A111505 Right half of Pascal's triangle (A007318) with zeros.

1, 0, 1, 0, 2, 1, 0, 0, 3, 1, 0, 0, 6, 4, 1, 0, 0, 0, 10, 5, 1, 0, 0, 0, 20, 15, 6, 1, 0, 0, 0, 0, 35, 21, 7, 1, 0, 0, 0, 0, 70, 56, 28, 8, 1, 0, 0, 0, 0, 0, 126, 84, 36, 9, 1, 0, 0, 0, 0, 0, 252, 210, 120, 45, 10, 1, 0, 0, 0, 0, 0, 0, 462, 330, 165

Offset: 0

Author

Philippe Deléham, Nov 16 2005

Keywords

• nonn
• tabl

Comments

A034869 is the version without zeros.

Examples

			Triangle begins:
1;
0, 1;
0, 2, 1;
0, 0, 3, 1;
0, 0, 6, 4, 1;
0, 0, 0, 10, 5, 1;
0, 0, 0, 20, 15, 6, 1;
0, 0, 0, 0, 35, 21, 7, 1;
0, 0, 0, 0, 70, 56, 28, 8, 1;
0, 0, 0, 0, 0, 126, 84, 36, 9, 1;
0, 0, 0, 0, 0, 252, 210, 120, 45, 10, 1;
0, 0, 0, 0, 0, 0, 462, 330, 165, 55, 11, 1;
0, 0, 0, 0, 0, 0, 924, 792, 495, 220, 66, 12, 1;
0, 0, 0, 0, 0, 0, 0, 1716, 1287, 715, 286, 78, 13, 1;
0, 0, 0, 0, 0, 0, 0, 3432, 3003, 2002, 1001, 364, 91, 14, 1;
		

Crossrefs

Cf. A007318, A034869, A094527, A111418.

Formula

Sum_{n, n>=k} T(n, k) = A001700(k).

Sum_{k =0..2*n} T(2*n, k) = A032443(n).

Sum_{k=0..2*n+1} T(2*n+1, k) = 4^n = A000302(n).

Sum_{k=0..2*n} T(2*n, k)^2 = A036910(n).

Sum_{k=0..2*n+1} T(2*n+1, k)^2 = C(4*n+1, 2*n) = A002458(n) . Paul D. Hanna