[seqfan] Re: plane partitions into unrestricted integers of 2 colors

[M. F. Hasler]

On Fri, Sep 21, 2018 at 11:02 PM Rick Shepherd <rlshepherd2 at gmail.com>
wrote:

> Hi David,
>
> My manual counts are 2, 10, 34, 122, 378, enough to see that this is not
> in the OEIS.


I get the same counts, as

a(1) = 2, a(2) = 2+2*2^2 = 10, a(3) = 2 + 2*2^2 + 3*2^3 = 34,

a(4) = 2 + 4*2^2 + 3*2^3 + 5*2^4 = 122

a(5) = 2 + 4*2^2 + 7*2^3 + 5*2^4 + 7*2^5 = 378
It's the sum of powers 2^k, 1 <= k <= n, multiplied by the number of plane
partitions of n with k parts, which is  A091298 <https://oeis.org/A091298>,
i.e., a(n) = sum( k=1..n, 2^k*A091298(n,k)),
which yields a = (2, 10, 34, 122, 378, 1242, 3690, 11266, 32666, 94994,
267202, ...)

For any other number of colors, we have the same formula with powers of 2
replaced by powers of the number of colors,
i.e., a(n,c) = sum( k=1..n, A091298(n,k)*c^k),
One could imagine a table T(n,c) = a(n,c)
column 1 = A000219 (# planar partitions of n),
col.2 = the above seq.; col.3 = 3 colors, etc.:
[ 1   2  3     4     5  .... ] = A000027 <https://oeis.org/A000027>
[ 3  10  21    36    55  ... ] = A014105 <https://oeis.org/A014105>
[ 6  34  102   228   430 ... ] = A067389 <https://oeis.org/A067389>
[13 122  525   1540  3605 ... ]
[24 378  2334  8964  25980 ...]
[48 1242 11100 56292 203280 ...]

(Maybe column 0 (all zeros) should be included ; row 0 (= 0^k) might be
less useful.)

- Maximilian


On Thu, Sep 20, 2018, 11:35 PM David Newman <davidsnewman at gmail.com> wrote:
>
> > I've tried this by hand and for n=1,2,3 I got the values 2, 10, 31 and
> I'm not even sure of those.
>
>