[seqfan] Re: Problems posed regarding A004736 and A003983.

[franktaw at netscape.net]

This product for A004736 is easy. Just do the calculation.

For A003983, the matrix you show is the primary representation of the 
sequence, as named "Array read by antidiagonals with T(n,k) = min(n,k)."

I don't think your M makes sense with any standard definition of an 
infinite matrix. Multiplication for infinite matrices is not 
well-defined in general, although there are a number of special cases 
where it does make sense.

Franklin T. Adams-Watters

-----Original Message-----
From: Ed Jeffery <lejeffery7 at gmail.com>

The two sequences are A004736 <https://oeis.org/A004736> and
A003983<https://oeis.org/A003983>.
First, A004736 is the triangle

1;
2,1;
3,2,1;
4,3,2,1;
5,4,3,2,1;
...,

in which the pattern for the n-th row continues as {n, n-1, ..., 2, 1}. 
Can
someone prove that this triangle, taken as an infinite lower-triangular
matrix, is equal to B^2, where

B=[1,0,...; 1,1,0,...; 1,1,1,0,...; ...] (here 0,... means 0,0,0,...)?

Second, A003983 is the triangle

1;
1,1;
1,2,1;
1,2,2,1;
1,2,3,2,1;
...,

in which you "...count up to ceiling(n/2) and back down again (repeating
the central term when n is even)," as described by Franklin T.
Adams-Watters.

I was trying to find a lower-triangular matrix whose square gives this
triangle but failed. I also got no clues from the generating function.
However, arranging the diagonals as rows and letting the sequence be 
read
from the antidiagonals of the resulting infinite (square) array

A=
1,1,1,1,1,...;
1,2,2,2,2,...;
1,2,3,3,3,...;
1,2,3,4,4,...;
1,2,3,4,5,...;
...,

can someone prove that A = M^2, where M is the infinite square matrix 
(if
that makes sense)

M = [0,...,0,1; 0,...,0,1,1; 0,...,0,1,1,1; ...]?

Finally, can someone prove that

B*M = [0,...,0,1; 0,...,0,1,2; 0,...,0,1,2,3; ...]?


LEJ

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