[seqfan] Re: Modular Partitions
Li-yao Xia
li-yao.xia at ens.fr
Sat May 3 15:35:39 CEST 2014
There is a link to necklaces when p and n are coprime which generalizes
William Keith's result.
By giving up primality, we lose the fact that rotation-equivalence of
profiles partitions them into equal-sized classes (i.e., the object of
paragraph 3, "If we add 1 to each part,...").
However, having p and n coprime still allows the next paragraph to be
generalized:
> (...) But adding 1 to each of n parts adds n to the total mod p, so
the residue of the profiles mod p are all distinct, and hence exactly
one of them is 0 mod p. (...)
In other words, equivalence classes of profiles are in bijection with
modular partitions.
Independently of coprimality of p and n, these equivalence classes are
in fact (in bijection with) necklaces, as is explained next.
A profile describes the number n(t) of parts of size t in a partition,
thus it is given by a sequence (n(0) ... n(p-1)), where n(t) is
nonnegative and n(0) + ... + n(p-1) = n (a total of n parts).
E.g., when (p, n) = (5, 4), the partition [0 3 3 4] is associated to
(1 0 0 2 1). (one 0, no 1, no 2, two 3s, one 4)
If we consider such sequences (n(t)) up to cyclic-shifts---which is the
same as adding a constant to all parts of a partition---they are in fact
necklaces with p white beads and n black beads, each n(t) indicating the
number of black beads between two successive white beads.
When p and n are not coprime, that function from partitions to their
profiles up to rotation is not a bijection. Shifting by p/gcd(n,p)
transforms a valid p-modular partition into another valid one, possibly
the same.
E.g., (p, n) = (6, 3), gcd(6, 3) = 3, T(6, 3) = 10
"Profiles" on the left those in the same class are gathered, partitions
on the right with their sums:
(3 0 0 0 0 0) | [0 0 0] 0
(0 0 3 0 0 0) | [2 2 2] 6
(0 0 0 0 3 0) | [4 4 4] 12
(1 1 0 0 0 1) | [0 1 5] 6
(0 1 1 1 0 0) | [1 2 3] 6
(0 0 0 1 1 1) | [3 4 5] 12
(1 0 0 2 0 0) | [0 3 3] 6
(0 0 1 0 0 2) | [2 5 5] 12
(0 2 0 0 1 0) | [1 1 4] 6
(1 0 1 0 1 0) | [0 2 4] 6
Impossible profiles:
(1 2 0 0 0 0)
(2 1 0 0 0 0)
(1 0 2 0 0 0)
(2 0 1 0 0 0)
(1 1 0 1 0 0)
(1 0 1 1 0 0)
Li-yao Xia
-------- Original Message --------
Subject: [seqfan] Re: Modular Partitions
From: William Keith <william.keith at gmail.com>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Date: 05/03/2014 07:19 AM
Chris' conjectures are accurate.
Suppose we are in the case n !==0 mod p. Diagram one of Jen's partitions
in Z/pZ into n parts with a Ferrers diagram, with parts of size 0 noted.
It will fit in a p-1 by n box. Choose any profile, to create a Ferrers
diagram; there are (n+p-1)-choose-n possible profiles.
If we add 1 to each part, rotating any new parts of size p to the end as
parts of size 0, we can do so p times and get back our original profile.
If p is prime and n !==0 mod p, then none of this group of p modpartitions
can possibly be the same: Suppose you had k parts of size 0, t additions
ago, then now you have k parts of size t. If this is the same partition as
the one t additions ago, then you must have had k parts of size t, so now
you have k parts of size 2t, etc, until you had k parts of each size to
begin with, i.e. p divides n.
So we can group the p resulting profiles. But adding 1 to each of n parts
adds n to the total mod p, so the residue of the profiles mod p are all
distinct, and hence exactly one of them is 0 mod p. Thus there are (1/p)
(n-1+p choose n) partitions of 0 mod p with n parts when p is prime, p not
dividing n.
When n = kp, then exactly the one partition listed above will be separate
from the groups; take (1/p)[(n+p-1 choose n) - 1] + 1. QED.
I think this diagram technique, and some sort of rotation counting, should
also be able to prove symmetry. I don't know anything about Molien
series. (Yet.) If something from that subject proves symmetry I'd be
interested in seeing it. But I would also like to see a map defined in
terms of the modpartitions themselves that gives symmetry in n and k.
William Keith
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