A153861 Triangle read by rows, binomial transform of triangle A153860.
1, 1, 1, 2, 3, 1, 3, 6, 4, 1, 4, 10, 10, 5, 1, 5, 15, 20, 15, 6, 1, 6, 21, 35, 35, 21, 7, 1, 7, 28, 56, 70, 56, 28, 8, 1, 8, 36, 84, 126, 126, 84, 36, 9, 1, 9, 45, 120, 210, 252, 210, 120, 45, 10, 1, 10, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1
Offset: 0
Examples
First few rows of the triangle are: 1; 1, 1; 2, 3, 1; 3, 6, 4, 1; 4, 10, 10, 5, 1; 5, 15, 20, 15, 6, 1; 6, 21, 35, 35, 21, 7, 1; 7, 28, 56, 70, 56, 28, 8, 1; 8, 36, 84, 126, 126, 84, 36, 9, 1; 9, 45, 120, 210, 252, 210, 120, 45, 10, 1; ...
Links
- G. C. Greubel, Table of n, a(n) for the first 46 rows
Crossrefs
Programs
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Mathematica
z = 10; c = 1; d = 1; p[0, x_] := 1 p[n_, x_] := x*p[n - 1, x] + 1; p[n_, 0] := p[n, x] /. x -> 0; q[n_, x_] := (c*x + d)^n t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x -> 0; w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, x_] := 1 g[n_] := CoefficientList[w[n, x], {x}] TableForm[Table[Reverse[g[n]], {n, -1, z}]] Flatten[Table[Reverse[g[n]], {n, -1, z}]] (* A193815 *) TableForm[Table[g[n], {n, -1, z}]] Flatten[Table[g[n], {n, -1, z}]] (* A153861 *) (* Clark Kimberling, Aug 06 2011 *)
Formula
Triangle read by rows, A007318 * A153860. Remove left two columns of Pascal's triangle and append (1, 1, 2, 3, 4, 5,...).
As a recursive operation by way of example, row 3 = (3, 6, 4, 1) =
[1, 1, 1, 0] * (flipped Pascal's triangle matrix) = [1, 3, 3, 1]
[1, 2, 1, 0]
[1, 1, 0, 0]
[1, 0, 0, 0].
T(n,k) = 2*T(n-1,k)+T(n-1,k-1)-T(n-2,k)-T(n-2,k-1), T(0,0) = T(1,0) = T(1,1) = T(2,2) = 1, T(2,0)=2, T(2,1)=3, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 15 2013
G.f.: (1-x+x^2+x^2*y)/((x-1)*(-1+x+x*y)). - R. J. Mathar, Aug 11 2015
Comments