A143088 Triangle T(n,m) = (2^(m+1) - 1) * (2^(n-m+1) - 1), read by rows, 0 <= m <= n.

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

Offset: 0

Author

Roger L. Bagula and Gary W. Adamson, Oct 16 2008

Keywords

• nonn
• tabl

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

Examples

			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)
		

Crossrefs

Cf. A000225 (first column), A068156 (second column).

Programs

Mathematica

Table[Table[(2^(m + 1) - 1)*(2^(n - m + 1) - 1), {m, 0, n}], {n, 0, 10}]; Flatten[%]

PARI

T(n,m) = (2^(m+1) - 1) * (2^(n-m+1) - 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