[seqfan] Re: A construction problem
franktaw at netscape.net
franktaw at netscape.net
Sat Oct 29 08:41:31 CEST 2011
Your numbers are off. The correct values are:
1, 2, 4, 10, 26, 76, 232, ...
This is A000085, the number of involutions.
Replace your a with 1, b with 2, etc., and you will find that you are
counting what are known as Young tableaux.
(In your example,
a d
b c
is not legal, so a(4) = 10, not 11.)
Franklin T. Adams-Watters
-----Original Message-----
From: Christopher Hunt gribble <cgribble263 at btinternet.com>
Dear Seqfans,
Let a(n) be the number of plane partitions that can be derived from a
linear
(ordinary) partition comprising n distinct parts.
Let these parts be labelled a, b, c, d, ... with a > b > c > d ..., then
a(n) = 1, 2, 4, 11, 26, 74, 198 ., not in the OEIS.
The number of different 2D shapes is the number of linear partitions on
n.
Plane partitions are such that its parts are non-increasing along rows
and
down columns.
For example, the possible plane partitions that can be derived from "a
b c
d"
are listed below.
a b c d
a c d
b
a b d
c
a b c
d
a c
b d
a d
b c
a b
c d
a d
b
c
a c
b
d
a b
c
d
a
b
c
d
Does anyone know of an algorithm, program or package that can generate
these "orderings"?
Is this linked to any other area of combinatorics?
Thanks,
Chris Gribble
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