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A209536 Number of partitions of 0 having positive part-sum <= n. 2

%I #21 Jun 09 2021 05:24:30

%S 1,5,14,39,88,209,434,918,1818,3582,6718,12647,22848,41073,72049,

%T 125410,213619,361844,601944,995073,1622337,2626341,4201366,6681991,

%U 10515755,16449851,25509951,39333475,60172700,91577516,138390480,208096281,310976730,462512830

%N Number of partitions of 0 having positive part-sum <= n.

%C A partition of 0 is a set {i(1), i(2),..., i(n)} of nonzero integers with sum 0. Such a set uniquely partitions into two multisets {x(1),..., x(j)} and {y(1),..., y(k)} where x(1)+x(2)+...+x(j) =-[y(1)+y(2)+...+y(k)] and x(i) > 0 and y(i) < 0 for every i. The number x(1)+x(2)+...+x(j) is the positive part-sum.

%C Let p(h) be the number of partitions of h>=1, as in A000041. There are p(h)^2 ways to choose each of the sets {x(1),...,x(j)} and {y(1),...,y(k)} having sum h. Consequently, there are p(1)^2+p(2)^2+...+p(n)^2 partitions of 0 having positive part-sum <= n.

%H Alois P. Heinz, <a href="/A209536/b209536.txt">Table of n, a(n) for n = 1..5000</a>

%F From _Alois P. Heinz_, Oct 21 2018: (Start)

%F a(n) = Sum_{j=1..n} A000041(j)^2.

%F a(n) = -1 + A259399(n). (End)

%e 0 = 1-1 = 2-2 = 2-(1+1) = (1+1)-2 = (1+1)-(1+1),

%e so that a(2) = 5.

%p a:= proc(n) option remember; `if`(n=0, 0,

%p combinat[numbpart](n)^2+a(n-1))

%p end:

%p seq(a(n), n=1..40); # _Alois P. Heinz_, Oct 21 2018

%t p[n_] := IntegerPartitions[n]; l[n_] := Length[p[n]];

%t s[n_] := Sum[l[k]^2, {k, 1, n}];

%t Table[s[n], {n, 1, 40}] (* A209536 *)

%t (* Second program: *)

%t a[n_] := a[n] = If[n == 0, 0, PartitionsP[n]^2 + a[n-1]];

%t Array[a, 40] (* _Jean-François Alcover_, Jun 09 2021, after _Alois P. Heinz_ *)

%Y Cf. A209535, A000041, A259399.

%K nonn

%O 1,2

%A _Clark Kimberling_, Mar 10 2012

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