|
|
A066185
|
|
Sum of the first moments of all partitions of n.
|
|
5
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,4
|
|
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
|
|
LINKS
|
|
|
FORMULA
|
G.f.: Sum_{k>=1} x^(2*k)/(1 - x^k)^3 / Product_{j>=1} (1 - x^j). - Ilya Gutkovskiy, Mar 05 2021
|
|
EXAMPLE
|
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.
|
|
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]:
|
|
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 *)
|
|
CROSSREFS
|
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|