answer to Wouter's question

[wouter meeussen]

Aha, *that* way 'round!

but we can't get a GF for [1,1,3,6,15,31..
from the (not-a-standard-GF) of these seq's
>%N A084610 Triangle, read by rows, where the n-th row lists the (2n+1)
>           coefficients of (1+x-x^2)^n.

BUT it turns out to be a 'fluke': more terms:
1,3,6,15,31,73,157,358,785,1762,3896,8702,19299,42995,95507
give no super' match.

Thanks all the same..

blushing (again),

W.

----- Original Message ----- 
From: "N. J. A. Sloane" <njas at research.att.com>
To: <seqfan at ext.jussieu.fr>; <wouter.meeussen at vandemoortele.com>; <wouter.meeussen at pandora.be>
Sent: Friday, December 30, 2005 6:52 PM
Subject: answer to Wouter's question



Superseeker is telling you that if G is
the g.f. for your sequence, then 1/G has coefficients
that match those two sequences.

And that is true, look:

> s1:=[1,1,3,6,15,31,73,157,358,784];

G is

1+1*x+3*x^2+6*x^3+15*x^4+31*x^5+73*x^6+157*x^7+358*x^8+784*x^9+O(x^10)

1/G is

1-1*x-2*x^2-1*x^3-2*x^4+1*x^5-1*x^6+3*x^7+5*x^9+O(x^10)

Neil