A284345 Number of partitions of n into squares dividing n.

1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 4, 1, 1, 1, 6, 1, 3, 1, 6, 1, 1, 1, 7, 2, 1, 4, 8, 1, 1, 1, 15, 1, 1, 1, 27, 1, 1, 1, 11, 1, 1, 1, 12, 6, 1, 1, 28, 2, 3, 1, 14, 1, 7, 1, 15, 1, 1, 1, 16, 1, 1, 8, 46, 1, 1, 1, 18, 1, 1, 1, 114, 1, 1, 4, 20, 1, 1, 1, 66, 11, 1, 1, 22, 1, 1, 1, 23, 1, 11, 1, 24, 1, 1, 1, 91, 1, 3, 12, 67

Offset: 0

Author

Ilya Gutkovskiy, Mar 25 2017

Keywords

• nonn

Examples

			a(8) = 3 because 8 has 4 divisors {1, 2, 4, 8} among which 2 are squares {1, 4} therefore we have [4, 4], [4, 1, 1, 1, 1] and [1, 1, 1, 1, 1, 1, 1, 1].
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..10000
• Index entries for related partition-counting sequences
• Index entries for sequences related to sums of squares

Crossrefs

Cf. A000290, A001156, A018818, A046951, A066882, A161148, A225244, A284289.

Programs

Maple

with(numtheory):
a:= proc(n) option remember; local b, l; l, b:=
      sort(select(issqr, [divisors(n)[]])),
      proc(m, i) option remember; `if`(m=0, 1, `if`(i<1, 0,
        b(m, i-1)+`if`(l[i]>m, 0, b(m-l[i], i))))
      end; b(n, nops(l))
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Mar 30 2017

Mathematica

Join[{1}, Table[d = Divisors[n]; Coefficient[Series[Product[1/(1 - Boole[Mod[DivisorSigma[0, d[[k]]], 2] == 1] x^d[[k]]), {k, Length[d]}], {x, 0, n}], x, n], {n, 1, 100}]]

Formula

a(n) = [x^n] Product_{d^2|n} 1/(1 - x^(d^2)).

a(n) = 1 if n is a squarefree.

a(n) = 2 if n is a square of prime.