A352700 G.f.: Sum_{n>=0} binomial((n+1)*(2*n+1),n)/(2*n+1) * x^n / C(x)^(n*(2*n+1)+1), where C(x) = 1 + x*C(x)^2 is the Catalan function (A000108).

1, 1, 12, 239, 7178, 296092, 15666162, 1014796995, 77899495174, 6919858148750, 698584345392968, 79022119891573410, 9902447587480555624, 1361894352334815968554, 203969111022547680433454, 33047362680815865252524643, 5759708920548423261284008230

Offset

0,3

Comments

Compare g.f. to: 1 = Sum_{n>=0} binomial((k+1)*(2*n+1),n)/(2*n+1) * x^n / C(x)^(k*(2*n+1)+1) which holds for fixed k, where C(x) = 1 + x*C(x)^2 is the Catalan function (A000108).

The g.f. A(x) seems to satisfy A(x)^3 = A(x^3) (mod 3); compare this to the congruence: C(x)^3 = C(x^3) (mod 3), where C(x) is the Catalan function.

Odd terms seem to occur only at positions 2^n-1 for n >= 0.

Conjectures: given g.f. A(x), let C(x) = (1 - sqrt(1-4*x))/(2*x) be the Catalan power series (A000108), then

(1) A(x)^3 = A(x^3) (mod 3),

(2) A(x) = C(x) + x^2*C(x)^3 (mod 3) = (2 - x)*C(x) - 1 (mod 3),

(3) A(x) = C(x) (mod 2),

(4) a(n) = binomial(2*n+1,n)/(2*n+1) + 3*binomial(2*n-1,n-2)/(2*n-1) (mod 3) for n >= 0,

(5) a(n) = 2*A000108(n) - A000108(n-1) (mod 3) for n >= 1,

(6) a(n) = A000108(n) (mod 2) for n >= 0.

Links

Table of n, a(n) for n=0..16.

Formula

G.f. A(x) satisfies:

(1) A(x-x^2) = Sum_{n>=0} binomial((n+1)*(2*n+1),n)/(2*n+1) * (x/(1-x))^n * (1-x)^((n+1)*(2*n+1)).

(2) A(x/(1+x)^2) = Sum_{n>=0} binomial((n+1)*(2*n+1),n)/(2*n+1) * x^n / (1+x)^((n+1)*(2*n+1)).

Example

G.f.: A(x) = 1 + x + 12*x^2 + 239*x^3 + 7178*x^4 + 296092*x^5 + 15666162*x^6 + 1014796995*x^7 + 77899495174*x^8 + 6919858148750*x^9 + ...
where
A(x) = 1/C(x) + 2*x/C(x)^4 + 21*x^2/C(x)^11 + 468*x^3/C(x)^22 + 16555*x^4/C(x)^37 + 812448*x^5/C(x)^56 + 51274146*x^6/C(x)^79 + 3965837928*x^7/C(x)^106 + ... + (n+1)*A299429(n)*x^n/C(x)^(n*(2*n+1)+1) + ...
and
C(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 + 429*x^7 + 1430*x^8 + 4862*x^9 + ... + A000108(n)*x^n + ...
Congruence modulo 3.
(1) It appears that A(x)^3 is congruent to A(x^3) modulo 3, where
A(x)^3 = 1 + 3*x + 39*x^2 + 790*x^3 + 23436*x^4 + 949701*x^5 + 49503687*x^6 + 3171679536*x^7 + 241578165750*x^8 + 21340270771814*x^9 + ...
and
(A(x)^3 - A(x^3))/3 = x + 13*x^2 + 263*x^3 + 7812*x^4 + 316567*x^5 + 16501225*x^6 + 1057226512*x^7 + 80526055250*x^8 + 7113423590525*x^9 + ...
(2) Also, g.f. A(x) seems to be congruent to C(x) + x^2*C(x)^3, where
C(x) + x^2*C(x)^3 = 1 + x + 3*x^2 + 8*x^3 + 23*x^4 + 70*x^5 + 222*x^6 + 726*x^7 + 2431*x^8 + 8294*x^9 + ... + (C(2*n,n)/(n+1) + C(2*n-1,n-2)*3/(2*n-1))*x^n + ...

Prog

(PARI) {a(n) = my(C = (1 - sqrt(1-4*x +O(x^(n+3))))/(2*x),
A = sum(m=0, n, binomial((m+1)*(2*m+1), m)/(2*m+1) * x^m/C^(m*(2*m+1)+1))); polcoeff(A, n)}
for(n=0, 20, print1(a(n), ", "))

Crossrefs

Cf. A298696, A299429, A000108, A000782.

Sequence in context: A280071 A279610 A222702 * A012351 A189883 A033469

Adjacent sequences: A352697 A352698 A352699 * A352701 A352702 A352703

Keyword

nonn

Author

Paul D. Hanna, Mar 29 2022

Status

approved