[seqfan] Re: Do we have this?

[Neil Sloane]

I created A274634 for n!*A003436(n).

(But I didn't give any context)

Best regards
Neil

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


On Thu, Jul 7, 2016 at 6:04 PM, Rob Pratt <Rob.Pratt at sas.com> wrote:

> The problem description doesn’t specify that there are exactly two copies
> of each version of the test.  For example, isn’t alternating versions 1 and
> 2 a legal distribution of tests, without using versions 3, 4, or 5?
>
> -----Original Message-----
> From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of
> israel at math.ubc.ca
> Sent: Thursday, July 07, 2016 5:29 PM
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] Re: Do we have this?
>
> I pressed "send" too soon: a(n)/n! is A003436. Note Andrew Howroyd's
> comment "Number of perfect matchings in the complement of C_{2n} where
> C_{2n} is the cycle graph on 2n vertices".
>
> Cheers,
> Robert
>
> On Jul 7 2016, israel at math.ubc.ca wrote:
>
> >I'm pretty sure it should be 744, not 696, for 8 perople.  And I get
> >35160 for 10.
> >
> > Let a(n) be the number of ways to distribute tests for 2n people. The
> > 2n people originally form the vertices of a cycle graph. At the k'th
> > stage, we take two non-adjacent vertices, assign the two copies of
> > version k to those, and delete those vertices. After the first stage,
> > each stage will have some disjoint collection of path graphs. Start by
> > assigning the two copies of the first test version to two non-adjacent
> > people, and deleting those people. This breaks up the cycle graph into
> > two intervals. Continue assigning copies of each test version. At each
> > stage the remaining people form some collection of interval graphs: we
> > take, in all possible ways, two non-adjacent vertices to assign the
> > next version of the test, remove those vertices, and calculate
> recursively. My Maple code is as follows.
> >
> >L:= proc(G) option remember;
> ># G is a list of integers, representing lengths of path graphs # k =
> >sum(G)/2 exam versions still to distribute # wlog G is nondecreasing
> >local i,j,ki,kj,nG,tot,Gp; nG:= nops(G); tot:= 0; for i from 1 to nG do
> >  for j from i to nG do
> ># take an elt of list i and of list j.
> >     if i=j then # same list, need non-adjacent
> >       for ki from 1 to G[i]-2 do
> >         for kj from ki+2 to G[i] do
> >            Gp:= sort(subs(0=NULL,subsop(i=(ki-1,kj-ki-1,G[i]-kj),G)));
> >            tot:= tot + procname(Gp);
> >         od
> >       od
> >     else # different lists
> >       for ki from 1 to G[i] do
> >         for kj from 1 to G[j] do
> >            Gp:=
> >sort(subs(0=NULL,subsop(i=(ki-1,G[i]-ki),j=(kj-1,G[j]-kj),G)));
> >            tot:= tot + procname(Gp);
> >         od
> >       od
> >     fi
> >   od
> >od;
> >tot
> >end proc;
> >L([]) := 1;
> >
> >F:= proc(n)
> ># this gives the number for the cycle graph C_{2n}
> >n*add(L(sort([i,2*n-i-2])),i=1..2*n-3) end proc;
> >
> >seq(F(n),n=1..10);
> >
> >0, 2, 24, 744, 35160, 2394720, 222712560, 27154350720, 4205374225920,
> >806700010233600
> >
> >It seems not to be in OEIS yet.
> >
> >Cheers,
> >Robert
> >
> >
> >On Jul 7 2016, Frank Adams-Watters wrote:
> >
> >>
> >>
> >>
> >> https://brilliant.org/practice/linear-recurrence-relations-level-4-5-
> >> challenges/?problem=are-you-ready-for-brimo&utm_medium=social&utm_sou
> >> rce=facebook&utm_campaign=adv_fb_posts_07062016&utm_content=qq_ch_lin
> >> ear_recurrence_l45
> >>
> >> (You have to register with them to see their answer; but see below,
> >> I'm sure that their answer is not correct.)
> >>
> >> There is an obvious sequence for, the same problem for 2n people. If
> >> I haven't made a mistake, this should start with 0*1!, 1*2!, 4*3!,
> >> 29*4! = 0, 2, 24, 696, which is not in the database. (The n! term
> >> comes from permuting the tests without changing the pattern in which
> >> they are handed out.) Looking only at 2, 24, 696 gives A030438 as the
> >> only match; while this seems similar, I don't think it is actually
> >> the correct sequence.
> >>
> >> The search dividing out the n! factor gives too many matches to look
> >> through.
> >>
> >> The linked site claims the next term is 1048580, but this can't be
> >> correct; it is not a multiple of 5!.
> >>
> >> So first, is my 29 correct? Second, is this sequence in the database?
> >> If not, can someone compute a few terms?
> >>
> >>Franklin T. Adams-Watters
> >>
> >>
> >>--
> >>Seqfan Mailing list - http://list.seqfan.eu/
> >>
> >>
> >
> >--
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> >
> >
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