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
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
Programs
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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.
Comments