[seqfan] "half-Stirling numbers of the first kind"

[Vladimir Shevelev]

Dear SeqFans,
 
Let us define recursively half-permanent (hperm) of a square nxn (n>=2) matrix. For 2x2 matrix A with usual 2-index numeration elements, we put  hperm(A)=a_{11}*a_{22}. For 3x3 matrix, by the expansion over its first row (without alternating signs),  and using the definition of half-permanent for 2x2 matrices, we define
 hperm(A)=a_{11}*a_{22}*a_{33}+a_{12}*a_{21}*a_{33}+a_{13}*a_{21}*a_{32}.
In the same way, using the definition of half-permanent for 3x3 matrices, we define hperm(A) for 4x4 matrix, etc. 
Consider permutations of (1,2), (1,2,3), (1,2,3,4), etc. given by the second indices in summands.  In case n=2, we have only permutation (1,2) with two cycles; in case
n=3, we have 3 permutations (1,2,3),(2,1,3),(3,1,2) with 3,2,1 cycles respectively; in case n=4 we have 12 permutations: one with 4 cycles, 3 with 3 cycles, 5 with 2 cycles and 3 with one cycle. Thus we obtain the triangle (number of permutations of n elements over number of cycles: 1,2,3,...)
(1)
0 1
1 1 1
3 5 3 1
....
with row sums n!/2, n>=2.
It is natural to call these numbers "half-Stirling of the first kind".
Problems: 1) To coninue the triangle; 2) To find a GF for half-Stirling numbers of the first kind; 3) Consider a permutation (k_1,...,k_n) of
numbers (1,...,n). To find a test when it is a permutation of the second indices in summands of hperm of square matrix A={a_(i,j)} of order n.
 
Regards,
Vladimir
 

 Shevelev Vladimir‎