A279789 Number of ways to choose a constant partition of each part of a constant partition of n.

1, 1, 3, 3, 8, 3, 17, 3, 30, 12, 41, 3, 130, 3, 137, 45, 359, 3, 656, 3, 1306, 141, 2057, 3, 5446, 36, 8201, 544, 18610, 3, 34969, 3, 72385, 2061, 131081, 165, 290362, 3, 524297, 8205, 1109206, 3, 2130073, 3, 4371490, 33594, 8388617, 3, 17445321, 132, 33556496

Offset: 0

Author

Gus Wiseman, Dec 18 2016

Keywords

• nonn

Comments

Also number of ways to choose a divisor d|n and then a sequence of n/d divisors of d.

Examples

			The a(6)=17 twice-constant partitions are:
((6)),
((3)(3)), ((33)),
((3)(111)), ((111)(3)),
((2)(2)(2)), ((222)),
((2)(2)(11)), ((2)(11)(2)), ((11)(2)(2)),
((2)(11)(11)), ((11)(2)(11)), ((11)(11)(2)),
((1)(1)(1)(1)(1)(1)), ((11)(11)(11)), ((111)(111)), ((111111)).
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..6643
• Gus Wiseman, Sequences enumerating triangles of integer partitions

Crossrefs

Cf: A000005 (tau), A000040, A018818, A063834, A260685, A279375.

Programs

Maple

with(numtheory):
a:= proc(n) option remember; `if`(n=0, 1,
      add(tau(n/d)^d, d=divisors(n)))
    end:
seq(a(n), n=0..70);  # Alois P. Heinz, Dec 20 2016

Mathematica

nn=20;Table[DivisorSum[n,Power[DivisorSigma[0,#],n/#]&],{n,nn}]

PARI

a(n)=if(n==0, 1, sumdiv(n, d, numdiv(n/d)^d)) \\ Andrew Howroyd, Aug 26 2018

Formula

a(n) = Sum_{d|n} tau(n/d)^d for n > 0. - Andrew Howroyd, Aug 26 2018

G.f.: 1 + Sum_{k>=1} tau(k)*x^k/(1 - tau(k)*x^k). - Ilya Gutkovskiy, May 23 2019

a(n) = 3 <=> n is prime <=> n in { A000040 }. - Alois P. Heinz, May 23 2019