A140998 Triangle G(n, k), read by rows, for 0 <= k <= n, where G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, and G(n+3, m) = G(n+1, m-1) + G(n+1, m) + G(n+2, m) for n >= 0 and m = 1..n+1.

1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 7, 5, 2, 1, 1, 12, 11, 5, 2, 1, 1, 20, 23, 12, 5, 2, 1, 1, 33, 46, 28, 12, 5, 2, 1, 1, 54, 89, 63, 29, 12, 5, 2, 1, 1, 88, 168, 137, 69, 29, 12, 5, 2, 1, 1, 143, 311, 289, 161, 70, 29, 12, 5, 2, 1, 1, 232, 567, 594, 367, 168, 70, 29, 12, 5, 2, 1

Offset: 0

Author

Juri-Stepan Gerasimov, Jul 08 2008

Keywords

• nonn
• tabl

Comments

From Petros Hadjicostas, Jun 10 2019: (Start)

According to the attached picture, the index of asymmetry here is s = 1 and the index of obliqueness (or obliquity) is e = 0.

In the picture, the equation G(n, e*n) = 1 becomes G(n, 0) = 1, while the equations G(n+x+1, n-e*n+e*x-e+1) = 2^x for 0 <= x < s = 1 become G(n+1, n+1) = 1 and G(n+2, n+1) = 2.

Also, in the picture, the recurrence G(n+s+2, k) = G(n+1, k-e*s+e-1) + Sum_{m=1..s+1} G(n+m, k-e*s+m*e-2*e) for k = 1..n+1 becomes G(n+3, k) = G(n+1, k-1) + G(n+1, k) + G(n+2, k) for k = 1..n+1.

Except for a shifting of the indices by 1, this array is a mirror image of array A140993. We have G(n, k) = A140993(n+1, n-k+1) for 0 <= k <= n. Triangular array A140993 has the same index of asymmetry (i.e., s = 1) but index of obliqueness e = 1.

(End)

Examples

			Triangle begins (with rows for n >= 0 and columns for k >= 0):
  1;
  1,   1;
  1,   2,   1;
  1,   4,   2,   1;
  1,   7,   5,   2,   1;
  1,  12,  11,   5,   2,   1;
  1,  20,  23,  12,   5,   2,   1;
  1,  33,  46,  28,  12,   5,   2,   1;
  1,  54,  89,  63,  29,  12,   5,   2,   1;
  1,  88, 168, 137,  69,  29,  12,   5,   2,   1;
  1, 143, 311, 289, 161,  70,  29,  12,   5,   2,   1;
		

Links

• G. C. Greubel, Rows n = 0..100 of triangle, flatten
• Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Crossrefs

Cf. A000071, A007318, A140992, A140993, A140994, A140995, A140996, A140997, A141020, A141021.

Programs

Mathematica

G[n_,k_] := G[n,k] = Which[k==0 || k==n, 1, k==n-1, 2, True, G[n-2,k-1] + G[n-2,k] + G[n-1,k]]; Table[G[n,k], {n,0,12}, {k,0,n}] (* Jean-François Alcover, Jun 09 2019 *)

PARI

G(n,k) = if(k==0 || k==n, 1, if(k==n-1, 2, G(n-1, k) + G(n-2, k) + G(n-2, k-1)));
for(n=0,12, for(k=0,n, print1(G(n,k), ", "))) \\ G. C. Greubel, Jun 09 2019

Sage

def G(n,k):
    if (k==0 or k==n): return 1
    elif (k==n-1): return 2
    else: return G(n-1, k) + G(n-2, k) + G(n-2, k-1)
[[G(n,k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jun 09 2019

Formula

From Petros Hadjicostas, Jun 10 2019: (Start)

G(n, k) = A140993(n+1, n-k+1) for 0 <= k <= n.

Let A(x,y) = Sum_{n,k >= 0} G(n, k)*x^n*y^k and B(x,y) = Sum_{n,k >= 1} A140993(n, k). Then A(x, y) = x^(-1) * B(x*y, y^(-1)). Thus, the g.f. of the current array is A(x, y) = (1 - x - x^2 + x^3*y)/((1 - x) * (1 - x*y) * (1 - x - x^2 - x^2*y)).

To find the g.f. of the k-th column (where k >= 0), we differentiate A(x, y) k times with respect to y, divide by k!, and substitute y = 0. For example, differentiating A(x, y) once w.r.t. y and setting y = 0, we get the g.f. of the k = 1 column: x/((1 - x)*(1 - x - x^2)). This is the g.f. of sequence (A000071(n+2): n >= 0) = (Fibonacci(n+2) - 1: n >= 0).

G.f. of column k = 2 is x^2*(1 - x + x^3)/((1 - x)*(1 - x - x^2)^2). Thus, column k = 2 is a shifted version of (A140992(n): n >= 0).

(End)

Extensions

Indices in the definition corrected by R. J. Mathar, Aug 02 2009

Name edited by Petros Hadjicostas, Jun 10 2019