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A352409 G.f. A(x) satisfies: [x^(n+1)] (1+x - x^2*A(x))^(n*(2*n+1)) = 0, for n >= 0.

%I A352409 #15 Mar 26 2022 15:11:44
%S A352409 1,3,45,1267,51597,2761539,182885885,14415019395,1316237331069,
%T A352409 136512958750979,15842506286290173,2033176597680449283,
%U A352409 285833727841312233725,43677225803116362273795,7207197437612731825348605,1277141936892060488486787075
%N A352409 G.f. A(x) satisfies: [x^(n+1)] (1+x - x^2*A(x))^(n*(2*n+1)) = 0, for n >= 0.
%C A352409 Conjectures:
%C A352409 (1) a(3*n) = A039969(n) (mod 3) = A000108(n) (mod 3) for n >= 0, where A000108 is the Catalan sequence.
%C A352409 (2) a(3*n+1) = 0 (mod 3) for n >= 0.
%C A352409 (3) a(3*n+2) = 0 (mod 3) for n >= 0.
%C A352409 (4) a(n) = 1 (mod 2) for n >= 0.
%H A352409 Paul D. Hanna, <a href="/A352409/b352409.txt">Table of n, a(n) for n = 0..300</a>
%F A352409 Conjectures: g.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
%F A352409 (1) A(x) = 1 + x^3*A(x)^2 (mod 3),
%F A352409 (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).
%F A352409 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
%e A352409 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 + ...
%e A352409 Related table.
%e A352409 The table of coefficients of x^k in (1+x - x^2*A(x))^(n*(2*n+1)) begins:
%e A352409 n=0: [1,  0,    0,     0,       0,        0,         0, ...];
%e A352409 n=1: [1,  3,    0,   -14,    -153,    -4059,   -162214, ...];
%e A352409 n=2: [1, 10,   35,     0,    -825,   -17758,   -642015, ...];
%e A352409 n=3: [1, 21,  189,   847,       0,   -55818,  -1835218, ...];
%e A352409 n=4: [1, 36,  594,  5772,   32715,        0,  -4524660, ...];
%e A352409 n=5: [1, 55, 1430, 23100,  252450,  1762706,         0, ...];
%e A352409 n=6: [1, 78, 2925, 69836, 1179672, 14597856, 122423756, 0, ...];
%e A352409 ...
%e A352409 in which a diagonal equals all zeros, illustrating that
%e A352409 [x^(n+1)] (1+x - x^2*A(x))^(n*(2*n+1)) = 0, for n >= 0.
%e A352409 Congruence modulo 3:
%e A352409 (1) The terms of this sequence appear to be divisible by 3 when the index is not divisible by 3:
%e A352409 a(3*n+k) = 0 (mod 3) for n >= 0 and k = 1 or 2.
%e A352409 (2) For the terms a(3*n), the residues modulo 3 begin:
%e A352409 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, ...],
%e A352409 which appears to be congruent to the Catalan sequence A000108 modulo 3; i.e.,
%e A352409 a(3*n) = binomial(2*n,n)/(n+1) (mod 3), for n >= 0.
%e A352409 The above conjectures have been verified for the initial 1201 terms of this sequence.
%o A352409 (PARI) {a(n) = my(A=[1]); for(i=1,n, A=concat(A,0); m=#A;
%o A352409 A[#A] = polcoeff( (1+x - x^2*Ser(A))^(m*(2*m+1)) / (m*(2*m+1)) ,m+1););A[n+1]}
%o A352409 for(n=0,10,print1(a(n),", "))
%Y A352409 Cf. A352408, A039969, A000108.
%K A352409 nonn
%O A352409 0,2
%A A352409 _Paul D. Hanna_, Mar 15 2022