A258367 a(n) is the smallest A (in absolute value) such that for p = prime(n), 2^{(p-1)/2} == +-1 + A*p (mod p^2), i.e., such that p is a near-Wieferich prime.

1, 1, 1, 3, 5, 2, 8, 3, 14, 3, 18, 9, 9, 22, 18, 4, 18, 5, 1, 28, 30, 24, 3, 20, 46, 22, 47, 21, 15, 9, 57, 42, 15, 48, 28, 41, 48, 60, 85, 25, 74, 25, 52, 11, 32, 51, 17, 13, 34, 113, 13, 71, 2, 16, 64, 130, 81, 35, 37, 29, 39, 147, 68, 60, 71, 96, 92, 99, 12

Offset: 2

Author

Felix Fröhlich, May 28 2015

Keywords

• nonn

Comments

p is in A001220 iff a(n) = 0. This is the case iff A014664(n) = A243905(n), which happens for n = 183 and n = 490.

Is a(n) = 0 for any other n, and, if yes, are there infinitely many such n?

Links

• Felix Fröhlich, Table of n, a(n) for n = 2..10000
• R. Crandall, K. Dilcher and C. Pomerance, A search for Wieferich and Wilson primes, Mathematics of Computation, 66 (1997), 433-449.

Crossrefs

Cf. A001220, A195988, A241014, A244801, A246568, A250406, A250407, A306885, A338646, A343410.

Programs

PARI

a(n,p=prime(n))=abs(centerlift(Mod(2,p^2)^((p-1)/2))\/p)
apply(p->a(0,p), primes(100)[2..100]) \\ Charles R Greathouse IV, Jun 15 2015

Formula

a(n) = min(b(n) mod p, -b(n) mod p) where p = prime(n) and b(n) = Sum_{i=1..ceiling((p-1)/4)} (2i-1)^(p-2). - Daniel Chen, Sep 01 2022