three binomial(n,k) definitions

[Michael Somos]

seqfan,

      Brendan McKay writes that :

> is adopted.  The reason is that the BINOMIAL THEOREM
> requires
>    binomial(n,k) = n*(n-1)*...*(n-k+1) / k!
> for all non-negative integers k, REGARDLESS of n and even
> if n is a complex number!  Vast tracts of mathematics,

and is perhaps overstating his case for dramatic emphasis.
However, I think that outside the Pascal triangle range
(that is n and k nonnegative integers) there is no unique
way to define binomial(n,k) which will satisfy all possible
uses. Just for example, here is yet another way to extend
the table to preserve a connection to Fibonacci sequence :

      fibonacci(n)=sum(binomial(n-k,k))

n\k  -4   -3   -2   -1    0    1    2    3
   +--------------------------------------
-4 |  1    0    0    0    0    0    0    0
-3 | -3    1    0    0    0    0    0    0
-2 |  3   -2    1    0    0    0    0    0
-1 | -1    1   -1    1    0    0    0    0
 0 |  0    0    0    0    1    0    0    0
 1 |  0    0    0    0    1    1    0    0
 2 |  0    0    0    0    1    2    1    0
 3 |  0    0    0    0    1    3    3    1

where, as is well-known, the sum along anti-diagonals gives
the Fibonacci numbers and with this extension also gives
the two-way infinite Fibonacci sequence. A version of this
debate also arises for the single case of 0^0 and for this
see the sci.math FAQ. I hope that I have demonstrated that
there are several possible extensions of binomial(n,k) and
one is not better than the other, but only more convenient
or useful for a particular purpose. The key here is to be
very clear and explicit about which extension is to be used.
I again repeat my call for discussion of two-way infinite
versus one-way infinite sequences and the implications for
OEIS entries. Shalom, Michael