Curious binomial-identity /A002720 (small correction of prev. post)

[Gottfried Helms]

Am 25.11.2006 22:53 schrieb Joseph Biberstine:
> The example you gave has an off-by-one row index as follows:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n+1)
> 
> Note that the row index is incremented.  Mathematica simplifies this
> fraction to LaguerreL[-(n+1), 1]/LaguerreL[n+1, -1].  This is not equal
> to Exp[1], but interestingly seems to approach Exp[1] monotonically from
> below as n approaches infinity.
> 
> I'm sure you meant:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n)
> 
> Mathematica simplifies this to LaguerreL[-(n+1), 1]/LaguerreL[n, -1].
> Although I can't get it to explicitly resolve that this is exactly
> Exp[1], the values do seem to bear this truth out.
> 
> So, I believe your example should read: e.g. for n=3,
> 
>     1/0! + 4/1! + 10/2! + 20/3! + 35/4! + ...  weighted col-sum
> e = -----------------------------------------  -------------------
>            1/0! + 3/1! + 3/2! + 1/3!           weighted row-sum
> 
> This does seem to hold for every col/row pair in the triangle.

Yes, your correction is right - I didn't look right at the
matrix.


The connection to the Laguerre-polynomials is not too
much surprising; I'm dealing with a ceratin classes
of lower-triangular matrices, which are derived from the
logic of the computing of the Pascal-matrix by matrix-
exponentiation.

While

  Pascalmatrix     = Pk(1)=  matrixexp( subdiagonal(1; [1,2,3,4,5,...]))

the Laguerre-matrix occurs from

  Laguerrematrix(*) = Pk(2) = matrixexp( subdiagonal(1; [1,2^2,3^2,4^2,5^2,...]))
           ((*) means: in a signed and simply scaled form)


I also have
                      Pk(0) = matrixexp( subdiagonal(1; [1,2^0,3^0,4^0,5^0,...]))

(where the vectors are filled in the first principal subdiagonal of an
infinite square-matrix)




Now the current matrix was created by a somehow routinely
"completion of my toolbox" of pascal-like-matrices, as a
hadamard-product of

      Pf =    Pk(1) (x) Pk(0)

         =    1/0!
              1/1!  1/0!
              1/2!  2/1!  1/0!
              1/3!  3/2!  3/1!  1/0!
              1/4!  4/3!  6/2!  4/1!   1/0!
                ....


Then with a powerseries-column-vector

      V(x) = [1,x,x^2,x^3,....] ~

and a diagonal factorial-matrix (used for scaling)

     dF(m) = diag(0! , 1!^m, 2!^m .... )

I got

   dF(1)  *   Pf  * V(1) = A002720       // A002720 read as column-vector
                                         // Due to the entry of Paul Berry

and
   dF(1) *    Pf~ * V(1) = A002720  *exp(1)  // added today



As a table it looks like:

                                                 Row-Sum
         =    1/0!                              =  1/0!    =  A(0)/0!
              1/1!  1/0!                        =  2/1!    =  A(1)/1!
              1/2!  2/1!  1/0!                  =  7/2!    =  A(2)/2!
              1/3!  3/2!  3/1!  1/0!            = 34/3!    =  A(3)/3!
              1/4!  4/3!  6/2!  4/1!   1/0!     =209/4!    =  A(4)/4!
                ....
             --------------------------------------------------------
Col-sum   e*[  1    2/1!  7/2! 34/3! 209/4! ]

Note, that using

   dF(1)  *   Pf  * V(2) =          ???
   dF(1) *    Pf~ * V(2) = exp(2) * ???

  in the second an exponentiation of 2 occurs, and generally for an s

   dF(1) *    Pf~ * V(s) = exp(s) * ???

 with unknown, but -heuristically- integer sequences of coefficients in ???
 for integer s>=0 .


------------------------------------------------------------------------


The Pascal- and (unsigned,scaled) Laguerre-matrix are also
related by

  Pk(0)                              // "base"-matrix
  Pk(1) =   dF(1)* Pk(0) * dF(-1)    // Pascal-matrix
  Pk(2) =   dF(1)* Pk(1) * dF(-1)    // unsigned,scaled Laguerrematrix
 ...
(which is due to the construction via matrix-exponentiation
of the related subdiagonal-matrices)
so finally the occurence of the relation to the Laguerre-matrix
using Mathematica is not too surprising...

Regards -

Gottfried Helms