[seqfan] Re: Guess the Triangle

[Rob Pratt]

Sorry, triangular only if the 0 appears first or last on the diagonal.  The correct characterization seems to be: differs from the identity matrix in exactly one or row or column, with 0 as the diagonal element of that column.

On Mar 29, 2013, at 8:52 PM, "Ron Hardin" <rhhardin at att.net> wrote:

> 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
>>>>> 
>>>>> _______________________________________________
>>>>> 
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