A209423 Difference between the number of odd parts and the number of even parts in all the partitions of n.
1, 1, 4, 4, 10, 13, 24, 30, 52, 68, 105, 137, 202, 264, 376, 485, 669, 864, 1162, 1486, 1968, 2501, 3256, 4110, 5285, 6630, 8434, 10511, 13241, 16417, 20505, 25273, 31344, 38438, 47346, 57782, 70746, 85947, 104663, 126594, 153386, 184793, 222865, 267452
Offset: 1
Keywords
Examples
The partitions of 5 are [5], [4,1], [3,2], [3,1,1], [2,2,1], [2,1,1,1], and [1,1,1,1,1], a total of 15 odd parts and 5 even parts, so that a(5)=10.
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Alois P. Heinz)
Programs
-
Maple
b:= proc(n, i) option remember; local m, f, g; m:= irem(i, 2); if n=0 then [1, 0, 0] elif i<1 then [0, 0, 0] else f:= b(n, i-1); g:= `if`(i>n, [0$3], b(n-i, i)); [f[1]+g[1], f[2]+g[2]+m*g[1], f[3]+g[3]+(1-m)*g[1]] fi end: a:= n-> b(n, n)[2] -b(n, n)[3]: seq(a(n), n=1..50); # Alois P. Heinz, Jul 09 2012 g := add(x^j/(1+x^j), j = 1 .. 80)/mul(1-x^j, j = 1 .. 80): gser := series(g, x = 0, 50): seq(coeff(gser, x, n), n = 0 .. 45); # Emeric Deutsch, Feb 08 2016
-
Mathematica
f[n_, i_] := Count[Flatten[IntegerPartitions[n]], i] o[n_] := Sum[f[n, i], {i, 1, n, 2}] e[n_] := Sum[f[n, i], {i, 2, n, 2}] Table[o[n], {n, 1, 45}] (* A066897 *) Table[e[n], {n, 1, 45}] (* A066898 *) %% - % (* A209423 *) b[n_, i_] := b[n, i] = Module[{m, f, g}, m = Mod[i, 2]; If[n==0, {1, 0, 0}, If[i<1, {0, 0, 0}, f = b[n, i-1]; g = If[i>n, {0, 0, 0}, b[n-i, i]]; {f[[1]] + g[[1]], f[[2]] + g[[2]] + m*g[[1]], f[[3]] + g[[3]] + (1-m)* g[[1]]}]]]; a[n_] := b[n, n][[2]] - b[n, n][[3]]; Table[a[n], {n, 1, 50}] (* Jean-François Alcover, Nov 16 2015, after Alois P. Heinz *)
Formula
G.f.: Sum_{j>0} x^j/(1+x^j)/Product_{k>0}(1 - x^k). - Emeric Deutsch, Feb 08 2016
a(n) = Sum_{i=1..n} (-1)^(i + 1)*A181187(n, i). - John M. Campbell, Mar 18 2018
a(n) ~ log(2) * exp(Pi*sqrt(2*n/3)) / (2^(3/2) * Pi * sqrt(n)). - Vaclav Kotesovec, May 25 2018
a(n) = Sum_{k=-floor(n/2)+(n mod 2)..n} k * A240009(n,k). - Alois P. Heinz, Oct 23 2018
a(n) = Sum_{k>0} k * A264398(n,k). - Alois P. Heinz, Aug 05 2020
Comments