[seqfan] Re: Guess the Triangle

[Ron Hardin]

Oh and of course that's equal to something simpler: apparently

T(n,k)=2*n*(1+k)^(n-1)-n

Should the formula be obvious somehow?

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



----- Original Message ----
> From: Ron Hardin <rhhardin at att.net>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Sent: Fri, March 29, 2013 9:09:41 AM
> Subject: [seqfan] Re: Guess the Triangle
> 
> 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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