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A214553
G.f. satisfies: A(x) = 1 + 4*x*A(x)^(5/2).
4
1, 4, 40, 520, 7680, 122360, 2050048, 35600400, 635043840, 11566760920, 214221455360, 4021962900592, 76374500966400, 1464312851075760, 28307243610931200, 551140224522544160, 10797908842864705536, 212721273248318069400, 4211238736846158561280
OFFSET
0,2
COMMENTS
Radius of convergence of g.f. A(x) is r = (3/5)^(5/2) / 6 where A(r) = 5/3.
The two-parameter Fuss-Catalan sequence is A_n(p,r) := r*binomial(n*p + r, n)/(n*p + r). This sequence is 4^n*A_n(5/2,1). - Peter Bala, Oct 15 2015
LINKS
FORMULA
a(n) = 4^n * binomial(5*n/2, n) / (3*n/2 + 1).
From Peter Bala, Oct 13 2015: (Start)
O.g.f. A(x) satisfies A(x) = C(4*x*sqrt(A)), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108.
Sqrt(A(x)) = 1/x * series reversion( x/sqrt(C(4*x)) ) is the o.g.f. for A245112. (End)
EXAMPLE
G.f.: A(x) = 1 + 4*x + 40*x^2 + 520*x^3 + 7680*x^4 + 122360*x^5 + 2050048*x^6 +... where A(x) = 1 + 4*x*A(x)^(5/2).
Radius of convergence: r = (3/5)^(5/2)/6 = 0.046475800...
Related expansions:
A(x)^(5/2) = 1 + 10*x + 130*x^2 + 1920*x^3 + 30590*x^4 + 512512*x^5 +...
A(x)^(1/2) = 1 + 2*x + 18*x^2 + 224*x^3 + 3230*x^4 + 50688*x^5 + 840420*x^6 + 14483456*x^7 + 256856886*x^8 +...
MAPLE
seq(4^n*binomial(5*n/2, n)/(3*n/2+1), n=0..50); # Robert Israel, Oct 18 2015
MATHEMATICA
m = 19; A[_] = 0;
Do[A[x_] = 1 + 4*x*A[x]^(5/2) + O[x]^m, {m}];
CoefficientList[A[x], x] (* Jean-François Alcover, Oct 20 2019 *)
PROG
(PARI) {a(n)=4^n*binomial(5*n/2, n)/(3*n/2+1)}
(PARI) {a(n)=local(A=1+x); for(i=1, n, A =1+4*x*(A+x*O(x^n))^(5/2)); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul D. Hanna, Jul 20 2012
STATUS
approved