OFFSET

0,2

COMMENTS

In the attached photograph, we see that the index of asymmetry is denoted by s and the index of obliqueness by e. The general recurrence is G(n+s+2, k) = G(n+1, k-e*s+e-1) + Sum_{1 <= m <= s+1} G(n+m, k-e*s+m*e-2*e) for n >= 0 with k = 1..(n+1) when e = 0 and k = (s+1)..(n+s+1) when e = 1. The initial conditions are G(n+x+1, n-e*n+e*x-e+1) = 2^x for x=0..s and n >= 0. There is one more initial condition, namely, G(n, e*n) = 1 for n >= 0.

For s = 0, we get Pascal's triangle A007318. For s = 1, we get A140998 (e = 0) and A140993 (e = 1). For s = 2, we get A140997 (e = 0) and A140994 (e = 1). For s = 3, we get A140996 (e = 0) and A140995 (e = 1). For s = 4, we have arrays A141020 (with e = 0) and A141021 (with e = 1). In some of these arrays, the indices n and k are shifted.

For the triangular array G(n, k) = A140995(n, k), we have G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, G(n+4, 3) = 8, and G(n+5, k) = G(n+1, k-3) + G(n+1, k-4) + G(n+2, k-3) + G(n+3, k-2) + G(n+4, k-1) for n >= 0 and k = 4..(n+4).

With G(n, k) = A140995(n, k), the current sequence (a(k): k >= 0) is defined by a(k) = lim_{n -> infinity} G(n, k) for k >= 0. Then a(k) = a(k-4) + 2*a(k-3) + a(k-2) + a(k-1) for k >= 4 with a(x) = 2^x for x = 0..3.

LINKS

FORMULA

a(0) = 1, a(1) = 2, a(2) = 4, a(3) = 8, and a(k) = a(k-1) + a(k-2) + 2*a(k-3) + a(k-4) for k >= 4.

G.f.: (x^2 + x + 1)/(-x^4 - 2*x^3 - x^2 - x + 1).

CROSSREFS

KEYWORD

nonn,easy

AUTHOR

Petros Hadjicostas, Aug 03 2019

STATUS

approved