[seqfan] Re: Counting hexagonal partitions of n
Allan Wechsler
acwacw at gmail.com
Thu Nov 8 20:34:33 CET 2018
I came up with a somewhat more reliable way of counting these, so I can now
present enough data to show that the proposed sequence is not in OEIS. In
parentheses, I list the actual partitions as digit strings.
a(1) = 1 (1)
a(2) = 2 (11, 2)
a(3) = 4 (111, 12, 21, 3)
a(4) = 4 (1111, 121, 22, 4)
a(5) = 6 (11111, 122, 221, 23, 32, 5)
a(6) = 7 (111111, 1221, 123, 222, 321, 33, 6)
a(7) = 7 (1111111, 1222, 2221, 232, 34, 43, 7)
a(8) = 9 (11111111, 12221, 1232, 2222, 2321, 233, 332, 44, 8)
a(9) = 12 (111111111, 12222, 12321, 1233, 22221, 234, 3321, 333, 432, 45,
54, 9)
On Thu, Nov 8, 2018 at 11:48 AM Allan Wechsler <acwacw at gmail.com> wrote:
> By "hexagonal partition" I mean an ordered partitions whose parts start at
> some value, then increase by one, up to some maximum, then stay constant at
> that maximum, for some number of steps, then decrease to an ending value.
>
> For n=9, for example, the hexagonal partitions are
>
> 1+1+1+1+1+1+1+1+1
> 1+2+2+2+2
> 1+2+3+2+1
> 1+2+3+3
> 2+2+2+2+1
> 2+3+4
> 3+3+2+1
> 3+3+3
> 4+5
> 5+4
> 9
>
> Unless I have missed some, there are 11 possibilities, so a(9) = 11.
>
> I have collected some values in which I have fairly little confidence, but
> I think that the number of such partitions of n is not in OEIS.
>
> There is an intimate connection to A116513. Each hexagon that A116513
> counts has at least one and as many as six "readings" as a hexagonal
> partition.
>
> Can someone provide some values with more confidence than I have been able
> to do?
>
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