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Column 0 of the matrix inverse of triangle A118185(n,k) = (4^k)^(n-k).
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%I #9 Jun 29 2021 06:32:54

%S 1,-1,3,-33,1407,-237057,158992383,-425715556353,4556004503093247,

%T -194971932801554579457,33370662957719457037287423,

%U -22845215336421444625717664940033,62557106610069521429900219032249827327,-685195337175488637158242110253091749621661697

%N Column 0 of the matrix inverse of triangle A118185(n,k) = (4^k)^(n-k).

%C The entire matrix inverse of triangle A118185 is determined by column 0 (this sequence): [A118185^-1](n,k) = a(n-k)*4^(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/(1-4^n*x) by applying the inverse transformation: c(n) = Sum_{k=0..n} a(n-k)*b(k)*4^(k*(n-k)).

%H G. C. Greubel, <a href="/A118188/b118188.txt">Table of n, a(n) for n = 0..55</a>

%F G.f.: 1 = Sum_{n>=0} a(n)*x^n/(1-4^n*x).

%F 0^n = Sum_{k=0..n} a(k)*4^(k*(n-k)) for n>=0.

%F G.f.: Sum_{n>=0} a(n)*x^n/2^(n^2) = 1/Sum_{n>=0} x^n/2^(n^2). - _Vladeta Jovovic_, Oct 14 2009

%F a(n) = (-1)*Sum_{j=0..n-1} 4^(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)*(4^0)^4 +a(1)*(4^1)^3 +a(2)*(4^2)^2 +a(3)*(4^3)^1 +a(4)*(4^4)^0

%e = 1*(4^0) - 1*(4^3) + 3*(4^4) - 33*(4^3) + 1407*(4^0).

%e The g.f. is illustrated by:

%e 1 = 1/(1-x) - 1*x/(1-4*x) + 3*x^2/(1-16*x) - 33*x^3/(1-64*x) +

%e 1407*x^4/(1-256*x) - 237057*x^5/(1-1024*x) + 158992383*x^6/(1-4096*x) +...

%t a[n_]:= a[n]= If[n<2, (-1)^n, -Sum[4^(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,(4^(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(4^(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. A118185 (triangle).

%K sign

%O 0,3

%A _Paul D. Hanna_, Apr 15 2006