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A255673 Coefficients of A(x), which satisfies: A(x) = 1 + x*A(x)^3 + x^2*A(x)^6.

%I A255673 #59 Jan 11 2025 03:28:30
%S A255673 1,1,4,21,127,833,5763,41401,305877,2309385,17739561,138197876,
%T A255673 1089276972,8670856834,69606939717,562879492551,4580890678781,
%U A255673 37490975387565,308369889858450,2547741413147700,21133987935358776,175947462569886786,1469656053534121804
%N A255673 Coefficients of A(x), which satisfies: A(x) = 1 + x*A(x)^3 + x^2*A(x)^6.
%C A255673 This sequence is the next after A001006 and A006605.
%H A255673 Alois P. Heinz, <a href="/A255673/b255673.txt">Table of n, a(n) for n = 0..1000</a>
%H A255673 Jun Yan, <a href="https://arxiv.org/abs/2501.01152">Lattice paths enumerations weighted by ascent lengths</a>, arXiv:2501.01152 [math.CO], 2025. See p. 13.
%F A255673 a(n+1) = Sum_{j=0..3*n+4} binomial(j,2*j-5*n-7) * binomial(3*n+4,j) / (3*n+4). (conjectured). [_Vladimir Kruchinin_, Mar 09 2013]
%F A255673 a(n) = 1/(3*n+1) * Sum_{k=0..n} (-1)^k * binomial(3*n+1, k) * binomial(6*n+2-2*k, n-k). (conjectured)
%F A255673 G.f. A(x) satisfies A(x) = G(x*A(x)), where G is g.f. of A006605.
%F A255673 G.f. A(x) satisfies A(x) = H(x*A(x)^2), where H is g.f. of A001006.
%F A255673 From _Peter Bala_, Jul 27 2023: (Start)
%F A255673 Define b(n) = [x^n] (1 + x + x^2)^(3*n). Then A(x)^3 = exp(Sum_{n >= 1} b(n)*x^n/n).
%F A255673 A(x^3) = (1/x) * series reversion of x/(1 + x^3 + x^6) = 1 + x^3 + 4*x^6 + 21*x^9 + 127*x^12 + .... (End)
%F A255673 a(n) = (1/(3*n+1)) * Sum_{k=0..floor(n/2)} binomial(n-k,k) * binomial(3*n+1,n-k). - _Seiichi Manyama_, Sep 02 2023
%e A255673 A(x) = 1 + x + 4*x^2 + 21*x^3 + 127*x^4 + 833*x^5 + 5763*x^6 ...
%p A255673 a:= n-> coeff(series(RootOf(1-A+x*A^3+x^2*A^6, A), x, n+1), x, n):
%p A255673 seq(a(n), n=0..30);  # _Alois P. Heinz_, Jul 15 2015
%p A255673 # second Maple program:
%p A255673 a:= proc(n) option remember; `if`(n<2, 1, 9*(((3*n-1))*
%p A255673      (2*n-1)*(3*n-2)*(9063*n^4-18126*n^3+8403*n^2+660*n-280)*a(n-1)
%p A255673      +(27*(n-1))*(3*n-1)*(3*n-4)*(3*n-2)*(3*n-5)*(57*n^2-2)*a(n-2))
%p A255673       /((5*(5*n+2))*(5*n-1)*(5*n+1)*(5*n-2)*n*(57*n^2-114*n+55)))
%p A255673     end:
%p A255673 seq(a(n), n=0..30);  # _Alois P. Heinz_, Jul 16 2015
%t A255673 m = 30; A[_] = 0;
%t A255673 Do[A[x_] = 1 + x A[x]^3 + x^2 A[x]^6 + O[x]^m, {m}];
%t A255673 CoefficientList[A[x], x] (* _Jean-François Alcover_, Oct 04 2019 *)
%o A255673 (PARI) a(n) = sum(k=0, n\2, binomial(n-k, k)*binomial(3*n+1, n-k))/(3*n+1); \\ _Seiichi Manyama_, Sep 02 2023
%Y A255673 Cf. A001006, A006605.
%Y A255673 Cf. A365183, A365189.
%K A255673 nonn
%O A255673 0,3
%A A255673 _Werner Schulte_, Jul 10 2015
%E A255673 More terms from _Alois P. Heinz_, Jul 15 2015