A128677 Least k>p such that (kp)^3 divides (p-1)^(kp)^2+1 for prime p = A000040(n).

19, 41, 29, 23, 79, 41617, 20939, 47, 40427, 4093, 4441, 2543, 1033, 659, 2612032921, 394502321, 14958421, 17957, 569, 14747, 12641, 167, 174263, 100493, 285629

Offset: 2

Author

Alexander Adamchuk, Mar 30 2007, Mar 31 2007, Apr 09 2007

Keywords

• hard
• more
• nonn

Comments

For every prime p>2, p^3 divides (p-1)^(p^2)+1 and furthermore p divides all numbers n>1 such that n^3 divides (p-1)^(n^2)+1.

a(27)>10^15. - Max Alekseyev, Nov 30 2017

Some further terms: a(28)-a(36) = {857, 3271, 7243979, 509, 263, 43019, 38921, 2683, 312055091}. a(38)-a(43) = {7499, 88588425539, 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(75) = {275884327, 719, 44041, 3122698559, 15161}. a(77)-a(80) = {907927, 202471, 5788837, 16361}.

a(n) <= A177996(n).

A000040(n) divides (a(n) - 1)/2. The quotients (a(n)-1)/2/A000040(n) are listed in A136374.

Examples

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

Crossrefs

Cf. A127263, A128678, A128679, A128680, A128681, A128682, A128683, A128684, A128685, A136374.

Programs

Mathematica

a[n_] := Module[{p, k}, p = Prime[n]; k = p + 1;
   While[! Divisible[(p - 1)^(k p)^2 + 1, (k p)^3], k++]; k];
Table[a[n], {n, 2, 15}] (* Robert Price, Mar 23 2020 *)

Formula

a(n) = smallest prime divisor of (p-1)^(p^2)+1 other than p, where p=A000040(n).

Extensions

a(16)-a(26), a(39), a(74) from Max Alekseyev, May 16 2010