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) = A141021(n, k), we have G(n+6, k) = G(n+1, k-4) + G(n+1, k-5) + G(n+2, k-4) + G(n+3, k-3) + G(n+4, k-2) + G(n+5, k-1) for n >= 0 and k = 5..(n+5) with G(n+x+1, x) = 2^x for x = 0..4 and n >= 0.
With G(n, k) = A141021(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-5) + 2*a(k-4) + a(k-3) + a(k-2) + a(k-1) for k >= 5 with a(x) = 2^x for x = 0..4.
LINKS
FORMULA
a(k) = a(k-5) + 2*a(k-4) + a(k-3) + a(k-2) + a(k-1) for k >= 5 with a(k) = 2^k for k = 0..4.
G.f.: -(x + 1)*(x^2 + 1)/(x^5 + 2*x^4 + x^3 + x^2 + x - 1).
CROSSREFS
KEYWORD
nonn
AUTHOR
Petros Hadjicostas, Jun 25 2019
STATUS
approved