1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 5, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 7, 1, 1, 1, 4, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 5, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 11, 1, 1, 1, 2, 1, 1, 1, 6, 1, 1, 2, 2, 1, 1, 1, 5, 5, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 7, 1, 2, 2, 4, 1, 1, 1, 3, 1, 1, 1
Offset: 1
a(1) = 1 since the trivial group {e} is the only group of order 1, and it is Abelian; alternatively, since the only factorization of 1 into prime powers is the empty product.
a(p) = 1 for any prime p, since the only factorization into prime powers is p = p^1, and (in view of Lagrange's theorem) there is only one group of prime order p; it is isomorphic to (Z/pZ,+) and thus Abelian.
From _Wolfdieter Lang_, Jul 22 2011: (Start)
a(8) = 3 because 8 = 2^3, hence a(8) = pa(3) = A000041(3) = 3 from the partitions (3), (2, 1) and (1, 1, 1), leading to the 3 factorizations of 8: 8, 4*2 and 2*2*2.
a(36) = 4 because 36 = 2^2*3^2, hence a(36) = pa(2)*pa(2) = 4 from the partitions (2) and (1, 1), leading to the 4 factorizations of 36: 2^2*3^2, 2^2*3^1*3^1, 2^1*2^1*3^2 and 2^1*2^1*3^1*3^1.
(End)
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