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

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, 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, 3, 3, 3, 3

Offset: 2

Author

Felix Fröhlich, Jul 08 2014

Keywords

• nonn

Comments

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).

Links

• R. J. McIntosh, On the converse of Wolstenholme's theorem, Acta Arith., Volume 71, Issue 4 (1995), 381-389.
• R. J. McIntosh and E. L. Roettger, A search for Fibonacci-Wieferich and Wolstenholme primes, Math. Comp., 76 (2007), 2087-2094.

Crossrefs

Cf. A034602, A088164.

Programs

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, ", ")))