Partition Function (Subscript[b, k])(n)

Author

Eric W. Weisstein
November 30, 2008

This notebook downloaded from http://mathworld.wolfram.com/notebooks/Combinatorics/PartitionFunctionb.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/PartitionFunctionb.html.

©2008 Wolfram Research, Inc. except for portions noted otherwise

Initialization

<<Combinatorica`

Code

PartitionsBList[k_,n_]:=Select[Partitions[n],!Or@@IntegerQ/@(#/k)&]
PartitionsB[k_,n_]:=Count[Partitions[n],_?(!Or@@IntegerQ/@(#/k)&)]

Examples

Partitions[5]

PartitionsBList[2,5]

Table[{PartitionsQ[n],PartitionsB[2,n]},{n,10}]

Table[PartitionsB[k,n],{k,2,5},{n,15}]//ColumnForm

Generating Function

Product[((1-(x^(k n)))/(1-(x^n))),{n,∞}]

SeriesCoefficient[((QPochhammer[(x^k)])/(QPochhammer[x])),{x,0,n}]

k==2

With[{k=2},SeriesCoefficient[((QPochhammer[(x^k),(x^k)])/(QPochhammer[x,x])),{x,0,n}]]

FullSimplify[((QPochhammer[(x^2)])/(QPochhammer[x]))]

FullSimplify[1/2 QPochhammer[-1,x]-QPochhammer[x^2]/QPochhammer[x]]

Table[With[{k=2},SeriesCoefficient[((QPochhammer[(x^k),(x^k)])/(QPochhammer[x,x])),{x,0,n}]],{n,0,10}]

Table[SeriesCoefficient[(1/2)QPochhammer[-1,x],{x,0,n}],{n,0,10}]

With[{k=2},Series[Product[((1-(x^(k n)))/(1-(x^n))),{n,100}],{x,0,10}]]

PartitionsB[2,#]&/@Range[0,10]

PartitionsQ[Range[0,10]]

k==3

With[{k=3},SeriesCoefficient[((QPochhammer[(x^k),(x^k)])/(QPochhammer[x,x])),{x,0,n}]]

With[{k=3},Series[Product[((1-(x^(k n)))/(1-(x^n))),{n,100}],{x,0,10}]]

PartitionsB[3,#]&/@Range[0,10]