A100076 E.g.f. A(x) satisfies: Sum_{k=0..n} (A(x)^n)_k/k! = [sqrt(5)^n] for all n>=0, where (A(x)^n)_k/k! is the coefficient of x^k in A(x)^n.

1, 1, 1, -3, 9, -33, 513, -10917, 155313, -869697, -27095391, 1126973331, -25370851671, 400873570911, -3945969886815, -19472448499317, 3355787673885537, -205870807636111233, 10635145244261722305, -447262563680813504349, 13896854240554592685081

Offset

0,4

Comments

See triangle A100075 of initial coefficients of successive powers of the e.g.f. for this sequence.

Links

Table of n, a(n) for n=0..20.

Example

List the coefficients of powers of e.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!:
A(x)^0: [1,__0,0,0,0,0,0,0,0,...],
A(x)^1: [1,1,__1,-3,9,-33,513,-10917,155313,...],
A(x)^2: [1,2,4,__0,0,-36,1080,-17928,181440,...],
A(x)^3: [1,3,9,15,__9,-99,1521,-17631,112401,...],
A(x)^4: [1,4,16,48,96,__-72,1296,-11664,31104,...],
A(x)^5: [1,5,25,105,345,555,__1305,-6705,-6255,...],
A(x)^6: [1,6,36,192,864,2772,6408,__648,-15552,...],...
then for each row n, Sum_{k=0..n} (A(x)^n)_k/k! = [sqrt(5)^n]:
[sqrt(5)^0] = 1 = 1
[sqrt(5)^1] = 1+1 = 2
[sqrt(5)^2] = 1+2+4/2! = 5
[sqrt(5)^3] = 1+3+9/2!+15/3! = 11
[sqrt(5)^4] = 1+4+16/2!+48/3!+96/4! = 25
[sqrt(5)^5] = 1+5+25/2!+105/3!+345/4!+555/5! = 55
[sqrt(5)^6] = 1+6+36/2!+192/3!+864/4!+2772/5!+6408/6! = 125

Prog

(PARI) {a(n)=local(A, C, F, G); if(n==0, A=1, F=sum(k=0, n-1, a(k)*x^k/k!); C=floor(sqrt(5)^n+1/10^15)-sum(k=0, n-1, polcoeff(F^n+x*O(x^k), k, x)); G=sum(k=0, n-1, polcoeff(F^n+x*O(x^k), k, x)*x^k); A=n!*polcoeff((G+C*x^n)^(1/n)+x*O(x^n), n, x)); A}

Crossrefs

Cf. A100075, A100064.

Sequence in context: A264237 A097677 A138769 * A213907 A377145 A148999

Adjacent sequences: A100073 A100074 A100075 * A100077 A100078 A100079

Keyword

sign

Author

Paul D. Hanna, Nov 03 2004

Status

approved