A100225 G.f. A(x) satisfies: 3^n - 1 = Sum_{k=0..n} [x^k]A(x)^n and also satisfies: (3+z)^n - (1+z)^n + z^n = Sum_{k=0..n} [x^k](A(x)+z*x)^n for all z, where [x^k]A(x)^n denotes the coefficient of x^k in A(x)^n.

1, 1, 2, 0, -4, 0, 16, 0, -80, 0, 448, 0, -2688, 0, 16896, 0, -109824, 0, 732160, 0, -4978688, 0, 34398208, 0, -240787456, 0, 1704034304, 0, -12171673600, 0, 87636049920, 0, -635361361920, 0, 4634400522240, 0, -33985603829760, 0, 250420238745600, 0, -1853109766717440

Offset

0,3

Comments

More generally, if g.f. A(x) satisfies: m^n-b^n = Sum_{k=0..n} [x^k]A(x)^n, then A(x) also satisfies: (m+z)^n - (b+z)^n + z^n = Sum_{k=0..n} [x^k](A(x)+z*x)^n for all z and A(x)=(1+(m-1)*x+sqrt(1+2*(m-2*b-1)*x+(m^2-2*m+4*b+1)*x^2))/2.

Links

Table of n, a(n) for n=0..40.

Formula

G.f.: (1+2*x+sqrt(1+8*x^2))/2. G.f.: A(x) = x/(series_reversion[x*(1-x)/(1-2*x-x^2)]). a(n) = -8*(n-3)*a(n-2)/n for n>2, with a(0)=1, a(1)=1, a(2)=2. a(2*n) = 2^n*(-1)^(n-1)*A000108(n-1), a(2*n+1)=0, for n>=1, where A000108=Catalan.

Example

From the table of powers of A(x) (A100226), we see that
3^n-1 = Sum of coefficients [x^0] through [x^n] in A(x)^n:
A^1=[1,1],2,0,-4,0,16,0,-80,...
A^2=[1,2,5],4,-4,-8,16,32,-80,...
A^3=[1,3,9,13],6,-12,-4,48,0,...
A^4=[1,4,14,28,33],8,-24,16,80,...
A^5=[1,5,20,50,85,81],10,-40,60,..
A^6=[1,6,27,80,171,246,197],12,-60,...
the main diagonal of which is A100227 = [1,5,13,33,81,197,477,...], where Sum_{n>=1} (A100227(n)/n)*x^n = log((1-x)/(1-2*x-x^2)).

Prog

(PARI) a(n)=if(n==0, 1, (3^n-1-sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j)^n+x*O(x^k), k)))/n)
(PARI) a(n)=if(n==0, 1, if(n==1, 1, if(n==2, 2, -8*(n-3)*a(n-2)/n)))
(PARI) a(n)=polcoeff((1+2*x+sqrt(1+8*x^2+x^2*O(x^n)))/2, n)

Crossrefs

Cf. A100226, A100227, A000108, A025225, A100223, A100228.

Sequence in context: A286776 A265829 A022896 * A007420 A362368 A019219

Adjacent sequences: A100222 A100223 A100224 * A100226 A100227 A100228

Keyword

sign

Author

Paul D. Hanna, Nov 28 2004

Status

approved