A066185 Sum of the first moments of all partitions of n with weights starting at 0.

0, 0, 1, 4, 12, 26, 57, 103, 191, 320, 537, 843, 1342, 2015, 3048, 4457, 6509, 9250, 13170, 18316, 25483, 34853, 47556, 64017, 86063, 114285, 151462, 198871, 260426, 338275, 438437, 564131, 724202, 924108, 1176201, 1489237, 1881273, 2365079, 2966620, 3705799

Offset: 0

Author

Wouter Meeussen, Dec 15 2001

Keywords

• easy
• nonn

Comments

The first element of each partition is given weight 0.

Consider the partitions of n, e.g., n=5. For each partition sum T(e-1) and sum all these. E.g., 5 -> T(4)=10, 41 -> T(3)+T(0)=6, 32 -> T(2)+T(1)=4, 311 -> T(2)+T(0)+T(0)=3, 221 -> T(1)+T(1)+T(0)=2, 21111 ->1 and 11111 ->0. Summing, 10+6+4+3+2+1+0 = 26 as desired. - Jon Perry, Dec 12 2003

Also equals the sum of f(p) over the partitions p of n, where f(p) is obtained by replacing each part p_i of partition p by p_i*(p_i-1)/2. See I. G. Macdonald: Symmetric functions and Hall polynomials 2nd edition, p. 3, eqn (1.5) and (1.6). - Wouter Meeussen, Sep 25 2014

Examples

			a(3)=4 because the first moments of all partitions of 3 are {3}.{0},{2,1}.{0,1} and {1,1,1}.{0,1,2}, resulting in 0,1,3; summing to 4.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A000337, A001788, A066183, A066184, A066186, A161680, A264034.

Programs

Maple

b:= proc(n, i) option remember; `if`(n=0 or i=1, [1, 0],
      b(n, i-1)+(h-> h+[0, h[1]*i*(i-1)/2])(b(n-i, min(n-i, i))))
    end:
a:= n-> b(n$2)[2]:
seq(a(n), n=0..50);  # Alois P. Heinz, Jan 29 2014

Mathematica

Table[ Plus@@ Map[ #.Range[ 0, -1+Length[ # ] ]&, IntegerPartitions[ n ] ], {n, 40} ]
b[n_, i_] := b[n, i] = If[n==0, {1, 0}, If[i<1, {0, 0}, If[i>n, b[n, i-1], b[n, i-1] + Function[h, h+{0, h[[1]]*i*(i-1)/2}][b[n-i, i]]]]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Oct 26 2015, after Alois P. Heinz *)

Formula

a(n) = 1/2*(A066183(n) - A066186(n)). - Vladeta Jovovic, Mar 23 2003

G.f.: Sum_{k>=1} x^(2*k)/(1 - x^k)^3 / Product_{j>=1} (1 - x^j). - Ilya Gutkovskiy, Mar 05 2021

a(n) = Sum_{k=0..A161680(n)} k * A264034(n,k). - Alois P. Heinz, Jan 20 2023

a(n) ~ 3 * zeta(3) * sqrt(n) * exp(Pi*sqrt(2*n/3)) / (sqrt(2) * Pi^3). - Vaclav Kotesovec, Jul 06 2025