A185522 a(n) equals the coefficient of x^n in the (n-1)-th iteration of x*(1+x)/(1-x) for n>=1.

1, 2, 12, 138, 2320, 51450, 1418004, 46736466, 1793145792, 78506270994, 3862498271324, 210975923301242, 12668208032568400, 829409050807729002, 58804315058897866020, 4488388292080413635362, 366956820758560590789376

Offset

1,2

Links

Table of n, a(n) for n=1..17.

Example

 Given G(x) = x*(1+x)/(1-x):
G(x) = x + 2*x^2 + 2*x^3 + 2*x^4 + 2*x^5 + 2*x^6 +...
then the initial coefficients of the n-th iterations of G(x) begin:
n=0: [(1), 0, 0, 0, 0, 0, 0, ...];
n=1: [1,(2), 2, 2, 2, 2, 2, 2, 2, ...];
n=2: [1, 4,(12), 32, 80, 196, 476, 1152, 2784, ...];
n=3: [1, 6, 30,(138), 602, 2542, 10518, 42994, ...];
n=4: [1, 8, 56, 368,(2320), 14216, 85368, 505312, ...];
n=5: [1, 10, 90, 770, 6370,(51450), 408202, 3194978, ...];
n=6: [1, 12, 132, 1392, 14272, 143372,(1418004), 13854368, ...];
n=7: [1, 14, 182, 2282, 27930, 335846, 3983518,(46736466), ...]; ...;
the coefficients in parenthesis form the initial terms of this sequence.

Prog

 (PARI) {a(n)=local(A=x, G=x*(1+x)/(1-x+x*O(x^n))); for(i=1, n-1, A=subst(G, x, A+x*O(x^n))); polcoeff(A, n)}

Crossrefs

Cf. A185523, A185524.

Sequence in context: A117513 A343439 A297078 * A119819 A330655 A093543

Adjacent sequences: A185519 A185520 A185521 * A185523 A185524 A185525

Keyword

nonn

Author

Paul D. Hanna, Jan 30 2011

Status

approved