I'd like some help - First Steps

[Paul D. Hanna]

Emeric, and Seqfans, 
     Here is a start to finding an answer to this very nice (and
insightful) question. 
Let's take some baby steps in finding the coefficients of z in 
G(1,1,x,z), where the coefficients are polynomials in x. 
I'll use the following PARI code. 
 
/* PARI code: */
{Q(n) =
  if(n==0,1,
    if(n%2==1,
      t*deriv(Q(n-1),t)+x*deriv(Q(n-1),s)+x*deriv(Q(n-1),x)+t*Q(n-1),
      x*deriv(Q(n-1),t)+s*deriv(Q(n-1),s)+x*deriv(Q(n-1),x)+s*Q(n-1)
    )
  )
}
for(n=0,12,print(Vec(subst(subst(Q(n),s,1),t,1)+x*O(x^(n\2)))))
/* END */

The program produces coefficients T(n,k) (of ascending powers of x):
[1]
[1]
[1, 1]
[2, 3]
[4, 9, 2]
[10, 30, 12]
[25, 100, 72, 6]
[75, 370, 372, 60]
[225, 1369, 1922, 600, 24]
[780, 5587, 9920, 4500, 360]
[2704, 22801, 51200, 33750, 5400, 120]
[10556, 101774, 273920, 234000, 55800, 2520]
[41209, 454276, 1465472, 1622400, 576600, 52920, 720]
... 
Notice that the row sums equals the Bell numbers (A000110). 
(This is surely not at all surprising to Emeric.) 
  
Compare coefficients in above triangle T(n,k) to Triangle A049020 : 
1;
1, 1;
2, 3, 1;
5, 10, 6, 1;
15, 37, 31, 10, 1;
52, 151, 160, 75, 15, 1; 
203, 674, 856, 520, 155, 21, 1; ...
 
to arrive at (by inspection): 
 
(*) T(n,k) = k!*A049020(n\2,k)*A049020((n+1)\2,k)
 
where offsets are chosen so that T(0,0)=1 and A049020(0,0)=1.
  

This result (*) yields an interesting recurrence for the Bell numbers:
 
(**) A000110(n) = Sum_{k=0..n} k!*A049020(n\2,k)*A049020((n+1)\2,k) .
 
Check: (PARI)
{A049020(n,k)=
  if(k<0|k>n,0,
    n!*polcoeff(polcoeff(exp((1+y)*(exp(x+x*O(x^n))-1)),n),k)
  ) 
} 
/* Demonstrate A000110(n) = Bell(n) for n=0..50: */
Bell(n)=if(n<0,0,n!*polcoeff(exp(exp(x+x*O(x^n))-1),n))
A000110(n) = sum(k=0,n,k!*A049020(n\2,k)*A049020((n+1)\2,k))
for(n=0,50,print1(A000110(n) - Bell(n),","))
/* END */
  

Perhaps someone can build on these first steps (or leap over them). 
 
But at least this demonstrates that the complexity of the e.g.f. of G(). 
    Paul
 
On Mon, 30 Oct 2006 22:21:02 -0500 (EST) Emeric Deutsch
<deutsch at duke.poly.edu> writes:
> Dear seqfans,
> 
> I have defined a sequence of polynomials Q[n](t,s,x) by the recurrence 
> relation
> Q[n]=t*dQ[n-1]/dt + x*dQ[n-1]/ds + x*dQ[n-1]/dx + tQ[n-1] if n is odd
and
> 
> Q[n]=x*dQ[n-1]/dt + s*dQ[n-1]/ds + x*dQ[n-1]/dx + sQ[n-1] if n is even;
> 
> Q[0]=1.
> 
> I would like to find the tetravariate e.g.f
> 
> G(t,s,x,z) = Q[0] + Q[1]*z/1! + Q[2]*z^2/2! + ...
[...]
> Emeric
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