[seqfan] Re: Guess the Triangle

[Ron Hardin]

Let me check...

Here's all 51 solutions for n=3 k=2
1..0..2....1..0..0....1..0..0....1..1..0....1..0..0....0..0..0....1..0..2
0..1..1....0..1..1....2..0..2....0..0..0....0..1..0....0..1..0....0..1..2
0..0..0....0..0..0....0..0..1....0..0..1....0..0..0....2..0..1....0..0..0

1..0..0....1..0..0....1..0..0....0..0..0....0..2..0....0..0..0....1..1..0
0..1..0....0..1..0....1..0..0....1..1..0....0..1..0....2..1..0....0..0..0
2..0..0....1..2..0....0..0..1....0..0..1....0..0..1....2..0..1....0..1..1

1..0..0....1..0..1....0..0..0....1..0..0....0..0..0....0..0..0....1..0..0
0..0..1....0..1..2....1..1..0....1..0..1....0..1..0....2..1..0....0..1..0
0..0..1....0..0..0....2..0..1....0..0..1....1..0..1....1..0..1....2..1..0

0..1..0....1..2..0....1..0..0....1..0..0....0..2..2....1..0..0....1..0..0
0..1..0....0..0..0....2..0..1....0..1..0....0..1..0....0..0..0....1..0..2
0..0..1....0..1..1....0..0..1....2..2..0....0..0..1....0..1..1....0..0..1

0..0..0....1..0..0....1..2..0....1..0..0....1..0..1....1..0..0....1..0..0
0..1..0....0..0..0....0..0..0....0..0..2....0..1..0....0..1..0....0..0..0
0..0..1....0..2..1....0..0..1....0..0..1....0..0..0....1..0..0....0..0..1

1..2..0....0..1..2....1..0..0....0..2..1....0..0..1....1..0..2....0..0..0
0..0..0....0..1..0....0..1..2....0..1..0....0..1..0....0..1..0....1..1..0
0..2..1....0..0..1....0..0..0....0..0..1....0..0..1....0..0..0....1..0..1

1..0..1....1..1..0....1..0..0....0..0..2....1..0..0....1..0..0....1..0..0
0..1..1....0..0..0....0..1..0....0..1..0....0..1..0....0..1..0....2..0..0
0..0..0....0..2..1....0..2..0....0..0..1....1..1..0....0..1..0....0..0..1

0..1..1....0..0..0
0..1..0....2..1..0
0..0..1....0..0..1

The e.g. third not being upper or lower triangular.



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



----- Original Message ----
> From: Rob Pratt <Rob.Pratt at sas.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Cc: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Sent: Fri, March 29, 2013 8:10:21 PM
> Subject: [seqfan] Re: Guess the Triangle
> 
> Well, idempotent implies that the eigenvalues are 0 and 1, with 0 and 1 on main  
>diagonal, and trace = rank.  So the main diagonal consists of n - 1 ones  and 1 
>zero.  There are n choices for the position of this zero.  If  the matrix is 
>upper triangular, the only nonzero elements appear in the column  corresponding 
>to the 0 diagonal element, and these n - 1 elements can be  arbitrary.  This 
>yields n(k+1)^(n-1) upper triangular matrices.   Similar argument yields the 
>same number of lower triangular matrices.  But  the n diagonal matrices have 
>been counted twice, so subtract n.  It remains  to show that such matrices must 
>be triangular so that these are the only  possibilities.
> 
> On Mar 29, 2013, at 9:27 AM, "Ron Hardin" <rhhardin at att.net> wrote:
> 
> > 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
> >>> 
> >>> _______________________________________________
> >>> 
> >>> Seqfan  Mailing list - http://list.seqfan.eu/
> >> 
> >>  _______________________________________________
> >> 
> >> Seqfan  Mailing  list - http://list.seqfan.eu/
> > 
> >  _______________________________________________
> > 
> > Seqfan Mailing  list - http://list.seqfan.eu/
> > 
> 
> 
> _______________________________________________
> 
> Seqfan  Mailing list - http://list.seqfan.eu/
>