[seqfan] Re: A095814

Wouter Meeussen wouter.meeussen at telenet.be
Fri Jun 7 21:14:23 CEST 2013


hi,

I was quite happy, back then, to be able to relate different counts upto 
symmetry for plane partitions.
In those days, all these sequences were called "symmetric plane partitions" 
without further detail:
see A048140, A048142, A048141, A005987.

Wouter.


-----Original Message----- 
From: Allan Wechsler
Sent: Friday, June 07, 2013 5:28 PM
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: A095814

I've tried various interpretations over the last hour and none of them
generate the numbers given. However, the numbers given *do *seem to satisfy
the formula, which is essentially ceil(A000041(n)/2).

What I think happened is that the author was aiming at A005987 and missed.
(By the way, I find the title there, "symmetric plane partitions", to be
confusing, but perhaps it is justified by the literature. Comments or
examples would help.)

If this interpretation is true, then the only reason for keeping A095814 is
that it's ceil(A000041(n)/2).



On Fri, Jun 7, 2013 at 3:34 AM, Tw Mike <mt.kongtong at gmail.com> wrote:

> Dear David,
> Maybe "nonisomorphic partitions" means "non-self-conjugate partition",so
> that a(10) = (42-2)/2 =20
> Yours Mike,
>
>
> 2013/6/7 David Newman <davidsnewman at gmail.com>
>
> > I'm still having problems with this sequence, but this time the problem
> is
> > not in the nature of a typographical error.
> >
> > We have four sources of knowledge about this sequence, the numbers in 
> > the
> > sequence itself, the comment, the formula, and the title.  The only ones
> > which seem to agree are the sequence and the formula.
> >
> > The title does not seem clear to me.  It is:  "Number of nonisomorphic
> > partitions of  n on the Ferres graph".  I take this to mean "the number
> of
> > unrestricted partitions of n up to conjugation," where conjugation is 
> > the
> > familiar operation of flipping the Ferrers graph so that rows become
> > columns and columns become rows"
> >
> > If this is the correct interpretation of the title, then a(n) should be
> the
> > number of self-conjugate partitions of n, plus one half the number of
> > partitions of n which are not self-conjugate.  To give an example:  The
> > number of unrestricted partitions of 10 is 42.  There are 2
> self-conjugate
> > partitions :52111 and 4321.  So a(10) should be (42-2)/2  +2=22.  But
> > A095814(10)=21.
> >
> > Let's try calculating a(10) using the comment.  The comment reads:
> > "partitions of n into at most ceil(n/2) parts and with at least one part
> > greater than or equal to n-floor(n/2)"  For n=10 this becomes 
> > "partitions
> > of 10 into at most 5 parts and with at least one part greater than or
> equal
> > to 5."
> >
> > Here is a list of such partitions:
> >
> > 1.     10
> > 2.      91
> > 3.      82
> > 4.      811
> > 5.      73
> > 6.      721
> > 7.      7111
> > 8.      64
> > 9.      631
> > 10.     622
> > 11.     6211
> > 12.     61111
> > 13.     55
> > 14.     541
> > 15.     532
> > 16.     5311
> > 17.     5221
> > 18.     52111
> >
> > This seems to give a(10)=18.
> >
> > _______________________________________________
> >
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>
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