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

%I A143088 #12 Sep 13 2024 08:06:46
%S A143088 1,3,3,7,9,7,15,21,21,15,31,45,49,45,31,63,93,105,105,93,63,127,189,
%T A143088 217,225,217,189,127,255,381,441,465,465,441,381,255,511,765,889,945,
%U A143088 961,945,889,765,511,1023,1533,1785,1905,1953,1953,1905,1785,1533,1023,2047
%N A143088 Triangle T(n,m) = (2^(m+1) - 1) * (2^(n-m+1) - 1), read by rows, 0 <= m <= n.
%C A143088 Row sums are A045618.
%C A143088 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
%F A143088 T(n,m) = T(n,n-m).
%F A143088 T(n,0) = T(n,n) = 2^(n+1) - 1. - _M. F. Hasler_, Sep 12 2024
%e A143088 1;
%e A143088 3, 3;
%e A143088 7, 9, 7;
%e A143088 15, 21, 21, 15;
%e A143088 31, 45, 49, 45, 31;
%e A143088 63, 93, 105, 105, 93, 63;
%e A143088 127, 189, 217, 225, 217, 189, 127;
%e A143088 255, 381, 441, 465, 465, 441, 381, 255;
%e A143088 511, 765, 889, 945, 961, 945, 889, 765, 511;
%e A143088 1023, 1533, 1785, 1905, 1953, 1953, 1905, 1785, 1533, 1023;
%e A143088 2047, 3069, 3577, 3825, 3937, 3969, 3937, 3825, 3577, 3069, 2047;
%e A143088 ...
%e A143088 From _M. F. Hasler_, Sep 12 2024: (Start)
%e A143088 Considered as a square array A(m,n), read by antidiagonals, with m, n >= 1, this represents the following matrix A:
%e A143088     m \ n: 1  |  2  |  3  |  4  |  5  | ...
%e A143088   -----+------+-----+-----+-----+-----+-----
%e A143088     1  |   1  |  3  |   7 |  15 |  31 | ...
%e A143088     2  |   3  |  9  |  21 |  45 |  93 | ...
%e A143088     3  |   7  | 21  |  49 | 105 | 217 | ...
%e A143088     4  |  15  | 45  | 105 | 225 | 465 | ...
%e A143088    ...
%e A143088 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)
%t A143088 Table[Table[(2^(m + 1) - 1)*(2^(n - m + 1) - 1), {m, 0, n}], {n, 0, 10}]; Flatten[%]
%o A143088 (PARI) T(n,m) = (2^(m+1) - 1) * (2^(n-m+1) - 1) \\ _M. F. Hasler_, Sep 12 2024
%Y A143088 Cf. A000225 (first column), A068156 (second column).
%K A143088 nonn,tabl
%O A143088 0,2
%A A143088 _Roger L. Bagula_ and _Gary W. Adamson_, Oct 16 2008