OFFSET
0,3
COMMENTS
The entire matrix inverse of triangle A118180 is determined by column 0 (this sequence): [A118180^-1](n,k) = a(n-k)*(3^k)^(n-k) for n>=k>=0. Any g.f. of the form: Sum_{k>=0} b(k)*x^k may be expressed as: Sum_{n>=0} c(n)*x^n/(1-3^n*x) by applying the inverse transformation: c(n) = Sum_{k=0..n} a(n-k)*b(k)*(3^k)^(n-k).
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..65
FORMULA
G.f.: 1 = Sum_{n>=0} a(n)*x^n/(1-3^n*x).
0^n = Sum_{k=0..n} a(k)*(3^k)^(n-k) for n>=0.
a(n) = (-1)*Sum_{j=0..n-1} 3^(j*(n-j))*a(j) with a(0) = 1 and a(1) = -1. - G. C. Greubel, Jun 29 2021
EXAMPLE
Recurrence at n=4:
0 = a(0)*(3^0)^4 +a(1)*(3^1)^3 +a(2)*(3^2)^2 +a(3)*(3^3)^1 +a(4)*(3^4)^0
= 1*(3^0) - 1*(3^3) + 2*(3^4) - 10*(3^3) + 134*(3^0).
The g.f. is illustrated by:
1 = 1/(1-x) -1*x/(1-3*x) +2*x^2/(1-9*x) -10*x^3/(1-27*x) +134*x^4/(1-81*x)
- 4942*x^5/(1-243*x) +505682*x^6/(1-729*x) -142838074*x^7/(1-2187*x) +...
MATHEMATICA
a[n_]:= a[n]= If[n<2, (-1)^n, -Sum[3^(j*(n-j))*a[j], {j, 0, n-1}]];
Table[a[n], {n, 0, 30}] (* G. C. Greubel, Jun 29 2021 *)
PROG
(PARI) {a(n)=local(T=matrix(n+1, n+1, r, c, if(r>=c, (3^(c-1))^(r-c)))); return((T^-1)[n+1, 1])}
(Sage)
@CachedFunction
def a(n): return (-1)^n if (n<2) else -sum(3^(j*(n-j))*a(j) for j in (0..n-1))
[a(n) for n in (0..30)] # G. C. Greubel, Jun 29 2021
CROSSREFS
KEYWORD
sign
AUTHOR
Paul D. Hanna, Apr 15 2006
STATUS
approved