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%I #10 Oct 23 2024 01:29:06
%S 487,4,1,1,46,1,0,1,11,1,2,1,0,2
%N a(n) = (prime(n)^(p-1) - 1)/p^2 mod p, where p is the first prime that divides (prime(n)^(p-1) - 1)/p.
%C a(n) = (prime(n)^(p-1) - 1)/p^2 mod p, where p = A174422(n) is the first Wieferich prime base prime(n).
%C a(n) = (prime(n)^(p-1) - 1)/p^2 mod p, where p is the first prime such that p^2 divides prime(n)^(p-1) - 1.
%C See references and additional comments, links, and cross-refs in A001220 and A039951.
%C a(15) > 2451011, a(16) = 1, a(17) = 4, a(18) = 1, a(19) = 5, a(20) = 2, a(21) = 0, a(22) = 6, a(23) = 1186, a(24) = 0, a(25) = 0, a(26) = 1, a(27) > 10^5, a(28) = 0, a(29) = 1, a(30) = 0, a(31) = 1, a(32) = 7, a(33) = 0, a(35) = 1, a(36) = 4, a(37) = 1, a(38) = 0, a(40) = 1, a(41) = 2, a(42) = 1, a(43) = 2, a(44) = 0, a(45) = 1, a(46) = 2, a(48) = 30, a(49) = 3, a(50) = 1. - _J.W.L. (Jan) Eerland_, Sep 27 2024
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Fermat_quotient#Generalized_Wieferich_primes">Generalized Wieferich primes</a>
%F a(n) = k mod 2, if prime(n) = 4k+1.
%F a(n) = A178814(prime(n)) .
%F a(1) = A178812(1).
%e Prime(2) = 3 and the first prime p that divides (3^(p-1) - 1)/p is 11, so a(2) = (3^10 - 1)/11^2 mod 11 = 488 mod 11 = 4.
%t Table[If[IntegerQ[s[[2]]],s,{s[[1]], "no solution in range 1 <= k <= 10^5"}], {s,Table[k = 1;Monitor[Parallelize[While[k <= 10^5,If[IntegerQ[((Prime[n]^(Prime[k] - 1) - 1)/Prime[k])/Prime[k]],Break[]]; k++];{n, Mod[(Prime[n]^(Prime[k] - 1) - 1)/Prime[k]^2, Prime[k]]}],k], {n, 1, 10}]}] (* _J.W.L. (Jan) Eerland_, Sep 27 2024 *)
%Y Cf. A001220, A039951, A174422, A178812, A178814.
%K hard,more,nonn
%O 1,1
%A _Jonathan Sondow_, Jun 17 2010