A285900 Sum of all parts of all partitions of all positive integers <= n into consecutive parts.

1, 3, 9, 13, 23, 35, 49, 57, 84, 104, 126, 150, 176, 204, 264, 280, 314, 368, 406, 446, 530, 574, 620, 668, 743, 795, 903, 959, 1017, 1137, 1199, 1231, 1363, 1431, 1571, 1679, 1753, 1829, 1985, 2065, 2147, 2315, 2401, 2489, 2759, 2851, 2945, 3041, 3188, 3338, 3542, 3646, 3752, 3968, 4188, 4300, 4528, 4644, 4762, 5002

Offset: 1

Author

Omar E. Pol, May 02 2017

Keywords

• nonn

Comments

a(n) is also the sum of all parts of all partitions of all positive integers <= n into an odd number of equal parts. - Omar E. Pol, Jun 05 2017

Examples

			For n = 15, there are four partitions of 15 into consecutive parts: [15], [8, 7], [6, 5, 4] and [5, 4, 3, 2, 1]. The sum of all parts of these four partitions is 60, and a(14) = 204, so a(15) = 204 + 60 = 264.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..5000

Crossrefs

Partial sums of A245579.

Cf. A001227, A060831, A204217, A237593, A285891, A285899.

Programs

Mathematica

a285900[n_] := Accumulate[Map[# DivisorSum[#, 1 &, OddQ] &, Range[n]]]
a285900[60] (* data *) (* Hartmut F. W. Hoft, Jun 06 2017 *)

PARI

a(n)=sum(i=1, n, i * sumdiv(i, d, d%2)); \\ Andrew Howroyd, Nov 06 2018

PARI

a(n)=sum(k=1, (n+1)\2, (2*k - 1)/2 * (n\(2*k - 1)) * (1 + n\(2*k - 1))); \\ Andrew Howroyd, Nov 06 2018

Formula

a(n) = Sum_{k=1..floor((n+1)/2)} (2*k-1)/2 * floor(n/(2*k-1)) * floor(1 + n/(2*k-1)). - Daniel Suteu, Nov 06 2018