NOTES ON A211226	Peter Bala, April 2012

The table entries are defined by

	T(n,k) := (floor(n/2))!/((floor(k/2))!*(floor((n-k)/2))!).

It is straightforward to obtain from this the explicit values

	T(2*n+1,2*k) = T(2*n+1,2*k+1) = T(2*n,2*k) = binomial(n,k)

and

	T(2*n,2*k+1) = n*binomial(n-1,k).

So the table entries are integers. An alternative method to prove that T(n,k)
is always an integer is to use Legendre's theorem on the order in which a prime
divides a factorial:

	the highest power of a prime p dividing n! equals

 	floor(n/p) + floor(n/p^2) + floor(n/p^3) + ....

We also require the following two simple properties of the floor function:

(1) For real x and y

	floor(x+y) >= floor(x) + floor(y).

(2) For positive integers m and n and real x

	floor(floor(x/m)/n) = floor(x/(m*n)).

PROPOSITION

	T(n,k) := (floor(n/2))!/((floor(k/2))!*(floor((n-k)/2))!)

	is an integer for any n,k >= 0.

PROOF

It is sufficient to show that the power to which an arbitrary prime p appears
in the rational number T(n, k) is always greater than or equal to zero. By
Legendre's theorem the order of a prime p dividing (floor(n/2))! equals

 floor(floor(n/2)/p) + floor(floor(n/2)/p^2) + floor(floor(n/2)/p^3) + ...

 = floor(n/(2*p)) + floor(n/(2*p^2)) + floor(n/(2*p^3)) + ... 

by (2).

It follows that the order of p dividing T(n,k) equals

 + {floor(n/(2*p)) - floor(k/(2*p)) - floor((n-k)/(2*p))}
 + {floor(n/(2*p^2)) - floor(k/(2*p^2)) - floor((n-k)/(2*p^2))}
 + {floor(n/(2*p^3)) - floor(k/(2*p^3)) - floor((n-k)/(2*p^3))} + ... >= 0 

by taking x = k/(2*p^r) and y = (n-k)/(2*p^r) in (1) and summing over r

END PROOF.

The advantage of this approach is that it extends to proving that the numbers 

	(floor(r*n))!/((floor(r*k))!*(floor(r*(n-k)))!)

are integers for any real r > 0 : first prove it for r = 1/positive integer,
then for r a positive rational and finally, by a continuity argument, for
arbitrary positive real r.

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