A274994 Primes p such that p^2 divides Sum_{k=1..(p-1)/2} (k^(p-2))*(k^(p-1)-1).

3, 1093, 3511, 9511, 13691

Offset: 1

Author

René Gy, Nov 11 2016

Keywords

• nonn
• more

Comments

Any prime p divides Sum_{k=1..(p-1)/2} (k^(p-2))*(k^(p-1)-1). But a restricted list of primes p are such that p^2 divides Sum_{k=1..(p-1)/2}(k^(p-2))*(k^(p-1)-1).

Also primes p such that (2^(p-1)-1)/p == 0 (mod p) or 2*((p-1)!+1)/p +(2^(p-1)-1)/p == 0 (mod p), because it can be shown that Sum_{k=1..(p-1)/2} (k^(p-2))*(k^(p-1)-1) == p*((2^(p-1)-1)/p)*(2*((p-1)!+1)/p +(2^(p-1)-1)/p) (mod p^2).

The Wieferich primes (A001220) belong to the sequence.

No more terms up to 2000000, because A280300 has no more terms up to 2000000, and A001220 has no other terms below 4.97*10^17 (see the comments in these sequences). - René Gy, Jan 01 2017

Links

• Amir Akbary and Sahar Siavashi, The Largest Known Wieferich Numbers, INTEGERS, 18(2018), A3. See Table 1 p. 5.

Crossrefs

Equals the union of A001220 and A280300.

Programs

Mathematica

p=3; While[p<20000, If[Mod[Sum[PowerMod[k,p-2,p^2]*(PowerMod[k,p-1,p^2]-1), {k,1,(p-1)/2}], p^2] == 0, Print [p]]; p=NextPrime[p]]

PARI

is(n)=if(!isprime(n), return(0)); my(m=n^2,e=n-2); sum(k=1,n\2, Mod(k,m)^e*(Mod(k,m)^(e+1)-1))==0 && n>2 \\ Charles R Greathouse IV, Nov 13 2016