OFFSET
0,2
COMMENTS
Row sums are A045618.
Considered as a square array A(m,n) = (2^m - 1)(2^n - 1), (m, n >= 1), read by rising antidiagonals, this gives the number of m X n matrices of rank 1 over the field F_2. For a different field F_q, that number would be A(m,n) = (q^m - 1)(q^n - 1)/(q - 1). It satisfies the recurrence relation A(m,n) = A(m,n-1)*q + A(m,1). - M. F. Hasler, Sep 12 2024
FORMULA
T(n,m) = T(n,n-m).
T(n,0) = T(n,n) = 2^(n+1) - 1. - M. F. Hasler, Sep 12 2024
EXAMPLE
1;
3, 3;
7, 9, 7;
15, 21, 21, 15;
31, 45, 49, 45, 31;
63, 93, 105, 105, 93, 63;
127, 189, 217, 225, 217, 189, 127;
255, 381, 441, 465, 465, 441, 381, 255;
511, 765, 889, 945, 961, 945, 889, 765, 511;
1023, 1533, 1785, 1905, 1953, 1953, 1905, 1785, 1533, 1023;
2047, 3069, 3577, 3825, 3937, 3969, 3937, 3825, 3577, 3069, 2047;
...
From M. F. Hasler, Sep 12 2024: (Start)
Considered as a square array A(m,n), read by antidiagonals, with m, n >= 1, this represents the following matrix A:
m \ n: 1 | 2 | 3 | 4 | 5 | ...
-----+------+-----+-----+-----+-----+-----
1 | 1 | 3 | 7 | 15 | 31 | ...
2 | 3 | 9 | 21 | 45 | 93 | ...
3 | 7 | 21 | 49 | 105 | 217 | ...
4 | 15 | 45 | 105 | 225 | 465 | ...
...
Here each row equals twice the previous row plus the first row, and likewise for columns. See my comment relating this to rank 1 matrices over F_2. (End)
MATHEMATICA
Table[Table[(2^(m + 1) - 1)*(2^(n - m + 1) - 1), {m, 0, n}], {n, 0, 10}]; Flatten[%]
PROG
(PARI) T(n, m) = (2^(m+1) - 1) * (2^(n-m+1) - 1) \\ M. F. Hasler, Sep 12 2024
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Roger L. Bagula and Gary W. Adamson, Oct 16 2008
STATUS
approved