OFFSET
0,2
COMMENTS
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..300
FORMULA
Conjectures: g.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
(1) A(x) = 1 + x^3*A(x)^2 (mod 3),
(2) A(x) = C(x^3) (mod 3) = C(x)^3 (mod 3), where C(x) = 1 + x*C(x)^2 is the Catalan function (A000108).
a(n) ~ c * 2^(3*n) * (n-1)! / (-LambertW(-2*exp(-2)) * (2 + LambertW(-2*exp(-2))))^n, where c = 0.87591174815917817179... - Vaclav Kotesovec, Mar 19 2022
EXAMPLE
G.f.: A(x) = 1 + 3*x + 45*x^2 + 1267*x^3 + 51597*x^4 + 2761539*x^5 + 182885885*x^6 + 14415019395*x^7 + 1316237331069*x^8 + ...
Related table.
The table of coefficients of x^k in (1+x - x^2*A(x))^(n*(2*n+1)) begins:
n=0: [1, 0, 0, 0, 0, 0, 0, ...];
n=1: [1, 3, 0, -14, -153, -4059, -162214, ...];
n=2: [1, 10, 35, 0, -825, -17758, -642015, ...];
n=3: [1, 21, 189, 847, 0, -55818, -1835218, ...];
n=4: [1, 36, 594, 5772, 32715, 0, -4524660, ...];
n=5: [1, 55, 1430, 23100, 252450, 1762706, 0, ...];
n=6: [1, 78, 2925, 69836, 1179672, 14597856, 122423756, 0, ...];
...
in which a diagonal equals all zeros, illustrating that
[x^(n+1)] (1+x - x^2*A(x))^(n*(2*n+1)) = 0, for n >= 0.
Congruence modulo 3:
(1) The terms of this sequence appear to be divisible by 3 when the index is not divisible by 3:
a(3*n+k) = 0 (mod 3) for n >= 0 and k = 1 or 2.
(2) For the terms a(3*n), the residues modulo 3 begin:
a(3*n) (mod 3) = [1, 1, 2, 2, 2, 0, 0, 0, 2, 2, 2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 1, 1, 1, 0, ...],
which appears to be congruent to the Catalan sequence A000108 modulo 3; i.e.,
a(3*n) = binomial(2*n,n)/(n+1) (mod 3), for n >= 0.
The above conjectures have been verified for the initial 1201 terms of this sequence.
PROG
(PARI) {a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A;
A[#A] = polcoeff( (1+x - x^2*Ser(A))^(m*(2*m+1)) / (m*(2*m+1)) , m+1); ); A[n+1]}
for(n=0, 10, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Mar 15 2022
STATUS
approved