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A181792 Positive integers such that, for each k=0,1,2, the count of its divisors congruent to k modulo 3 is congruent to k modulo 3.

%I A181792 #14 Mar 30 2012 17:27:19
%S A181792 28,52,76,84,124,148,156,172,175,228,244,252,268,292,316,325,372,388,
%T A181792 412,436,444,468,475,508,516,525,556,604,628,652,684,700,724,732,756,
%U A181792 772,775,796,804,844,847,876,892,916,925,948,964,975,1075,1084,1108,1116,1132,1164
%N A181792 Positive integers such that, for each k=0,1,2, the count of its divisors congruent to k modulo 3 is congruent to k modulo 3.
%C A181792 Inspired by the positive perfect squares (cf. A000290), the analogous sequence for modulo 2.  (This sequence contains an infinite number of squares, the first of which is 9604.)  No analogous sequence exists for any even modulus greater than 2. (For n>1, if the number of 2n's divisors congruent to k mod 2n is congruent to k mod 2n for each k coprime to 2n, then the number of divisors congruent to n mod 2n must be congruent to 0 mod 2n.) Is there an analogous sequence for any odd modulus > 3?
%C A181792 It appears that a(n) < A000290(n) for all n>=22, despite this sequence's having 3 modular requirements for its divisors rather than 2.
%C A181792 n belongs to the sequence if and only if 3n does.
%F A181792 If the prime factorization of n is Product_ p(i)^e(i), these are the positive integers n such that:
%F A181792 a) For primes congruent to 1 modulo 3, an odd number of e(i) are congruent to 1 modulo 3, and none is congruent to 2 modulo 3.
%F A181792 b) For primes congruent to 2 modulo 3, all e(i) are congruent to 0 modulo 2, and at least one is congruent to 2 modulo 6.
%e A181792 Of 28's six divisors, four of them (1, 4, 7, and 28) are congruent to 1 mod 3; two of them (2 and 14) are congruent to 2 mod 3; and none of them are congruent to 0 mod 3.  Note that 4, 2 and 0 are congruent to 1 mod 3, 2 mod 3 and 0 mod 3 respectively. 28 therefore belongs to the sequence.
%t A181792 Reap[Do[d = Divisors[n];
%t A181792    c0 = Length[Select[d, Mod[#, 3] == 0 &]];
%t A181792    c1 = Length[Select[d, Mod[#, 3] == 1 &]];
%t A181792    c2 = Length[Select[d, Mod[#, 3] == 2 &]];
%t A181792    If[Mod[c0, 3] == 0 && Mod[c1, 3] == 1 && Mod[c2, 3] == 2, Sow[n]], {n, 1164}]][[2, 1]]
%Y A181792 Cf. A181793.
%K A181792 nonn
%O A181792 1,1
%A A181792 _Matthew Vandermast_, Nov 13 2010
%E A181792 Changed "number" to "count" in name so as to hopefully clarify what is being counted, and that mod 3 is performed at two steps in the process.