[seqfan] Re: A class of partitions.

[Wouter Meeussen]

William,

I overlooked the evident redundancy you pointed out. My somewhat denser 
coding:

Table[Coefficient[ Series[1/Product[(1 - x^k), {k, n}], {x, 0, 2 n + 1}], 
x^(2 n + 1)], {n, 64}]

{1, 3, 8, 18, 37, 71, 131, 230, 393, 653, 1060, 1686, 2637, 4057,
6158, 9228, 13671, 20040, 29098, 41869, 59755, 84626, 118991, 166187,
230647, 318199, 436534, 595694, 808795, 1092876, 1470028, 1968738,
2625726, 3488072, 4616049, 6086556, 7997494, 10473078, 13670680,
17789019, 23078578, 29854292, 38511347, 49544424, 63571724, 81363984,
103880472, 132312710, 168138130, 213185155, 269713036, 340508206,
429001955, 539412651, 676918338, 847864940, 1060018609, 1322869014,
1647995783, 2049508702, 2544577387, 3154066342, 3903297786, 4822963731}

Wouter.

-----Original Message----- 
From: William Keith
Sent: Sunday, April 28, 2013 6:00 PM
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: A class of partitions.

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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