[seqfan] Re: Guess the Triangle

[Ron Hardin]

Great (thanks also to Maximilian Hasler)

Then empirical: T(n,k) = 2*n*sum{ binomial(n-1,i)*k^(n-1-i), i=0..(n-1) } - n

matches all my data so far (which doesn't extend very far into k for big n, e.g. 
n=7,k=5; n=8,k=1).

I wonder if more available factorizations of a large k would turn up affecting 
the empirical formula.



 rhhardin at mindspring.com
rhhardin at att.net (either)



----- Original Message ----
> From: William Keith <william.keith at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Sent: Fri, March 29, 2013 5:07:04 AM
> Subject: [seqfan] Re: Guess the Triangle
> 
> > T(n,k)=Number of idempotent nXn 0..k matrices of rank n-1
> >
> >  Rows n=1..6 match
> > a(k) = 1
> > a(k) = 4*k + 2
> > a(k) = 6*k^2 +  12*k + 3
> > a(k) = 8*k^3 + 24*k^2 + 24*k + 4
> > a(k) = 10*k^4 + 40*k^3  + 60*k^2 + 40*k + 5
> > a(k) = 12*k^5 + 60*k^4 + 120*k^3 + 120*k^2 + 60*k +  6
> 
> Pascal's triangle times 2n, except for the last coefficient, which  is
> just times n.
> 
> William  Keith
> 
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