[seqfan] Re: A class of partitions.

William Keith william.keith at gmail.com
Sun Apr 28 18:00:31 CEST 2013


Wouter, Edson:

     This number can be quickly calculated by a package as the coefficient
of q^(2n+1) in the initial segment of the Euler product for the partition
function, \prod_{k=1}^n 1/(1-q^k).  The restriction that there must be at
least 3 parts is redundant since to partition 2n+1 with parts up to size n,
you must have at least three parts (being greedy, you must use at least
n+n+1).

To any desired length:

RhomTable = Table[0, {i, 1, 40}];
For[n = 1, n < 41, n++,
 RhomTable[[n]] =
   Coefficient[
    Series[Product[1/(1 - q^k), {k, 1, n}], {q, 0, 2 n + 3}],
    q^(2 n + 1)];
 ]
RhomTable

     I strongly doubt that there is any closed formula.  To build a
somewhat complicated recursion, I would keep track of the largest part j
and call these numbers p_r(n,j).  Then a desired partition of 2n+1 is the
sum over j from 1 to n, and each p_r(n,j) arises by appending 1 <= k <=
floor((2n+1)/j) parts of size j to the total number of previous such
partitions of 2n+1-kj with largest parts of any size, plus the desired
partitions with exactly 3 parts, of which there are a small number.  You
will note that this also obliges us to keep track of a parallel series for
the even numbers, so there is much room if I think little chance for
improvement here.

     I would, of course, be cheerfully congratulatory if proved wrong.

Cordially,
William Keith


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