three binomial(n,k) definitions
Michael Somos
somos at grail.cba.csuohio.edu
Sat Dec 14 20:00:31 CET 2002
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
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