I've got some mail, where the author considered
the case from the view of generation-functions,
which may be interesting for a needed proof of
the method.
Here some exchange:
Next mail:
Then I did not understand, how the generation-function was
related to the bivariate array:
Final mail:
the case from the view of generation-functions,
which may be interesting for a needed proof of
the method.
Here some exchange:
Code:
It seems that e.g.f. of your triangle is:
(exp(y*(exp(y*(exp(x)-1))-1))-1)/y^2
which gives
1,
1,1,1,
1,3,4,3,1,
1,7,13,19,13,6,1,
...Code:
And the e.g.f. connected to your coefficient a_3() is
(exp(y*(exp(y*(exp(y*(exp(x)-1))-1))-1))-1)/y^3
which gives triangle
1,
1,1,1,1,
1,3,4,6,4,3,1,
1,7,13,26,31,31,25,13,6,1,
...
And so on.related to the bivariate array:
Code:
ยด
> However, I did not understand how you arrived
> via the generation-function process at the
> actual K-matrices. Would you mind to explain
> this to me?
>
> GottfriedCode:
An example:
Substituting t = exp(y) in your U_t(x,2) we get
exp(y*(exp(y*(exp(x)-1))-1))-1 =
y^2*x+(1/2*y^2+1/2*y^3+1/2*y^4)*x^2+(1/6*y^2+1/2*y^3+2/3*y^4+1/2*y^5+1/6*y^6)*x^3+(1/24*y^2+7/24*y^3+13/24*y^4+19/24*y^5+13/24*y^6+1/4*y^7+1/24*y^8)*x^4+...
.
Gottfried Helms, Kassel

