[seqfan] Re: A101428 compared to A101409, polyminoes and trees

[Max Alekseyev]

There is a trivial identity:

G28 = G09 + 1

or in terms of t,z,g:

(1-t*z*g^2)/(1-t*z*g-t*z*g^2) = t*z*g/(1-t*z*g-t*z*g^2) + 1.

So, these generating functions are different only in a constant term,
which is not taken into account in A101428 and A101409 anyway.
I suggest to remove one of these sequences as a duplicate of the other.

Regards,
Max


On Sun, Jan 25, 2009 at 12:09 PM, Richard Mathar
<mathar at strw.leidenuniv.nl> wrote:
>
> Just by comparing numbers it seems that A101428 and A101409 are the same.
> Also by evaluation of the two different generating functions (see the Maple
> code below), individually along the rows and columns, we get the same arrays.
>  However, I do not manage to demonstrate equality by converting the two
> generating functions in analytical terms (?) There ought be somewhere a trick
> (math is a science, let's call it a methodology then) to convert some property
> of the g-term (inverse trigonometric sine essentially) to establish this with
> rigor...
>
> http://research.att.com/~njas/sequences/?q=id:A101428|id:A101409
>
> A101428 :=proc(n,k)
>        g:=2*sin(arcsin(3*sqrt(3*z)/2)/3)/sqrt(3*z):
>        G28:= (1-t*z*g^2)/(1-t*z*g-t*z*g^2):
>        coeftayl(G28,z=0,n) ;
>        coeftayl(%,t=0,k) ;
> end:
>
> A101409 :=proc(n,k)
>        g:=2*sin(arcsin(3*sqrt(3*z)/2)/3)/sqrt(3*z):
>        G09:= t*z*g/(1-t*z*g-t*z*g^2):
>        coeftayl(G09,z=0,n) ;
>        coeftayl(%,t=0,k) ;
> end:
>
> for n from 1 to 10 do
>        for k from 1 to n do
>                print(n,k,A101428(n,k),A101409(n,k)) ;
>        od:
> od:
>
> Richard J. Mathar
>
>
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