P. Stanica's answer re: A007318 - Pascal's triangle and its powers.

[Antti Karttunen]

Pantelimon Stanica asked me to forward the following message...

 (I appologize for my repeated message on Pascal triangle, but it seems
that
  my address is rejected, although I am a member of SeqFan (perhaps with
a
  different address). I would appreciate it, if you could forward my
message
  to the list. I have to figure out how to unsubscribe and then
subscribe
  again with the new address. Thank you.

  Pante(limon) Stanica
  Auburn University Montgomery)



-------- Original Message --------
Subject: Re: A007318 - Pascal's triangle and its powers. Example
corrected.
Date: Mon, 5 Nov 2001 11:41:38 -0600 (CST)
From: Pantelimon Stanica <stanica at strudel.aum.edu>
To: Antti Karttunen <karttu at megabaud.fi>
CC: seqfan at ext.jussieu.fr, juha.oikkonen at helsinki.fi



Dear all,

Regarding Antti's matrices: it is easy to prove
that the q-th power of Pascal triangle has entries

$q^(i-j) binom{i-1}{j-1}$

In fact, I proved (as a sequel to a paper of mine with Rhodes Peele, 
which will appear in Fib. Quart.) that any matrix satisfying a 4
term-recurrence (and by taking some zeros as coefficients, we get the
Pascal triangle) preserves the recurrence, in the sense that the
coefficients of the recurrence for any power of the matrix
are terms in a binary sequence.

e.g. powers of the matrix $A=(binom{i-1}{n-j})_{i,j}$
satisfy a 4-term recurrence where the coefficients are
Fibonacci numbers. 
The entries $a_{i,j}^{(e)}$ of the $e$-th power of $A$
satisfy the relation
\[
F_{e-1} a_{i,j}^{(e)}=F_{e} a_{i-1,j}^{(e)}+
F_{e+1} a_{i-1,j-1}^{(e)}-F_{e} a_{i,j-1}^{(e)},
\]
where $F_e$ is the Fibonacci sequence.


I conjectured that any $(p^2)$-term recurrence
is preserved, with coefficients, which are elements in a 
p-term recurrence. If anybody has some interest on that, I'll be happy
to
share what I have so far....

Pante Stanica
Auburn University Montgomery
Montgomery, AL 36124-4023

On Mon, 5 Nov 2001, Antti Karttunen wrote:

> 
> 
> Antti Karttunen wrote:
> 
> > Cher(e)s Fanaticien(ne)s,
> >
> > just realized that when the Pascal's triangle is considered as
> > an infinite lower triangular matrix, e.g. as
> >
> > (1 0 0 0 0 0 0 0 0 ...)
> > (1 1 0 0 0 0 0 0 0 ...)
> > (1 2 1 0 0 0 0 0 0 ...)
> > (1 3 3 1 0 0 0 0 0 ...)
> > (1 4 6 4 1 0 0 0 0 ...)
> >
> > then its nth power (n in Z, thus including also n=0, the identity
> > matrix,
> > and n=-1, A007318's inverse) seems to be given by the recursive
> > construction
> > T(0,0) = 1,
> > T(n,k) = T(n-1,k-1) + n*T(n-1,k)
> >
> 
> Again my personal sin, the overuse of variables. Let's have n for exponent
> and r for row, and the above is much clearer:
> 
> T(0,0) = 1,
> T(r,k) = T(r-1,k-1) + n*T(r-1,k)
> 
> 
> >
> > (e.g. see A007318's "square" A038207 and the "cube" A027465).
> >
> > What about any other triangles generated with such simple recursive
> > rules,
> > e.g. the Delannoy numbers A008288 (T(n,k) = T(n-1,k-1) + T(n-1,k) +
> > T(n-2,k-1))
> > or the Catalan's triangle A009766 (T(n,k) = T(n,k-1) + T(n-1,k), T(n,0)
> > = 1) (or rows reverted version: A033184)
> > do any of these have such "meta-rules" ???
> 
> I think I need to read Wolfdieter's Lang paper "On Generalizations of the
> Stirling Number Triangles"
> at http://www.research.att.com/~njas/sequences/JIS/VOL3/LANG/lang.html
> 
> 
> -- Same
> 
> >
> >
> > Terveisin,
> >
> > Antti Karttunen
>