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%I #27 Jun 04 2015 15:57:47
%S 1093,23,5,3,3457,72673,13,67,67
%N Smallest prime p such that p^2 divides n^(p-1)-(n-1)^(p-1).
%C According to the generalized Fermat's little theorem, if a prime p does not divide n or n-1, then p must divide n^(p-1)-(n-1)^(p-1), so we can ask that whether there exists a Wieferich-type prime to the "base" (n, n-1), that is, a prime p such that p^2 divides n^(p-1)-(n-1)^(p-1).
%C a(n) is currently unknown for n = {11, 36, 49, 52, 66, 70, 84, 89, 102, 112, 133, 142, 148, ...}
%C a(11) > 6.5*10^10.
%H Eric Chen, <a href="/A253234/a253234.txt">Table of n, a(n) for n = 2..1000 status (test limit: 6.5*10^10)</a>
%H Richard Fischer, <a href="http://www.fermatquotient.com/FermatQuotienten/PQuotient_Sort.txt">Primes p such that (n+1)^(p-1)-n^(p-1) == 0 (mod p^2)</a>
%F a(9k+5) = 3.
%t f[n_] := Block[{k = 1}, While[k < 10^8 && PowerMod[n, Prime[k] - 1, Prime[k]^2] != PowerMod[n - 1, Prime[k] - 1, Prime[k]^2], k++ ]; If[k == 10^8, 0, Prime[k]]]; Table[ f[n], {n, 2, 75}]
%o (PARI) for(n=2, 10, forprime(p=2, 1e8, if(Mod(n, p^2)^(p-1)==Mod(n-1, p^2)^(p-1), print1(p, ", "); next({2}))); print1("--, "))
%Y Cf. A039951, A174422.
%K nonn,more,hard
%O 2,1
%A _Eric Chen_, May 17 2015