[seqfan] Re: Modular Partitions

Sun May 4 00:54:07 CEST 2014

PS I take it back. That g.f. was not right. Stay tuned!

On Sat, May 3, 2014 at 2:56 PM, Neil Sloane <njasloane at gmail.com> wrote:

> > Is this a proof that the number of "modular partitions" is equinumerous
> with the number of necklaces?
>
> I believe I have a proof of this using generating functions.
> That is, there is a g.f. for A(n,k) = number of ways of writing 0
> as a sum of k terms in Z/nZ which shows that this number is equal to
> U(n,k) := (1/n)* Sum_{ d | gcd(n,k)}  phi(d) binomial(n/d, k/d) (see
> A047996).
> The trick is to consider G(x,t) = Prod_{j=1..n} (1+t*x^j).
> Write G(x,t) = Sum_{i}  a_{i}(t) x^i.
> Then a_0(t) + a_n(t) + a_{2n}(t) + ... = Sum_k A(n,k)*t^k.
> I haven't written out all the details, but I think it works.
>
> By the way, the formula in A241926 wasn't quite right.
> The correct formula is
> T(n,k) := (1/(n+k))* Sum_{ d | gcd(n,k)}  phi(d) binomial((n+k)/d, k/d)
>  it follows at once that T(n,k) = T(k,n), so this shows that Jens Voss's
> original table is indeed symmetric.
>
> Neil
>
>
>
> On Sat, May 3, 2014 at 12:17 PM, <franktaw at netscape.net> wrote:
>
>> Is this a proof that the number of "modular partitions" is equinumerous
>> with the number of necklaces?
>>
>> Right now that is, as a conjecture, awaiting review for A047996.
>>
>>
>> Franklin T. Adams-Watters
>>
>> -----Original Message-----
>> From: Li-yao Xia <li-yao.xia at ens.fr>
>> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
>> Sent: Sat, May 3, 2014 11:05 am
>> Subject: [seqfan]  Re: Modular Partitions
>>
>>
>> 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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>
>
> --
> Dear Friends, I have now retired from AT&T. New coordinates:
>
> Neil J. A. Sloane, President, OEIS Foundation
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
>
>

-- 
Dear Friends, I have now retired from AT&T. New coordinates:

Neil J. A. Sloane, President, OEIS Foundation
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com