# [seqfan] Re: 1,6,110,2562,66222,...

rhhardin at att.net rhhardin at att.net
Mon Feb 9 11:15:59 CET 2009

```More terms
0 1
1 6
2 110
3 2562
4 66222
5 1815506
6 51697802
7 1511679210
8 45076309166
9 1364497268946
10 41800229045610
11 1292986222651646
12 40317756506959050
13 1265712901796074842
14 39965073938276694002
15 1268208750951634765562
16 40419340092267053380782
17 1293151592990764737265490
18 41512921146114663782643914
19 1336696804525969269347753334
20 43158316470769422985036007722
21 1396894744060840361583526359534
22 45313952186387344032141424880310
23 1472935673743661698205554658491142
24 47967219502930046234923103653158602
25 1564763324432611139054569034910940506
26 51125575601254146187206660714592557842

--
rhhardin at mindspring.com
rhhardin at att.net (either)

-------------- Original message ----------------------
From: Edwin Clark <eclark at math.usf.edu>
>
>
> Perhaps some one can extend this sequence:
>
> 1, 6, 110, 2562, 66222, ...
>
> The sequence arises in this paper mentioned today on the NMBRTHRY list:
>
> J.-M. Couveignes, T. Ezome and R. Lercier. Elliptic periods and
> primality proving, (2008)
> http://www.math.univ-toulouse.fr/~couveig/publi/arxiv3.pdf
>
> See section 8.6. The enumeration problem is:
>
> Find the number of integer sequences of length d = 2n+1 such that
> the sum of the terms is 0 and the sum of the absolute values of the terms
> is d-1.
>
> As the authors state the sum of the positive terms = sum of
> absolute values of the negative terms = (d-1)/2.
> So the largest interger in a desirable sequence is (d-1)/2.
> I found the above terms for d = 1,3,5,7, 9  by brute force. Can someone do
> better?
>
> The numbers appear in the array T(n,k) at
> http://www.research.att.com/~njas/sequences/A103881
> It looks like T(2n,n) works (if we define T(0,0)=1) but I don't see how to
> prove it since I don't understand the definition of T(n,k).
>
>
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/

```