A279784 Twice partitioned numbers where the latter partitions are constant.

1, 1, 3, 5, 12, 18, 40, 60, 121, 186, 344, 524, 955, 1432, 2484, 3756, 6352, 9493, 15750, 23414, 38128, 56513, 90406, 133312, 211194, 309657, 484214, 708267, 1097159, 1597290, 2454245, 3560444, 5430091, 7854174, 11894335, 17151394, 25838413, 37145198, 55648059

Offset

0,3

Comments

Also number of ways to choose a divisor of each part of an integer partition of n.

Links

Alois P. Heinz, Table of n, a(n) for n = 0..5000

Gus Wiseman, Sequences enumerating triangles of integer partitions

Formula

G.f.: exp(Sum_{k>=1} Sum_{j>=1} d(j)^k*x^(j*k)/k), where d(j) is the number of the divisors of j (A000005). - Ilya Gutkovskiy, Jul 17 2018

From Vaclav Kotesovec, Jul 28 2018: (Start)

a(n) ~ c * 2^(n/2), where

c = 203.986136154799274492709451797084688042886818134781591... if n is even and

c = 201.491703180375661735217350021245093454724452720559762... if n is odd.

In closed form, a(n) ~ ((2 + sqrt(2)) * Product_{k>=3} (1/(1 - tau(k) / 2^(k/2))) + (-1)^n * (2 - sqrt(2)) * Product_{k>=3} (1/(1 - (-1)^k * tau(k) / 2^(k/2)))) * 2^(n/2 - 1), where tau() is A000005. (End)

Example

The a(4)=12 twice-partitions are:
((4)), ((3)(1)), ((2)(2)), ((22)),
((2)(1)(1)), ((2)(11)), ((11)(2)),
((1)(1)(1)(1)), ((11)(1)(1)), ((11)(11)), ((111)(1)), ((1111)).

Maple

b:= proc(n, i) option remember; `if`(n=0 or i=1, 1,
      b(n, i-1)+`if`(i>n, 0, numtheory[tau](i)*b(n-i, i)))
    end:
a:= n-> b(n$2):
seq(a(n), n=0..50);  # Alois P. Heinz, Dec 20 2016

Mathematica

nn=20; CoefficientList[Series[Product[1/(1-DivisorSigma[0, n]x^n), {n, nn}], {x, 0, nn}], x]

Crossrefs

Cf. A000005, A063834, A145515, A270995.

Sequence in context: A295360 A197988 A025088 * A198301 A323866 A082740

Adjacent sequences: A279781 A279782 A279783 * A279785 A279786 A279787

Keyword

nonn

Author

Gus Wiseman, Dec 18 2016

Status

approved