A363308 Expansion of g.f. C(x*C(x)^3), where C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).

1, 1, 5, 26, 141, 790, 4542, 26668, 159333, 966038, 5930678, 36801660, 230491410, 1455283172, 9253674120, 59209786992, 380961295445, 2463303690790, 15998687418030, 104325569140156, 682768883525830, 4483232450501492, 29527005540912660, 195006621974036808

Offset: 0

Author

Paul D. Hanna, May 28 2023

Keywords

• nonn

Comments

Compare the g.f. A(x) = C(x*C(x)^3) to the identity C(-x*C(x)^3) = 1/C(x), where C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).

Conjecture: a(n) is odd iff n is a power of 2 or n = 0.

Examples

			G.f.: A(x) = 1 + x + 5*x^2 + 26*x^3 + 141*x^4 + 790*x^5 + 4542*x^6 + 26668*x^7 + 159333*x^8 + 966038*x^9 + 5930678*x^10 + ...
such that A(x) = C(x*C(x)^3), where
C(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + ... + A000108(n)*x^n + ...
x*C(x)^3 = x + 3*x^2 + 9*x^3 + 28*x^4 + 90*x^5 + ... + A000245(n)*x^n + ...
Note that x*C(x)^3 = (C(x) - 1)*(1-x)/x - 1.
Also, the g.f. of related sequence A033296 begins
B(x) = 1 + x + 6*x^2 + 42*x^3 + 326*x^4 + 2706*x^5 + 23526*x^6 + ...
where A(x) = B(x/A(x)), B(x) = A(x*B(x)) = C(x*B(x)*C(x*B(x))^3).
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..400

Crossrefs

Cf. A127632, A153294, A033296, A000108 (C(x)), A000245 (x*C(x)^3).

Cf. A363309, A363111.

Programs

PARI

{a(n) = if(n==0,1, sum(k=1,n, 3*k* binomial(2*k+1,k) * binomial(2*n+k,n-k) / ((2*k+1)*(2*n+k)) ) )}
for(n=0, 30, print1(a(n), ", "))

PARI

/* G.f. A(x) = C(x*C(x)^3), where C(x) = 1 + x*C(x)^2 */
{a(n) = my(C = (1 - sqrt(1 - 4*x +x^2*O(x^n)))/(2*x)); polcoeff( subst(C, x, x*C^3), n)}
for(n=0, 30, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n may be defined as follows; here, C(x) is the g.f. of the Catalan numbers (A000108).

(1) A(x) = C(x*C(x)^3), where C(x) = (1 - sqrt(1-4*x))/(2*x).

(2) A(x) = B(x/A(x)) where B(x) = A(x*B(x)) = C( x*B(x) * C(x*B(x))^3 ) is the g.f. of A033296.

(3) a(n) = Sum_{k=1..n} 3*k* binomial(2*k+1,k) * binomial(2*n+k,n-k) / ((2*k+1)*(2*n+k)) for n > 0, with a(0) = 1.

D-finite with recurrence 4*n*(n-1)*(21687905*n [phone])*(n+1)*a(n) +2*n*(n-1) *(43375810*n^2 -6022811713*n [phone])*a(n-1) -(n-1) *(15429963345*n^3 -200018809315*n^2 [phone]*n -632905646028)*a(n-2) +(102558230760*n^4 -1409936457473*n^3 [phone]*n^2 -14414518702669*n [phone])*a(n-3) +(-212869593020*n^4 [phone]*n^3 -20069314453381*n^2 [phone]*n -52146873039204)*a(n-4) +4*(2*n-11) *(10773532140*n^3 -171459615587*n^2 [phone]*n -1572525214995)*a(n-5) +4*(n-6) *(208500820*n -955419151)*(2*n-11) *(2*n-13)*a(n-6)=0. - R. J. Mathar, Nov 22 2024