[seqfan] Re: A sort of composition

[Frank Adams-Watters]

And those two, A103446 and A218482, are the solution. There is some 
question whether David's sequence should have a(0) = 0 or 1, so either 
one could be considered correct.

Take a partition of m, and use it to determine the row lengths. There 
are then C(n-1,m-1) ways to distribute values totaling n into those m 
boxes. Thus a(n) = sum(p(m)*C(n-1,m-1)), and t his is the binomial 
transform of the partition numbers, as specified in A103446.

For example, n=3, and m=2, there are 2 partitions of 2: [2] and [1,1]. 
[2] gives us a single row of length 2, and the C(3-1,2-1) = C(2,1) = 2 
ways to fill it are 2,1 and 1,2. Likewise, partition [1,1] gives two 
rows of length 1, and we fill it with

2
1

or

1
2

. So the term for these values of n and m is p(2) * C(2,1) = 2*2 = 4.

Franklin T. Adams-Watters

-----Original Message-----
From: Rob Pratt <Rob.Pratt at sas.com>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Sent: Mon, Aug 18, 2014 9:19 pm
Subject: [seqfan] Re: A sort of composition


The next two terms are 344 and 856, and that is enough to reduce the 
candidates
to the two essentially identical ones out of the four.

-----Original Message-----
 From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of David 
Newman
Sent: Monday, August 18, 2014 9:53 PM
To: Sequence Fanatics Discussion list
Subject: [seqfan] A sort of composition

Count the number of compositions of n arranged in an array for which 
each row
length is less than or equal to the length of the row above it.
(Similar to a plane partition,but with no restriction that the summands 
be
decreasing along rows and columns.)

For example: for 3 there are 8 such compositions
3

2,1

1,2

2
1

1
2

1,1,1

1,1
1

1
1
1

The first few values that I get are 1,3,8,21,54,137

This matches four sequences in the OEIS, but it doesn't match any of 
the
definitions, as far as I can see .  Can someone out there tell me if 
it's the
same as one of the matches?

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