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,

%T 3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,

%U 3,3,3,3

%N For odd prime p, largest k such that binomial(2p-1, p-1) is congruent to 1 modulo p^k.

%C Wolstenholme's theorem implies that k >= 3 for all p > 3. The prime p is a Wolstenholme prime if and only if k > 3. For the primes up to 10^9 this holds only for p = 16843 and p = 2124679, where in each case a(n) = 4 (i.e. a(1944) = 4 and a(157504) = 4).

%H R. J. McIntosh, <a href="http://matwbn.icm.edu.pl/ksiazki/aa/aa71/aa7144.pdf">On the converse of Wolstenholme's theorem</a>, Acta Arith., Volume 71, Issue 4 (1995), 381-389.

%H R. J. McIntosh and E. L. Roettger, <a href="http://dx.doi.org/10.1090/S0025-5718-07-01955-2">A search for Fibonacci-Wieferich and Wolstenholme primes</a>, Math. Comp., 76 (2007), 2087-2094.

%o (PARI) forprime(p=3, 10^3, k=1; while(Mod(binomial(2*p-1, p-1), p^k)==1, j=k; k++); if(Mod(binomial(2*p-1, p-1), p^k)!=1, print1(j, ", ")))

%Y Cf. A034602, A088164.

%K nonn

%O 2,1

%A _Felix Fröhlich_, Jul 08 2014

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