%I #17 Aug 24 2024 06:02:23

%S 3,4,8,10,13,14,17,18,19,24,25,28,32,39,42,43,47,48,50,52,54,55,62,67,

%T 69,73,74,75,76,78,83,84,87,88,89,90,95,99,101,103,105,108,109,112,

%U 113,118,119,123,125,127,130,132,134,138,140,143,144,147,149,153,154,157

%N Numbers k such that A114912(k) = 1. Numbers k such that A000009(k) == 2 (mod 4).

%C All the terms are the sum of a generalized pentagonal number A001318 and a square A000290.

%C Let 24*k+1 = p_1^e_1 * ... * p_r^e_r * q_1^f_1 * ... * q_s^f_s, where the p_i's are distinct primes == 1, 5, 7, or 11 (mod 24) and the q_i's are distinct primes == 13, 17, 19, or 23 (mod 24). Then k belongs to the sequence iff all of the f_i's are even and all but one of the e_i's are even and the one e_i which is odd is == 1 (mod 4). - _Dean Hickerson_, Jan 19 2006

%H Amiram Eldar, <a href="/A114913/b114913.txt">Table of n, a(n) for n = 1..10000</a>

%H Krishnaswami Alladi, <a href="https://doi.org/10.1090/S0002-9947-97-01831-X">Partition Identities Involving Gaps and Weights</a>, Transactions of the American Mathematical Society, Vol. 349, No. 12 (Dec 1997), pp. 5001-5019.

%t q[n_] := Module[{f = FactorInteger[n], f1, f2}, f1 = Select[f, MemberQ[{1, 5, 7, 11}, Mod[First[#], 24]] &]; f2 = Select[f, MemberQ[{13, 17, 19, 23}, Mod[First[#], 24]] &]; AllTrue[f2[[;;, 2]], EvenQ] && Count[f1[[;;, 2]], _?OddQ] == 1]; Select[Range[160], q[24 * # + 1] &] (* _Amiram Eldar_, Aug 24 2024 *)

%Y Cf. A000009, A000290, A001318, A114912, A114914.

%Y A111174 is a subsequence.

%Y See comments in A113780 for explanation.

%K nonn

%O 1,1

%A _Christian G. Bower_, Jan 06 2006