A118196 Column 0 of the matrix inverse of triangle A117401(n,k) = (2^k)^(n-k).

1, -1, 1, -1, -1, 31, -449, 4223, 377087, -79232513, 13509592063, -74458331137, -5113818652864513, 11766261105083555839, -22128578595003966668801, -88147548436159218430476289, 3787186430286106428653327941631, -103331603080469761480767867413463041

Offset: 0

Author

Paul D. Hanna, Apr 15 2006

Keywords

• sign

Comments

The entire matrix inverse of triangle A117401 is determined by column 0 (this sequence): [A117401^-1](n,k) = a(n-k)*2^(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/(1-2^n*x) by applying the inverse transformation: c(n) = Sum_{k=0..n} a(n-k)*b(k)*2^(k*(n-k)).

Examples

			Recurrence at n=4:
0 = a(0)*(2^0)^4 +a(1)*(2^1)^3 +a(2)*(2^2)^2 +a(3)*(2^3)^1 +a(4)*(2^4)^0 = 1*(2^0) - 1*(2^3) + 1*(2^4) - 1*(2^3) - 1*(2^0).
The g.f. is illustrated by:
1 = 1/(1-x) -1*x/(1-2*x) +1*x^2/(1-4*x) -1*x^3/(1-8*x) -1*x^4/(1-16*x) +31*x^5/(1-32*x) -449*x^6/(1-64*x) + 4223*x^7/(1-128*x) +...
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..75

Crossrefs

Cf. A117401, A118197, A335330, A006125.

Programs

Mathematica

(* First program *)
m = 17;
M = Table[If[k <= n, 2^((n-k)k), 0], {n, 0, m}, {k, 0, m}];
Inverse[M][[All, 1]] (* Jean-François Alcover, Jun 13 2019 *)
(* Second program *)
a[n_]:= a[n]= If[n<2, (-1)^n, -Sum[2^(j*(n-j))*a[j], {j, 0, n-1}]];
Table[a[n], {n, 0, 30}] (* G. C. Greubel, Jun 30 2021 *)

PARI

{a(n) = local(T=matrix(n+1,n+1,r,c,if(r>=c,(2^(c-1))^(r-c)))); return((T^-1)[n+1,1])};

Sage

def A118196_list(len):
    R, C = [1], [1]+[0]*(len-1)
    for n in (1..len-1):
        for k in range(n, 0, -1):
            C[k] = C[k-1] / (2^k)
        C[0] = -sum(C[k] for k in (1..n))
        R.append(C[0]*2^(n*(n+1)/2))
    return R
print(A118196_list(18)) # Peter Luschny, Feb 20 2016

Sage

@CachedFunction
def a(n): return (-1)^n if (n<2) else -sum(2^(j*(n-j))*a(j) for j in (0..n-1))
[a(n) for n in (0..30)] # G. C. Greubel, Jun 30 2021

Formula

G.f.: 1 = Sum_{n>=0} a(n)*x^n/(1-2^n*x).

0^n = Sum_{k=0..n} a(k)*2^(k*(n-k)) for n>=0.

a(n) = (-1)*Sum_{j=0..n-1} 2^(j*(n-j))*a(j) with a(0) = 1 and a(1) = -1. - G. C. Greubel, Jun 30 2021

From Geoffrey Critzer, May 08 2023: (Start)

a(n) = Sum_{k=0..n} (-1)^k*A335330(n,k).

Sum_{n>=0} a(n)*z^n/A006125(n) = 1/E(z) where E(z) = Sum_{n>=0} z^n/A006125(n). (End)