# [seqfan] Re: Superseeker question + two interesting patterns

Neil Sloane njasloane at gmail.com
Sun Jul 22 19:58:16 CEST 2012

```No, this is an example where Superseeker's reply is useless,
I'm afraid. (But there are many many examples where
Superseeker does discover something useful. I use
it all the time.)

Neil

On Mon, Jul 16, 2012 at 4:34 AM, Johan de Ruiter
<johan.de.ruiter at gmail.com>wrote:

> Dear Seqfans,
>
> I have a beginners question regarding the Superseeker.
>
> For a sequence starting with
> [2,16,128,1156,10952,107584,1083392,11115556,115702472], it returned
> [4096 - 2048 a(n) + 384 a(n)^2 - 32 a(n)^3 + a(n)^4 , lgdegf]. That
> polynomial happens to equal (a(n)-8)^4.
> Does this mean that it suggested the generating function f(x) =
> f'(x)/8, so f(x) = c * e^(8x) and hence the coefficients
> of the exponential generating function should equal c * powers of 8?
> This is not the case.
> I know the response email warned it might only be an approximation,
> but maybe I misunderstood something important?
>
> ---
>
> Some context, for those who might be interested: I was wondering how
> many configurations I could end up with when starting with a row of
> piles of (say) playing cards and creating a new row of piles of
> playing cards while only moving a card when it is currently on top of
> a pile and moving each card exactly once. In particular I want to fix
> the heights of the piles.
>
> Example 1: From 2 piles of height 2, to 4 piles of height 1, there are
> 24 possibilities. (all permutations)
> Example 2: From 4 piles of height 1, to 2 piles of height 2, there are
> 24 possibilities. (symmetrical with Example 1; this symmetry always
> applies)
> Example 3: From 2 piles of height 2, to 2 piles of height 2, there are
> 16 possibilities. (no directed cycles allowed)
>
> When I computed a table of values for configurations with an equal
> number of piles in the start and end position, all with the same
> height, I noticed 2 remarkable patterns.
>
> Pattern 1) For 2 piles of height h in both the start and end
> configuration, the number of possible end configurations seems to be
> either a square or twice a square, depending on h being odd or even. I
> don't know why yet. Here are the first 19 terms:
>
> 2 = 2  * 1^2
> 16 = 4^2
> 128 = 2  * 8^2
> 1156 = 34^2
> 10952 = 2  * 74^2
> 107584 = 328^2
> 1083392 = 2  * 736^2
> 11115556 = 3334^2
> 115702472 = 2  * 7606^2
> 1218289216 = 34904^2
> 12948910592 = 2  * 80464^2
> 138708574096 = 372436^2
> 1495661223968 = 2  * 864772^2
> 16218468710656 = 4027216^2
> 176727219273728 = 2  * 9400192^2
> 1933956651447076 = 43976774^2
> 21243204576601928 = 2  * 103061158^2
> 234121111199439424 = 483860632^2
> 2587943032046002688 = 2 *  1137528688^2
>
> Pattern 2) If all piles are of height 2, the number of possible
> configurations for n piles is (2n)! - n^2(2n-2)! =
> (3n-2)/(4n-2)*(2n)!.
> I don't have an elegant interpretation of this elegant formula. I do
> have a nice recursive definition, but I still need to formally prove
> this is equivalent to the closed formula.
>
> Two forms of the recursion are:
> a_k = sum for t=0..k-1 of a_t*k!(k-1)!*(2^(2k-2t-1)-1)/(t!)^2
> a_k = sum for t=0..k-1 of a_t(k!!(k-1)!!/(t!!)^2 - k!(k-1)!/(t!)^2)
>
> Best regards,
> Johan de Ruiter
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>

--
Dear Friends, I have now retired from AT&T. New coordinates:

Neil J. A. Sloane, President, OEIS Foundation
11 South Adelaide Avenue, Highland Park, NJ 08904, USA