G.f.: (2/sqrt(3*x))*sin((1/3)*arcsin(sqrt(27*x/4))).
E.g.f.: hypergeom([1/3, 2/3], [1, 3/2], 27/4*x).
Integral representation as n-th moment of a positive function on [0, 27/4]: a(n) = Integral_{x=0..27/4} (x^n*((1/12) * 3^(1/2) * 2^(1/3) * (2^(1/3)*(27 + 3 * sqrt(81 - 12*x))^(2/3) - 6 * x^(1/3))/(Pi * x^(2/3)*(27 + 3 * sqrt(81 - 12*x))^(1/3)))), n >= 0. This representation is unique. (End)
G.f. A(x) satisfies A(x) = 1+x*A(x)^3 = 1/(1-x*A(x)^2) [Cyvin (1998)]. -
Ralf Stephan, Jun 30 2003
a(n) = n-th coefficient in expansion of power series P(n), where P(0) = 1, P(k+1) = 1/(1 - x*P(k)^2).
G.f. Rev(x/c(x))/x, where c(x) is the g.f. of
A000108 (Rev=reversion of). -
Paul Barry, Mar 26 2010
Let M = the production matrix:
1, 1
2, 2, 1
3, 3, 2, 1
4, 4, 3, 2, 1
5, 5, 4, 3, 2, 1
...
a(n) = upper left term in M^n. Top row terms of M^n = (n+1)-th row of triangle
A143603, with top row sums generating
A006013: (1, 2, 7, 30, 143, 728, ...). (End)
Recurrence: a(0)=1; a(n) = Sum_{i=0..n-1, j=0..n-1-i} a(i)a(j)a(n-1-i-j) for n >= 1 (counts ternary trees by subtrees of the root). -
David Callan, Nov 21 2011
G.f.: 1 + 6*x/(Q(0) - 6*x); Q(k) = 3*x*(3*k + 1)*(3*k + 2) + 2*(2*(k^2) + 5*k +3) - 6*x*(2*(k^2) + 5*k + 3)*(3*k + 4)*(3*k + 5)/Q(k+1); (continued fraction). -
Sergei N. Gladkovskii, Nov 27 2011
D-finite with recurrence: 2*n*(2n+1)*a(n) - 3*(3n-1)*(3n-2)*a(n-1) = 0. -
R. J. Mathar, Dec 14 2011
G.f.: F([2/3,4/3], [3/2], 27/4*x) / F([2/3,1/3], [1/2], (27/4)*x) where F() is the hypergeometric function. -
Joerg Arndt, Sep 01 2012
0 = a(n)*(-3188646*a(n+2) + 20312856*a(n+3) - 11379609*a(n+4) + 1437501*a(n+5)) + a(n+1)*(177147*a(n+2) - 2247831*a(n+3) + 1638648*a(n+4) - 238604*a(n+5)) + a(n+2)*(243*a(n+2) + 31497*a(n+3) - 43732*a(n+4) + 8288*a(n+5)) for all integer n. -
Michael Somos, Jun 03 2016
a(n) ~ 3^(3*n + 1/2)/(sqrt(Pi)*4^(n+1)*n^(3/2)). -
Ilya Gutkovskiy, Nov 21 2016
Given g.f. A(x), then A(1/8) = -1 + sqrt(5), A(2/27) = (-1 + sqrt(3))*3/2, A(4/27) = 3/2, A(3/64) = -2 + 2*sqrt(7/3), A(5/64) = (-1 + sqrt(5))*2/sqrt(5), etc. A(n^2/(n+1)^3) = (n+1)/n if n > 1. -
Michael Somos, Jul 17 2018
A(x) = exp( Sum_{n >= 1} (1/3)*binomial(3*n,n)*x^n/n ).
The sequence defined by b(n) := [x^n] A(x)^n =
A224274(n) for n >= 1 and satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 3. Cf.
A060941. (End)
G.f.: 1/sqrt(B(x)+(1-6*x)/(9*B(x))+1/3), with B(x):=((27*x^2-18*x+2)/54-(x*sqrt((-(4-27*x))*x))/(2*3^(3/2)))^(1/3). -
Vladimir Kruchinin, Sep 28 2021
a(n) = hypergeom([1 - n, -2*n], [2], 1). Row sums of
A108767. -
Peter Bala, Aug 30 2023
G.f.: z*exp(3*z*hypergeom([1, 1, 4/3, 5/3], [3/2, 2, 2], (27*z)/4)) + 1.
G.f.: hypergeometric([1/3, 2/3], [3/2], (3^3/2^2)*x). See the e.g.f. above. -
Wolfdieter Lang, Feb 04 2024
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