A172405 G.f. satisfies: A(x) = G(x*A(x)) where G(x) = A(x/G(x)) = Sum_{n>=0} x^(n(n+1)/2)*(1+x)^n.

1, 1, 2, 5, 15, 47, 153, 515, 1782, 6293, 22576, 82043, 301417, 1117693, 4177687, 15723545, 59538258, 226656336, 866983080, 3330496250, 12843380569, 49700905874, 192942347560, 751191150163, 2932439491143, 11475503589091

Offset

0,3

Links

Table of n, a(n) for n=0..25.

Formula

G.f.: A(x) = Sum_{n>=0} [x*A(x)]^(n(n+1)/2) * (1 + x*A(x))^n.

Let q = x*A(x), then g.f. A(x) satisfies:

(1) A(x) = Sum_{n>=0} (1+q)^n*q^n*Product_{k=1..n} (1 - (1+q)*q^(2k-1))/(1 - (1+q)*q^(2k)),

(2) A(x) = 1/(1- q*(1+q)/(1+ q*(1-q)*(1+q)/(1- q^3*(1+q)/(1+ q^2*(1-q^2)*(1+q)/(1- q^5*(1+q)/(1+ q^3*(1-q^3)*(1+q)/(1- q^7*(1+q)/(1+ q^4*(1-q^4)*(1+q)/(1-...))))))))) (continued fraction).

These formulas are due to (1) a q-series identity and (2) a partial elliptic theta function expression.

Example

G.f.: A(x) = 1 + x + 2*x^2 + 5*x^3 + 15*x^4 + 47*x^5 + 153*x^6 +...
where G(x) = A(x/G(x)) = Sum_{n>=0} x^(n(n+1)/2)*(1+x)^n is illustrated by:
G(x) = 1 + x*(1+x) + x^3*(1+x)^2 + x^6*(1+x)^3 + x^10*(1+x)^4 +...
which has the binomial coefficients of the flattened Pascal's triangle:
G(x) = 1 + (x + x^2) + (x^3 + 2*x^4 + x^5) + (x^6 + 3*x^7 + 3*x^8 + x^9) +...
G.f. A(x) = G(x*A(x)) satisfies:
A(x) = 1 + x*A(x)*(1 + x*A(x)) + x^3*A(x)^3*(1 + x*A(x))^2 + x^6*A(x)^6*(1 + x*A(x))^3 + x^10*A(x)^10*(1 + x*A(x))^4 +...

Prog

(PARI) {a(n)=local(G=sum(m=0, (sqrtint(8*n+1)+1)\2, x^(m*(m+1)/2)*(1+x)^m)+x*O(x^n)); polcoeff(G^(n+1)/(n+1), n)}
(PARI) /* Using a q-series expansion: */
{a(n)=local(A=1+x*O(x^n)); for(i=1, n, q=x*A; A=sum(m=0, n, (1+q)^m*q^m*prod(k=1, m, (1-(1+q)*q^(2*k-1))/(1-(1+q)*(q+x*O(x^n))^(2*k))))); polcoeff(A, n)}

Crossrefs

Sequence in context: A350434 A364161 A148364 * A149915 A149916 A149917

Adjacent sequences: A172402 A172403 A172404 * A172406 A172407 A172408

Keyword

nonn

Author

Paul D. Hanna, Feb 06 2010

Status

approved