# [seqfan] Re: decomposing partitions, is this new?

Meeussen Wouter (bkarnd) wouter.meeussen at vandemoortele.com
Fri Apr 19 14:32:41 CEST 2013

```Sequel:  Odd Parts: Excess Even Positions (over odd) in partitions of n=0..29
table is easy to extend: (zero's not shown)

-4	-3	-2	-1	0	1	2	3 = excess
1
1
2
2		1
3
3		2
1		4
4		3
2		5
5		4
3		6		1
6		5
4		7		2
7		6
5		8		3
1		8		7
6		9		4
2		9		8
7		10		5
3		10		9
8		11		6
4		11		10		1
9		12		7
5		12		11		2
10		13		8
6		13		12		3
11		14		9
7		14		13		4
1		12		15		10
8		15		14		5

etc.
number of entries>=0 in row n = 2*Floor[1 + (-1 + Sqrt[1 + 8*n])/4]
row n has first non-zero entry in column -Floor[1/2 + (-1 + Sqrt[1 + 8*n])/4]
Replace each integer z in the table by A000712(z) and you get the excess in
even positions over odd positions of the odd parts in (all) the partitions of n.
Row sums then sum to p(n).

Example:
n=6; p(6)=11;
row including zero's: {1,0,4,0} at columns {-2,-1,0,1}
replace z by A000712(z) :  {1,0,10,0}
so there are 10 partitions of 6 whose odd parts have zero excess (even over odd position)
and 1 with an excess of -2 (= 2 more odd parts in odd position than in even position
since {3,2,1} has its odd parts on positions 1 and 3 and none on even positions.

This can be coded by left-aligning row n (by right-shifting Floor[1 + (-1 + Sqrt[1 + 8*n])/4] places) as:

{1}
{1,0}
{0,2}
{0,2,0,1}
{0,0,3,0}
{0,3,0,2}
{1,0,4,0}
{0,4,0,3}
{2,0,5,0}
{0,5,0,4}
{0,3,0,6,0,1}
{0,0,6,0,5,0}
{0,4,0,7,0,2}
{0,0,7,0,6,0}
{0,5,0,8,0,3}
{1,0,8,0,7,0}
{0,6,0,9,0,4}
{2,0,9,0,8,0}
{0,7,0,10,0,5}
{3,0,10,0,9,0}
{0,8,0,11,0,6}
{0,4,0,11,0,10,0,1}
{0,0,9,0,12,0,7,0}

unless someone stops me, I'll submit it this weekend;

fumblingly yours,

Wouter.

-----Original Message-----
From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Wouter Meeussen
Sent: zondag 31 maart 2013 20:35
To: seqfan
Subject: [seqfan] decomposing partitions, is this new?

look at A000712(k) =partitions of k into parts of 2 kinds :

EXAMPLE:  Assume there are integers of two kinds k and k' then a(3) = 10 since 3 has the following partitions into parts of two kinds: 111, 111', 11'1', 1'1'1', 12, 1'2, 12', 1'2', 3, and 3'. [W. Edwin Clark, Jun 24 2011]

A000712 has a simple GF:  1/prod(m>=1, 1-x^m )^2

I found that the # of partitions of n can be decomposed into terms of
A000712(k)  with values of k <= Floor[n/2]+1  as:

n=1; k={1},
n=2; k={2},
n=3; k={2, 1},
n=4; k={3},
n=5; k={3, 2},
...
n=33;k={17, 16, 10, 7},
n=34;k={18, 15, 13, 4},
n=35;k={18, 17, 11, 8},
n=36;k={19, 16, 14, 5, 1}

producing
p(1)=1,
p(2)=2,
p(3)=2 +1 ,
p(3)=5,
p(3)=5+ 2,
...
p(33)=5822+ 3956+ 300+ 65,
p(34)=8470+ 2665+ 1165+ 10,
p(35)=8470+ 5822+ 481+ 110,
p(36)=12230+ 3956+ 1770+ 20+ 1

The count of terms is surprisingly low: (for n=1..36) {1, 1, 2, 1, 2, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 3, 3, 3, 3, 4, 3, 4, 3, 4, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5}

Question:  is this somewhere implied in the descriptions of A000712 ?
(I don't see it)

Wouter.

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