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

[Vladimir Shevelev]

Dear Olivier,

You are right, they are equivalent problems, and your calculation is excellent. However, concretely in this problem, I am not able to describe this constraint on the  relative position of two first symbols, such that the problem 3) remains open.

Best regards,
Vladimir

----- Original Message -----
From: Olivier Gerard <olivier.gerard at gmail.com>
Date: Tuesday, September 25, 2012 2:22
Subject: [seqfan] Re: "half-Stirling numbers of the first kind"
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> Dear Vladimir,
> 
> If I understand your idea correctly, this is equivalent to the 
> followingproblem:
> 
> 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
> >
> > _______________________________________________
> >
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
> 
> _______________________________________________
> 
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> 

 Shevelev Vladimir‎