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A118188
Column 0 of the matrix inverse of triangle A118185(n,k) = (4^k)^(n-k).
3
1, -1, 3, -33, 1407, -237057, 158992383, -425715556353, 4556004503093247, -194971932801554579457, 33370662957719457037287423, -22845215336421444625717664940033, 62557106610069521429900219032249827327, -685195337175488637158242110253091749621661697
OFFSET
0,3
COMMENTS
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)).
LINKS
FORMULA
G.f.: 1 = Sum_{n>=0} a(n)*x^n/(1-4^n*x).
0^n = Sum_{k=0..n} a(k)*4^(k*(n-k)) for n>=0.
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
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
EXAMPLE
Recurrence at n=4:
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
= 1*(4^0) - 1*(4^3) + 3*(4^4) - 33*(4^3) + 1407*(4^0).
The g.f. is illustrated by:
1 = 1/(1-x) - 1*x/(1-4*x) + 3*x^2/(1-16*x) - 33*x^3/(1-64*x) +
1407*x^4/(1-256*x) - 237057*x^5/(1-1024*x) + 158992383*x^6/(1-4096*x) +...
MATHEMATICA
a[n_]:= a[n]= If[n<2, (-1)^n, -Sum[4^(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, (4^(c-1))^(r-c)))); return((T^-1)[n+1, 1])}
(Sage)
@CachedFunction
def a(n): return (-1)^n if (n<2) else -sum(4^(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
Cf. A118185 (triangle).
Sequence in context: A012487 A188387 A113111 * A342170 A194889 A126675
KEYWORD
sign
AUTHOR
Paul D. Hanna, Apr 15 2006
STATUS
approved