A352408 a(n) = [x^(n+1)] (1+x - x^2*C(x^3))^(n*(2*n+1)) / (n*(2*n+1)), for n >= 1, where C(x) = 1 + x*C(x)^2 is the Catalan function (A000108).

0, 3, 105, 4521, 247509, 16743276, 1358620365, 129050741220, 14072813590533, 1734613925513373, 238643675398832787, 36267490290699148842, 6035968171216586764461, 1092097033048311239037276, 213473299061006718358241931, 44838030511923231603476358240

Offset: 1

Author

Paul D. Hanna, Mar 16 2022

Keywords

• nonn

Comments

This sequence explores yet another property of the Catalan sequence (A000108).

Conjecture: all terms are divisible by 3. Showing that a(n) = 0 (mod 3) for n >= 1 would consequently prove conjectures involving the g.f. G(x) of A352409, namely: G(x) = 1 + x^3*G(x)^2 (mod 3) and G(x) = C(x^3) (mod 3), where C(x) = 1 + x*C(x)^2.

Note: C(x^3) = C(x)^3 (mod 3), thus 1+x - x^2*C(x^3) = 1+x - x^2*C(x)^3 (mod 3) = 1+x + x*(C(x) - C(x)^2) (mod 3) = 2+x - (1-x)*C(x) (mod 3), where C(x) = 1 + x*C(x)^2.

Examples

			Given C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the Catalan function of A000108, which starts 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 + ... + binomial(2*n,n)/(n+1) * x^n + ...
then
1+x - x^2*C(x^3) = 1 + x - x^2 - x^5 - 2*x^8 - 5*x^11 - 14*x^14 - 42*x^17 - 132*x^20 - 429*x^23 - 1430*x^26 - ...
and the table of coefficients of x^k in (1+x - x^2*C(x^3))^(n*(2*n+1)) begin:
n=1: [1,   3,    0,    -5,       0,        0,        -7,          3, ...];
n=2: [1,  10,   35,    30,    -105,     -238,         0,        270, ...];
n=3: [1,  21,  189,   910,    2205,      378,    -13321,     -33324, ...];
n=4: [1,  36,  594,  5880,   38115,   162756,    407862,     175068, ...];
n=5: [1,  55, 1430, 23265,  263835,  2193191,  13612995,   62337330, ...];
n=6: [1,  78, 2925, 70070, 1201200, 15633540, 159772613, 1305975528, ...];
...
from which the terms a(n) = [x^(n+1)] (1+x - x^2*C(x^3))^(n*(2*n+1))/(n*(2*n+1)) can be derived, for n >= 1, as illustrated by:
a(1) = [x^2] (1+x - x^2*C(x^3))^3 / 3 = 0 / 3 = 0;
a(2) = [x^3] (1+x - x^2*C(x^3))^10 / 10 = 30 / 10 = 3;
a(3) = [x^4] (1+x - x^2*C(x^3))^21 / 21 = 2205 / 21 = 105;
a(4) = [x^5] (1+x - x^2*C(x^3))^36 / 36 = 162756 / 36 = 4521;
a(5) = [x^6] (1+x - x^2*C(x^3))^55 / 55 = 13612995 / 55 = 247509;
a(6) = [x^7] (1+x - x^2*C(x^3))^78 / 78 = 1305975528 / 78 = 16743276;
...
Apparently, all terms computed this way are divisible by 3 (verified for the initial 1201 terms).
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 1..300

Crossrefs

Cf. A352409.

Programs

PARI

{a(n) = my(C3 = (1 - sqrt(1 - 4*x^3 + O(x^(n+3)) ))/(2*x^3));
polcoeff( (1+x - x^2*C3)^(n*(2*n+1)) / (n*(2*n+1)), n+1)}
for(n=1,21, print1(a(n),", " ))

Formula

a(n) ~ 2^(n - 1/2) * exp(n - 1/4) * n^(n - 3/2) / sqrt(Pi). - Vaclav Kotesovec, Mar 18 2022