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

[Olivier Gerard]

Dear Vladimir,

If I understand your idea correctly, this is equivalent to the following
problem:

Counting permutations by cycles with a constraint on the relative position
of two first symbols (as
you are starting from a 2x2 matrix). Permutations with the pattern ... 1
... 2 .... are easy to compute.

This gives the following triangle:

0 | 0

1 | 0 1

3 | 1 1 1

12 | 3 5 3 1

60 | 12 24 17 6 1

360 | 60 134 110 45 10 1

2520 | 360 870 799 375 100 15 1

20160 | 2520 6474 6489 3409 1050 196 21 1


It is not currently in the OEIS. If you agree, I propose to enter it in the
Encyclopedia with reference to this mail.


With my best regards,


Olivier GERARD



On Tue, Sep 25, 2012 at 2:42 PM, Vladimir Shevelev <shevelev at bgu.ac.il>wrote:

> 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
>
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