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


%S 3,1093,3511,9511,13691

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

%C 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).

%C 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).

%C The Wieferich primes (A001220) belong to the sequence.

%C 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

%H Amir Akbary and Sahar Siavashi, <a href="http://math.colgate.edu/~integers/s3/s3.Abstract.html">The Largest Known Wieferich Numbers</a>, INTEGERS, 18(2018), A3. See Table 1 p. 5.

%t 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]]

%o (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

%Y Equals the union of A001220 and A280300.

%K nonn,more

%O 1,1

%A _René Gy_, Nov 11 2016

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Last modified January 27 13:38 EST 2022. Contains 350607 sequences. (Running on oeis4.)