[seqfan] Re: Looking for a simpler pattern for that matrix

[Gottfried Helms]

Dear Alexander / dear Frank -

 thank you for your help! Meanwhile I tried it myself hard and found
 a solution which can be nicely expressed in terms of row- and column-
 indexes.


 Let's denote the coefficient-matrix in my previous posting A.

 The first column is easy and possibly can have many different
 descriptions. So I looked at the second column first, using always
 positive values only.
 Because of the approximation of the quotient of subsequent entries
 (very likely!) to q=2 I assumed that a recursion with a factor 2 is
 involved.
 I wrote down the following differences

 D = {d}_ =

       0       .      .     .  .  .  .  .  .  .  .  .
       3       3      .     .  .  .  .  .  .  .  .  .
      22      16     10     .  .  .  .  .  .  .  .  .
     108      64     32    12  .  .  .  .  .  .  .  .
     432     216     88    24  0  .  .  .  .  .  .  .
    1520     656    224    48  0  0  .  .  .  .  .  .
    4896    1856    544    96  0  0  0  .  .  .  .  .
   14784    4992   1280   192  0  0  0  0  .  .  .  .
   42496   12928   2944   384  0  0  0  0  0  .  .  .
  117504   32512   6656   768  0  0  0  0  0  0  .  .
  314880   79872  14848  1536  0  0  0  0  0  0  0  .
  822272  192512  32768  3072  0  0  0  0  0  0  0  0

Here the entries d_{r,1} are the absolute values of the original sequence

In the second column D_{,2} are the weighted first order differences

   d_{r,2} = d_{r,1}-2*d_{r-1,1}

and in general I defined

   d_{r,c} = d_{r,c-1}-2*d_{r-1,c-1}


Luckily I found, that this weighted-differences scheme terminates after
the third-order differences and explicitely after solving the recursions,
introducing formal unknowns a,b,c,d,   I got the composition:

     0  = 1*a                                       ==> a=0
     3  = 2*a +  1* 1*b                             ==> b=3
    22  = 4*a +  2* 2*b +   1* 1*c                  ==> c=10
   108  = 8*a +  3* 4*b +   3* 2*c +  1* 1*d        ==> d=12
   432  =16*a +  4* 8*b +   6* 4*c +  4* 2*d
  1520  =32*a +  5*16*b +  10* 8*c + 10* 4*d
  4896  =64*a +  6*36*b +  15*16*c + 20* 8*d
  ...

where we find a simple composition by powers of 2 and binomials.


In fact, assuming the pascalmatrix P as lower triangular matrix,
then its second power, P^2 has precisely the coefficients of the
above scheme, so we can simply formulate the following matrix-formula.

We denote the second column of the matrix A as A_{,2}, and the
columnvector containing the found coefficients (a,b,c,d) as X
such that
           X=[a,b,c,d,0,0,0,....] // as columnvector

then the generation-rule for the given sequence is

        A_{,2} = P^2 * X

This is the general idea. Then the other columns have the same
scheme, I got for each column terminating difference-schemes to
the same order, only for the different weighting factors 1,2,3,...
involved by the various powers of P:

   A_{,1} = P^1 * [2,3, 2, 0, ...                    ]
   A_{,2} = P^2 * [0,3,10,12,  0, ...                ]
   A_{,3} = P^3 * [0,0, 4,21, 36, 0, ,...            ]
   A_{,4} = P^4 * [0,0, 0, 5, 36, 80, 0, ,...        ]
   A_{,5} = P^5 * [0,0, 0, 0,  6, 55,150, 0, ,...    ]



The pattern in the columnvectors in the rhs is very simple,
we can write them as

  2*1  3*1   2*1
   0   3*1   5*2  3*4
   0    0    4*1  7*3  4*9
   0    0     0   5*1  9*4  5*16
   0    0     0    0   6*1  11*5  6*25
  ...

or more compact and then it is easily visible how they depend
on the column-index for the original matrix A:

  2*1   3*1  2* 1       for A_{,1}
  3*1   5*2  3* 4       for A_{,2}
  4*1   7*3  4* 9       for A_{,3}
  5*1   9*4  5*16       for A_{,4}
  6*1  11*5  6*25       for A_{,5}

After that general solution we can now even "complete" the
matrix A in my previous post by one row at the top and one
column at the left introducing
  1*1   1*0  1* 0       for A_{,0}
getting

   A_{,0} = P^0 * [1, 0, ...                           ]
   A_{,1} = P^1 * [0,2,3, 2, 0, ...                    ]
   A_{,2} = P^2 * [0,0,3,10,12,  0, ...                ]
   A_{,3} = P^3 * [0,0,0, 4,21, 36, 0, ,...            ]
   A_{,4} = P^4 * [0,0,0, 0, 5, 36, 80, 0, ,...        ]
   A_{,5} = P^5 * [0,0,0, 0, 0,  6, 55,150, 0, ,...    ]


I do not yet know, whether the matrix A would be interesting
for the OEIS, I think this depends on the results which I
can now produce for my original question under study.

Thanks again to the correspondents!

Gottfried Helms

Am 06.05.2014 07:18 schrieb Alexander P-sky:
> I noticed that (or rather it appears that) for all rows starting from the
> row {-5, 3}
(...)