Permanents of m x n (0,1)-matrices with m and m-1 zeros

[Jaap Spies]

Jaap Spies wrote:

> Further
> 
> b(m,n)=(n-1)*b(m-1,n-1) + (m-1)*b(m-2,n-2),
> with appropriate initial values b(0,k)=1, b(1,k)=k-1.
> 


b(1,1)=0, b(1,n)=n-1, b(2,2)=1 and b(2,n)=(n-1)*b(1,n-1) + 1*b(0,n-2)=(n-1)*(n-2)+1=
n^2-3n+3. This is in accordance with result (3.3) with m=2, found in
Seok-Zun Song, et al. Extremes of permanents of (0,1)-matrices, Lin. Algebra and its Applic.
373 (2003), p.207.
b(3,n)=(n-1)*b(2,n-1) + 2*b(1,n-2)=n^3 - 6n^2 + 14n -13, See (3.5) on p. 208 of the same
publication.


Jaap