[seqfan] Re: guess the relation

[Aai]

 From the divided by 2 row, take the differences giving:

4: 1 1
5: 1 1 1 2
6: 1 1 1 1 1 2 4
7: 1 1 1 1 1 2 1 1 2 4 8

The next difference row then will be:

8: 1 1 1 1 1 2 1 1 2 4 1 1 2 4 8 16

Which then gives us:

1 2 3 4 5 6 8 9 10 12 16 17 18 20 24 32 48

But of course I'm not sure this is the right continuation.



On 19-01-14 03:50, D. S. McNeil wrote:
> While trying to find a formula to count certain binary matrices for a
> problem that came up on SO, I found plausible empirical generating
> functions for the rectangular generalization.  [See
> http://stackoverflow.com/questions/21149014/finding-a-better-way-to-count-matricesif
> you're interested.]
>
> One of the reasons they're so believable is how nicely they factor:
>
> 4  (x - 6) * (x - 4) * (x - 2)
> 5  (x - 12) * (x - 8) * (x - 6) * (x - 4) * (x - 2)
> 6  (x - 24) * (x - 16) * (x - 12) * (x - 10) * (x - 8) * (x - 6) * (x - 4)
> * (x - 2)
> 7  (x - 48) * (x - 32) * (x - 24) * (x - 20) * (x - 18) * (x - 16) * (x -
> 12) * (x - 10) * (x - 8) * (x - 6) * (x - 4) * (x - 2)
>
> Should be able to get 8 as well, but beyond that seems unlikely.
>
> Dividing out the factors of 2, we get
>
> 4  [1, 2, 3]
> 5  [1, 2, 3, 4, 6]
> 6  [1, 2, 3, 4, 5, 6, 8, 12]
> 7  [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 16, 24]
>
> as the numbers to be reproduced, and I feel like there's some
> divisibility-related pattern there I'm missing.  Any numerologists have a
> guess?
>
>
> Doug
>
> PS: If anyone can just look at the problem and write down a combinatorial
> formula at once, I'm sure the OP would appreciate that too. :^)
>
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@@i = Arie Groeneveld