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

0,3

Comments

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

Links

Alois P. Heinz, Table of n, a(n) for n = 0..6643

Gus Wiseman, Sequences enumerating triangles of integer partitions

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

Example

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)).

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}]

Prog

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

Crossrefs

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

Sequence in context: A029614 A143615 A199337 * A249389 A016606 A341616

Adjacent sequences: A279786 A279787 A279788 * A279790 A279791 A279792

Keyword

nonn

Author

Gus Wiseman, Dec 18 2016

Status

approved