OFFSET
1,2
COMMENTS
It appears that a(n) = 1 (mod 2) iff n = 2*4^k - 1 for k>=0.
Radius of convergence is 1/4 with A(1/4) = 2*(sqrt(2) - 1).
LINKS
Paul D. Hanna, Table of n, a(n) for n = 1..520
FORMULA
G.f. A(x) satisfies:
(1) A(x) = (sqrt(1 + 4*C(x)^2) - 1)/(2*x), where C(x) = (1 - sqrt(1 - 4*x))/2.
(2) A(x) = (1/x) * Series_Reversion( sqrt(x^2 + x^4) - (x^2 + x^4) )^2.
a(n) ~ 2^(2*n - 1/2) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Jun 23 2016
EXAMPLE
G.f.: A(x) = x + 2*x^2 + 4*x^3 + 10*x^4 + 30*x^5 + 96*x^6 + 313*x^7 + 1038*x^8 + 3512*x^9 + 12100*x^10 + 42286*x^11 + 149440*x^12 + +...
such that x*A(x) + x^2*A(x)^2 = C(x)^2, where
A(x)^2 = x^2 + 4*x^3 + 12*x^4 + 36*x^5 + 116*x^6 + 392*x^7 + 1350*x^8 + 4696*x^9 + 16500*x^10 + 58572*x^11 + 209824*x^12 +...+ A055395(n+2)*x^n +...
C(x)^2 = x^2 + 2*x^3 + 5*x^4 + 14*x^5 + 42*x^6 + 132*x^7 + 429*x^8 + 1430*x^9 + 4862*x^10 + 16796*x^11 +...+ A000108(n)*x^n +...
PROG
(PARI) /* From A(x) = (sqrt(1 + 4*C(x)^2) - 1)/(2*x) */
{a(n) = my(A=x, C=(1 - sqrt(1 - 4*x +x*O(x^n)))/2); A = (sqrt(1 + 4*C^2) - 1)/(2*x); polcoeff(A, n)}
for(n=1, 40, print1(a(n), ", "))
(PARI) /* From A(x) = (1/x)*Series_Reversion( sqrt(x^2 + x^4) - (x^2 + x^4) )^2 */
{a(n) = my(A=x); A = ( serreverse( sqrt(x^2 + x^4 +x^3*O(x^n)) - (x^2 + x^4) ) )^2/x; polcoeff(A, n)}
for(n=1, 40, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jun 10 2016
STATUS
approved