OFFSET
0,2
COMMENTS
Radius of convergence of g.f. A(x) is r = 1/27 where A(r) = sqrt(3/2).
FORMULA
a(n) = 4^n * binomial((6*n - 1)/2, n) / (4*n + 1).
G.f. A(x) satisfies A(x) = sqrt(1 + 4*x^2*A(x)^10) + 2*x*A(x)^5.
G.f.: A(x) = sqrt(D(4*x)), where D(x) is the g.f. of A001764. - Werner Schulte, Aug 10 2015
From Karol A. Penson, Mar 19 2024: (Start)
a(n) = 4^n*binomial(3*n+1/2,n)/(6*n+1).
G.f.: 3F2([1/6, 1/2, 5/6], [3/4, 5/4], 27*x).
G.f.: sqrt(2)*sqrt((-(sqrt(1 - 27*x) + 3*i*sqrt(3)*sqrt(x))^(1/3) + (sqrt(1 - 27*x) - 3*i*sqrt(3)*sqrt(x))^(1/3))*i)*3^(3/4)/(6*x^(1/4)), where i is the imaginary unit.
a(n) = Integral_{x=0...27} x^n*W(x), where W(x) = h1(x) + h2(x) + h3(x) and
h1(x) = 2^(2/3)*3F2([-1/12, 1/6, 5/12], [1/3, 2/3], x/27)/(4*Pi*x^(5/6));
h2(x) = -3F2([1/4, 1/2, 3/4], [2/3, 4/3], x/27)/(12*Pi*sqrt(x));
h3(x) = -2^(1/3)*3F2([7/12, 5/6, 13/12], [4/3, 5/3], x/27)/(576*Pi*x^(1/6)).
This integral representation is unique as W(x) is the solution of the Hausdorff power moment problem on x = (0, 27). Using only the definition of a(n), W(x) can be proven to be positive. W(x) is singular at x = 0 and for x > 0 is monotonically decreasing to zero at x = 27. For x -> 27, W'(x) tends to -infinity. (End)
EXAMPLE
G.f.: A(x) = 1 + 2*x + 22*x^2 + 340*x^3 + 6118*x^4 + 120060*x^5 + ...
where A(x)^2 = 1 + 4*x*A(x)^6:
A(x)^2 = 1 + 4*x + 48*x^2 + 768*x^3 + 14080*x^4 + 279552*x^5 + ...
A(x)^6 = 1 + 12*x + 192*x^2 + 3520*x^3 + 69888*x^4 + 1462272*x^5 + ...
Related series:
A(x)^5 = 1 + 10*x + 150*x^2 + 2660*x^3 + 51750*x^4 + 1068012*x^5 + ...
A(x)^10 = 1 + 20*x + 400*x^2 + 8320*x^3 + 179200*x^4 + 3969024*x^5 + ...
where A(x) = sqrt(1 + 4*x^2*A(x)^10) + 2*x*A(x)^5.
MAPLE
A245113:=n->4^n*binomial((6*n-1)/2, n)/(4*n+1): seq(A245113(n), n=0..30); # Wesley Ivan Hurt, Aug 11 2015
MATHEMATICA
Table[4^n*Binomial[(6 n - 1)/2, n]/(4 n + 1), {n, 0, 20}] (* Wesley Ivan Hurt, Aug 11 2015 *)
PROG
(PARI) /* From A(x)^2 = 1 + 4*x*A(x)^6 : */
{a(n) = local(A=1+x); for(i=1, n, A=sqrt(1 + 4*x*A^6 +x*O(x^n))); polcoeff(A, n)}
for(n=0, 20, print1(a(n), ", "))
(PARI) {a(n) = 4^n * binomial((6*n - 1)/2, n) / (4*n + 1)}
for(n=0, 20, print1(a(n), ", "))
(PARI) /* From A(x) = sqrt(1 + 4*x^2*A(x)^10) + 2*x*A(x)^5 : */
{a(n) = local(A=1+x); for(i=1, n, A = sqrt(1 + 4*x^2*A^10 +x*O(x^n)) + 2*x*A^5); polcoeff(A, n)}
for(n=0, 20, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul D. Hanna, Jul 31 2014
STATUS
approved