%I #31 Nov 11 2015 12:03:40

%S 0,-1,0,-2,0,-4,0,-7,1,-11,3,-18,6,-28,13,-42,24,-64,41,-96,69,-141,

%T 112,-208,175,-303,271,-437,410,-629,609,-898,896,-1271,1302,-1792,

%U 1868,-2510,2660,-3493,3752,-4839,5248,-6666,7293,-9131,10065,-12454

%N Imbalance of the number of partitions of n.

%C 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.

%H Alois P. Heinz, <a href="/A194795/b194795.txt">Table of n, a(n) for n = 1..1000</a>

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

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

%F a(n) = (-1)^(n+1) * (A240690(n+1) - A240690(n)) - 1. - _Vaclav Kotesovec_, Nov 11 2015

%F a(n) ~ (-1)^(n+1) * Pi * exp(Pi*sqrt(2*n/3)) / (24*sqrt(2)*n^(3/2)). - _Vaclav Kotesovec_, Nov 11 2015

%e For n = 6 the illustration of the three views of the shell model with 6 shells shows an imbalance (see below):

%e ------------------------------------------------------

%e Partitions Tree Table 1.0

%e of 6. A194805 A135010

%e ------------------------------------------------------

%e 6 6 6 . . . . .

%e 3+3 3 3 . . 3 . .

%e 4+2 4 4 . . . 2 .

%e 2+2+2 2 2 . 2 . 2 .

%e 5+1 1 5 5 . . . . 1

%e 3+2+1 1 3 3 . . 2 . 1

%e 4+1+1 4 1 4 . . . 1 1

%e 2+2+1+1 2 1 2 . 2 . 1 1

%e 3+1+1+1 1 3 3 . . 1 1 1

%e 2+1+1+1+1 2 1 2 . 1 1 1 1

%e 1+1+1+1+1+1 1 1 1 1 1 1 1

%e ------------------------------------------------------

%e .

%e . 6 3 4 2 1 3 5

%e . Table 2.0 . . . . 1 . . Table 2.1

%e . A182982 . . . 2 1 . . A182983

%e . . 3 . . 1 2 .

%e . . . 2 2 1 . .

%e . . . . . 1

%e ------------------------------------------------------

%e 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.

%p with(combinat):

%p a:= proc(n) option remember;

%p (-1)^n *(numbpart(n-1)-numbpart(n)) +`if`(n>1, a(n-1), 0)

%p end:

%p seq(a(n), n=1..70); # _Alois P. Heinz_, Apr 09 2012

%t 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_ *)

%t 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 *)

%t 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 *)

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

%K sign

%O 1,4

%A _Omar E. Pol_, Feb 02 2012