A118183 - OEIS

%I #7 Jun 29 2021 06:33:24

%S 1,-1,2,-10,134,-4942,505682,-142838074,108933186230,-210663798566302,

%T 812745803173573538,6022271614633142122646,

%U -2489044042602910169970590746,996768343710992528631250678460690,-928936693384587466168289179772677376782

%N Column 0 of the matrix inverse of triangle A118180.

%C 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).

%H G. C. Greubel, <a href="/A118183/b118183.txt">Table of n, a(n) for n = 0..65</a>

%F G.f.: 1 = Sum_{n>=0} a(n)*x^n/(1-3^n*x).

%F 0^n = Sum_{k=0..n} a(k)*(3^k)^(n-k) for n>=0.

%F 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

%e Recurrence at n=4:

%e 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

%e = 1*(3^0) - 1*(3^3) + 2*(3^4) - 10*(3^3) + 134*(3^0).

%e The g.f. is illustrated by:

%e 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)

%e - 4942*x^5/(1-243*x) +505682*x^6/(1-729*x) -142838074*x^7/(1-2187*x) +...

%t a[n_]:= a[n]= If[n<2, (-1)^n, -Sum[3^(j*(n-j))*a[j], {j,0,n-1}]];

%t Table[a[n], {n, 0, 30}] (* _G. C. Greubel_, Jun 29 2021 *)

%o (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])}

%o (Sage)

%o @CachedFunction

%o def a(n): return (-1)^n if (n<2) else -sum(3^(j*(n-j))*a(j) for j in (0..n-1))

%o [a(n) for n in (0..30)] # _G. C. Greubel_, Jun 29 2021

%Y Cf. A118180.

%K sign

%O 0,3

%A _Paul D. Hanna_, Apr 15 2006