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A118196 Column 0 of the matrix inverse of triangle A117401(n,k) = (2^k)^(n-k). 1
1, -1, 1, -1, -1, 31, -449, 4223, 377087, -79232513, 13509592063, -74458331137, -5113818652864513, 11766261105083555839, -22128578595003966668801, -88147548436159218430476289, 3787186430286106428653327941631, -103331603080469761480767867413463041 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,6

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).

LINKS

Table of n, a(n) for n=0..17.

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.

EXAMPLE

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) +...

MAPLE

{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])}

PROG

(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

CROSSREFS

Cf. A117401, A118197.

Sequence in context: A084234 A187622 A187630 * A160443 A125466 A142280

Adjacent sequences:  A118193 A118194 A118195 * A118197 A118198 A118199

KEYWORD

sign

AUTHOR

Paul D. Hanna, Apr 15 2006

STATUS

approved

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Last modified April 24 22:26 EDT 2019. Contains 322446 sequences. (Running on oeis4.)