cp's OEIS Frontend

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

%I A128677 #13 Mar 23 2020 19:03:31
%S A128677 19,41,29,23,79,41617,20939,47,40427,4093,4441,2543,1033,659,
%T A128677 2612032921,394502321,14958421,17957,569,14747,12641,167,174263,
%U A128677 100493,285629
%N A128677 Least k>p such that (kp)^3 divides (p-1)^(kp)^2+1 for prime p = A000040(n).
%C A128677 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.
%C A128677 a(27)>10^15. - _Max Alekseyev_, Nov 30 2017
%C A128677 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}.
%C A128677 a(n) <= A177996(n).
%C A128677 A000040(n) divides (a(n) - 1)/2. The quotients (a(n)-1)/2/A000040(n) are listed in A136374.
%F A128677 a(n) = smallest prime divisor of (p-1)^(p^2)+1 other than p, where p=A000040(n).
%e A128677 a(2) = A127263(3)/3 = 57/3 = 19.
%t A128677 a[n_] := Module[{p, k}, p = Prime[n]; k = p + 1;
%t A128677    While[! Divisible[(p - 1)^(k p)^2 + 1, (k p)^3], k++]; k];
%t A128677 Table[a[n], {n, 2, 15}] (* _Robert Price_, Mar 23 2020 *)
%Y A128677 Cf. A127263, A128678, A128679, A128680, A128681, A128682, A128683, A128684, A128685, A136374.
%K A128677 hard,more,nonn
%O A128677 2,1
%A A128677 _Alexander Adamchuk_, Mar 30 2007, Mar 31 2007, Apr 09 2007
%E A128677 a(16)-a(26), a(39), a(74) from _Max Alekseyev_, May 16 2010