[seqfan] Re: a sequence to check

[David Newman]

Thanks.  I see my mistake now.

On Tue, May 10, 2011 at 3:49 PM, <franktaw at netscape.net> wrote:

> Also working basically by hand, but cross-checking A000219 and A098859, I
> get a(6) = 32, not 31. I also get a(7) = 68, so the sequence starting at n=0
> is:
>
> 1,1,3,4,11,20,32,68
>
> For n=6, by partition: [6]->1, [3^2]->2, [4,1^2]->3, [2^3]->3, [3,1^3]->5,
> [2,1^4]->7, [1^6]->11.
>
> Franklin T. Adams-Watters
>
>
> -----Original Message-----
> From: David Newman <davidsnewman at gmail.com>
>
> I'd like someone to check (and extend) the following.
>
> What is the number of plane partitions each of whose summands is used a
> different number of times.
>
> As an example for n=3 there are 4 such partitions
>
> 3
>
> 111
>
> 11
> 1
>
> and
>
> 1
> 1
> 1.
>
> The plane partitions
>
> 21 and
>
> 2
> 1
>
> do not appear on the list because each of the summands "2" and "1" is used
> the same number of times.
>
>
> The numbers that I've gotten, for n=1 until n=6, by hand calculations are:
> 1,3,4,11,20, and 31
>
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