OFFSET
0,4
COMMENTS
The diagonal bisections of this triangle T forms the diagonals of T^2 and T*U, where U = SHIFT_UP(T) indicates that U results from shifting each column of T up 1 row, dropping the main diagonal of all 1's.
FORMULA
T(2n-k,k) = Sum_{j=k..n} T(n,j)*T(j,k) = [T^2](n,k) for n>=k; odd-indexed diagonals: T(2n+1-k,k) = Sum_{j=k..n} T(n,j)*T(j+1,k) = [T*U](n,k) for n>=k; with T(n+1,n)=n+1, T(n,n)=1.
EXAMPLE
Triangle T begins:
1;
1, 1;
2, 2, 1;
3, 4, 3, 1;
6, 8, 6, 4, 1;
9, 14, 15, 8, 5, 1;
16, 28, 24, 24, 10, 6, 1;
26, 44, 57, 36, 35, 12, 7, 1;
44, 86, 84, 96, 50, 48, 14, 8, 1;
73, 130, 192, 136, 145, 66, 63, 16, 9, 1;
116, 250, 270, 356, 200, 204, 84, 80, 18, 10, 1;
191, 364, 567, 476, 590, 276, 273, 104, 99, 20, 11, 1;
294, 696, 780, 1060, 760, 906, 364, 352, 126, 120, 22, 12, 1; ...
The matrix square of T, T^2, equals the even-indexed
diagonal bisection of T, or T^2 = A118040 =
1;
2, 1;
6, 4, 1;
16, 14, 6, 1;
44, 44, 24, 8, 1;
116, 130, 84, 36, 10, 1;
294, 364, 270, 136, 50, 12, 1;
748, 990, 780, 476, 200, 66, 14, 1; ...
Let U = SHIFT_UP(T), which shifts each column of T up 1 row
and drops the main diagonal, so that U =
1;
2, 2;
3, 4, 3;
6, 8, 6, 4;
9, 14, 15, 8, 5;
16, 28, 24, 24, 10, 6; ...
Then the matrix product T*U equals the odd-indexed
diagonal bisection of T, or T*U = A118045 =
1;
3, 2;
9, 8, 3;
26, 28, 15, 4;
73, 86, 57, 24, 5;
191, 250, 192, 96, 35, 6;
500, 696, 567, 356, 145, 48, 7;
1234, 1824, 1683, 1060, 590, 204, 63, 8; ...
Thus interleaving diagonals of T^2 and T*U forms T.
MAPLE
{T(n, k)=if(n<k|k<0, 0, if(n==k, 1, if(n==k+1, n, sum(i=k, (n+k)\2, T((n+k)\2, i)*T(i+(n+k)%2, k) ))))}
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Paul D. Hanna, Apr 10 2006
STATUS
approved