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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).
4
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
0,3
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.
LINKS
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.
EXAMPLE
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).
PROG
(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), ", "))
CROSSREFS
Cf. A127632, A153294, A033296, A000108 (C(x)), A000245 (x*C(x)^3).
Sequence in context: A081911 A081187 A182401 * A104498 A045379 A053487
KEYWORD
nonn
AUTHOR
Paul D. Hanna, May 28 2023
STATUS
approved