A007318 - Pascal's triangle and its powers. Example corrected.
Antti Karttunen
karttu at megabaud.fi
Mon Nov 5 17:04:34 CET 2001
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
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