A110681 A convolution triangle of numbers based on A071356.

1, 2, 1, 6, 4, 1, 20, 16, 6, 1, 72, 64, 30, 8, 1, 272, 260, 140, 48, 10, 1, 1064, 1072, 636, 256, 70, 12, 1, 4272, 4480, 2856, 1288, 420, 96, 14, 1, 17504, 18944, 12768, 6272, 2320, 640, 126, 16, 1, 72896, 80928, 57024, 29952, 12192, 3852, 924, 160, 18, 1

Offset: 0

Author

Philippe Deléham, Sep 14 2005

Keywords

• nonn
• tabl

Examples

			Triangle starts:
   1;
   2,  1;
   6,  4,  1;
  20, 16,  6,  1;
  72, 64, 30,  8,  1;
  ...
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..11324 (rows 0 <= n <= 150).
• G. Kreweras, Sur les hiérarchies de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #20 (1973), p. 32-33.
• G. Kreweras, Sur les hiérarchies de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #20 (1973). (Annotated scanned copy)

Crossrefs

Cf. A071356 (1st column), A071357 (2nd column).

Cf. A052705 (diagonal sums), A001003 (row sums).

Programs

Mathematica

T[n_, k_] := T[n, k] = Which[n == k == 0, 1, n == 0, 0, k == 0, 0, k > n, 0, True, T[n - 1, k - 1] + 2 T[n - 1, k] + 2 T[n - 1, k + 1]]; Table[T[n, k], {n, 0, 10}, {k, n}] // Flatten (* Michael De Vlieger, Nov 05 2017 *)

Formula

T(0, 0) = 1; T(n, k) = 0 if k<0 or if k>n; T(n, k) = T(n-1, k-1) + 2*T(n-1, k) + 2*T(n-1, k+1).

Sum_{k, k>=0} T(m, k)*T(n, k)*2^k = T(m+n, 0) = A071356(m+n).

Sum_{k, k>=0} T(n, k)*(2^(k+1) - 1) = 5^n.

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