A239241 Number of partitions of n into distinct parts for which (number of odd parts) = (number of even parts).
1, 0, 0, 1, 0, 2, 0, 3, 0, 4, 1, 5, 2, 6, 5, 7, 8, 8, 14, 9, 20, 11, 30, 13, 40, 17, 55, 23, 70, 32, 91, 45, 112, 65, 140, 91, 169, 128, 206, 177, 245, 241, 295, 323, 350, 429, 419, 559, 499, 722, 600, 921, 721, 1162, 874, 1452, 1062, 1800, 1299, 2210
Offset: 0
Examples
a(9) = 4 counts these partitions: 81, 72, 63, 54.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
Programs
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Maple
b:= proc(n, i, t) option remember; `if`(n>i*(i+1)/2, 0, `if`(n=0, `if`(t=0, 1, 0 ), b(n, i-1, t)+`if`(i>n, 0, b(n-i, i-1, t+`if`(irem(i, 2)=1, 1, -1))))) end: a:= n-> b(n$2, 0): seq(a(n), n=0..60); # Alois P. Heinz, Mar 15 2014
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Mathematica
z = 55; p[n_] := p[n] = IntegerPartitions[n]; d[u_] := d[u] = DeleteDuplicates[u]; g[u_] := g[u] = Length[u]; Table[g[Select[Select[p[n], d[#] == # &], Count[#, ?OddQ] < Count[#, ?EvenQ] &]], {n, 0, z}] (* A239239 *) Table[g[Select[Select[p[n], d[#] == # &], Count[#, ?OddQ] <= Count[#, ?EvenQ] &]], {n, 0, z}] (* A239240 *) Table[g[Select[Select[p[n], d[#] == # &], Count[#, ?OddQ] == Count[#, ?EvenQ] &]], {n, 0, z}] (* A239241 *) Table[g[Select[Select[p[n], d[#] == # &], Count[#, ?OddQ] > Count[#, ?EvenQ] &]], {n, 0, z}] (* A239242 *) Table[g[Select[Select[p[n], d[#] == # &], Count[#, ?OddQ] >= Count[#, ?EvenQ] &]], {n, 0, z}] (* A239243 *) (* Peter J. C. Moses, Mar 10 2014 *) b[n_, i_, t_] := b[n, i, t] = If[n > i*(i+1)/2, 0, If[n==0, If[t==0, 1, 0], b[n, i-1, t] + If[i>n, 0, b[n-i, i-1, t + If[Mod[i, 2]==1, 1, -1]]]]]; a[n_] := b[n, n, 0]; Table[a[n], {n, 0, 60}] (* Jean-François Alcover, Dec 27 2015, after Alois P. Heinz *)
Formula
a(n) = [x^n y^0] Product_{i>=1} 1+x^i*y^(2*(i mod 2)-1). - Alois P. Heinz, Apr 03 2014
Comments