# [seqfan] Re: a partition sequence

Nathaniel Johnston nathaniel at nathanieljohnston.com
Mon Oct 3 00:12:24 CEST 2011

```>
> There are two 3s in this partition, and the list of frequencies of 3s
> in each column is 1 and 1.

Presumably he meant to write the second 3 underneath (not beside) the first,
which fixes the problem.

All the best,

- Nathaniel Johnston

On Sun, Oct 2, 2011 at 6:09 PM, William Keith <william.keith at gmail.com>wrote:

> On Sun, Oct 2, 2011 at 10:22 PM, David Newman <davidsnewman at gmail.com
> >wrote:
>
> > I'd like someone to check my numbers before I propose this sequence.
> >
> > Let a(n) be the number of planar partitions whose summands are
> > distinct going across rows and have distinct frequencies going down
> > columns.
> >
>
> You'd better be more precise.  I understand this as near as I can to mean
> that each row consists of distinct summands, and the list of frequencies of
> summands in each column (other than an infinite set of 0s) is a list of
> distinct numbers for each summand.  However:
>
> 3 3
> >
>
> There are two 3s in this partition, and the list of frequencies of 3s in
> each column is 1 and 1.
>
> So either there's something off with your calculations or I didn't