A118183 Column 0 of the matrix inverse of triangle A118180.
1, -1, 2, -10, 134, -4942, 505682, -142838074, 108933186230, -210663798566302, 812745803173573538, 6022271614633142122646, -2489044042602910169970590746, 996768343710992528631250678460690, -928936693384587466168289179772677376782
Offset: 0
Keywords
Examples
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) +...
Links
- G. C. Greubel, Table of n, a(n) for n = 0..65
Crossrefs
Cf. A118180.
Programs
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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 *)
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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])}
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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
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
Comments