2,1

p^3 divides (p-1)^(p^2)+1 for all prime p>2. Prime p>2 divides all numbers n>1 such that n^3 divides (p-1)^(n^2)+1. Conjecture: All terms are primes.

a(17)-a(18) = {394502321, 14958421}. a(24)-a(26) = {174263, 100493, 285629}. a(28)-a(36) = {857, 3271, 7243979, 509, 263, 43019, 38921, 2683, 312055091}. a(38) = 7499. a(40)-a(43) = {9689, 359, 1087, 383}. a(45)-a(61) = {931417, 40597, 2111, 2677, 14983, 261061, 1302937, 479, 17935703, 503, 4227137, 39398453, 2153, 1627, 1109, 28663, 1699}. a(63)-a(69) = {1229, 1867, 78877, 500861, 1987, 62683, 2777}. a(71)-a(73) = {275884327, 719, 44041}. a(75) = 15161. a(77)-a(80) = {907927, 202471, 5788837, 16361}. a(n) is currently not known for n = {16, 27, 37, 39, 44, 62, 70, 74, 76, ...}.

Conjecture: p = prime[n] divides (a(n) - 1)/2. The quotients ((a(n) - 1)/2)/prime[n] are listed in A136374(n) = {3,4,2,1,3,1224,551,1,697,66,60,31,12,7,...}. - Alexander Adamchuk (alex(AT)kolmogorov.com), Dec 27 2007

a(2) = A127263(3)/3 = 57/3 = 19.

Cf. A127263 = numbers n such that n^3 divide 2^(n^2)+1. Cf. A128678, A128679, A128680, A128681, A128682, A128683, A128684, A128685.

Cf. A136374 = ((A128677(n) - 1)/2)/Prime[n].

Sequence in context: A041720 A098850 A164078 this_sequence A065349 A085778 A090265

Adjacent sequences: A128674 A128675 A128676 this_sequence A128678 A128679 A128680

hard,more,nonn

Alexander Adamchuk (alex(AT)kolmogorov.com), Mar 30 2007, Mar 31 2007, Apr 09 2007