[seqfan] Re: A class of partitions.

William Keith william.keith at gmail.com
Mon Apr 29 00:42:13 CEST 2013

On Sun, Apr 28, 2013 at 5:08 PM, L. Edson Jeffery <lejeffery2 at gmail.com>wrote:

> But the number of
> partitions of N for which no part is greater than k is equal to the number
> of partitions of N with precisely k parts, N = 1, 2, ..., 1 <= k <= N.

The sum over these k gives partitions of N with no part greater than N, yes.

> Then
> apparently the number of partitions of 2*n+1 for which no part is greater
> than n should be given by
> a(n) = sum_{k=1,...,n} A008284(2*n+1,k)
>      = coefficient of x^(2*n+1) in series expansion of
> 1/((1-x)*(1-x^2)*...*(1-x^n)),
> unless I made a mistake.

Both of those expressions are correct.

> It appears further that the series expansion of
> x^(N-1)/((1-x)*(1-x^2)*...*(1-x^k))
> generates column k of A008284.

> If true, then that generating function is
> not included in the formula section of A008284.

This is essentially Wolfdieter Lang's comment from 2000.  The power of x in
the numerator only changes the offset slightly.

William Keith

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