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.

28, 52, 76, 84, 124, 148, 156, 172, 175, 228, 244, 252, 268, 292, 316, 325, 372, 388, 412, 436, 444, 468, 475, 508, 516, 525, 556, 604, 628, 652, 684, 700, 724, 732, 756, 772, 775, 796, 804, 844, 847, 876, 892, 916, 925, 948, 964, 975, 1075, 1084, 1108, 1116, 1132, 1164

Offset: 1

Author

Matthew Vandermast, Nov 13 2010

Keywords

• nonn

Comments

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?

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.

n belongs to the sequence if and only if 3n does.

Examples

			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.
		

Crossrefs

Cf. A181793.

Programs

Mathematica

Reap[Do[d = Divisors[n];
   c0 = Length[Select[d, Mod[#, 3] == 0 &]];
   c1 = Length[Select[d, Mod[#, 3] == 1 &]];
   c2 = Length[Select[d, Mod[#, 3] == 2 &]];
   If[Mod[c0, 3] == 0 && Mod[c1, 3] == 1 && Mod[c2, 3] == 2, Sow[n]], {n, 1164}]][[2, 1]]

Formula

If the prime factorization of n is Product_ p(i)^e(i), these are the positive integers n such that:

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.

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.

Extensions

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.