A194795 Imbalance of the number of partitions of n.

0, -1, 0, -2, 0, -4, 0, -7, 1, -11, 3, -18, 6, -28, 13, -42, 24, -64, 41, -96, 69, -141, 112, -208, 175, -303, 271, -437, 410, -629, 609, -898, 896, -1271, 1302, -1792, 1868, -2510, 2660, -3493, 3752, -4839, 5248, -6666, 7293, -9131, 10065, -12454

Offset: 1

Author

Omar E. Pol, Feb 02 2012

Keywords

• sign

Comments

Consider the three-dimensional structure of the shell model of partitions version "tree". Note that only the parts > 1 produce the imbalance. The 1's are located in the central columns. Note that every column contains exactly the same parts, the same as a periodic table (see example). For more information see A135010.

Examples

			For n = 6 the illustration of the three views of the shell model with 6 shells shows an imbalance (see below):
------------------------------------------------------
Partitions                Tree             Table 1.0
of 6.                    A194805            A135010
------------------------------------------------------
6                   6                     6 . . . . .
3+3                   3                   3 . . 3 . .
4+2                     4                 4 . . . 2 .
2+2+2                     2               2 . 2 . 2 .
5+1                         1   5         5 . . . . 1
3+2+1                       1 3           3 . . 2 . 1
4+1+1                   4   1             4 . . . 1 1
2+2+1+1                   2 1             2 . 2 . 1 1
3+1+1+1                     1 3           3 . . 1 1 1
2+1+1+1+1                 2 1             2 . 1 1 1 1
1+1+1+1+1+1                 1             1 1 1 1 1 1
------------------------------------------------------
.
.                   6 3 4 2 1 3 5
.     Table 2.0     . . . . 1 . .     Table 2.1
.      A182982      . . . 2 1 . .      A182983
.                   . 3 . . 1 2 .
.                   . . 2 2 1 . .
.                   . . . . 1
------------------------------------------------------
The number of partitions with parts on the left hand side is equal to 7 and the number of partitions with parts on the right hand side is equal to 3, so a(6) = -7+3 = -4. On the other hand; for n = 6 the first n terms of A002865 (with positive indices) are 0, 1, 1, 2, 2, 4 therefore a(6) = 0-1+1-2+2-4 = -4.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000041, A002865, A135010, A182703, A187219, A194796, A194797, A194805, A194809.

Programs

Maple

with(combinat):
a:= proc(n) option remember;
      (-1)^n *(numbpart(n-1)-numbpart(n)) +`if`(n>1, a(n-1), 0)
    end:
seq(a(n), n=1..70); # Alois P. Heinz, Apr 09 2012

Mathematica

a[n_] := a[n] = (-1)^n*(PartitionsP[n-1]-PartitionsP[n]) + If[n>1, a[n-1], 0]; Table[a[n], {n, 1, 70}] (* Jean-François Alcover, Nov 11 2015, after Alois P. Heinz *)
nmax = 60; Rest[CoefficientList[Series[x/(1-x) - (1+x)/(1-x) * Product[1/((1 + x^(2*k-1))*(1 - x^(2*k))), {k, 1, nmax}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Nov 11 2015 *)
nmax = 60; Rest[CoefficientList[Series[-x/(1+x) - (1-x)/(1+x) * Product[1/(1-x^k), {k, 1, nmax}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Nov 11 2015 *)

Formula

a(n) = Sum_{k=1..n} (-1)^(k-1)*(p(k)-p(k-1)), where p(k) is the number of partitions of k.

a(n) = Sum_{k=1..n} (-1)^(k-1)*A002865(k).

a(n) = (-1)^(n+1) * (A240690(n+1) - A240690(n)) - 1. - Vaclav Kotesovec, Nov 11 2015

a(n) ~ (-1)^(n+1) * Pi * exp(Pi*sqrt(2*n/3)) / (24*sqrt(2)*n^(3/2)). - Vaclav Kotesovec, Nov 11 2015