

A066357


Number of ordered (i.e., planar) trees on 2n edges with every subtree at the root having an even number of edges.


15



1, 1, 6, 53, 554, 6362, 77580, 986253, 12927170, 173452334, 2370742868, 32892031042, 462030186916, 6557906929108, 93909078262808, 1355087936016957, 19684187540818866, 287612514032460070, 4224238030616082948, 62329883931236020470, 923519220367120779820
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OFFSET

0,3


COMMENTS

a(n) is the number of walks from (0,0) to (2n,2n) using steps (0,1) and (1,0) which never stray below the line y=x and which avoid the points (m,m) m odd.  Paul Boddington, Mar 14 2003
Series reversion of Sum_{n>0} a(n)(x)^n is g.f. of A005900.
a(n) is the number of linear extensions of the onelevel grid poset G[(0^n), (1^(n1)), (1^(n1))]. The definition of a onelevel grid poset can be found in the Pan links.  Ran Pan, Jul 05 2016
These numbers have the same parity as the Catalan numbers C(n), that is, a(n) is even except when n has the form 2^m  1. This follows immediately from the formula a(n) = C(2*n+1) + 2*C(2*n)  2^(2*n + 1)*C(n) given below by Callan. We conjecture that a(n) and C(n) have the same 2adic valuation (checked up to n = 100).  Peter Bala, Aug 02 2016


LINKS

Nantel Bergeron, Cesar Ceballos, Vincent Pilaud, Hopf dreams, arXiv:1807.03044 [math.CO], 2018. See p. 17.
D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), 307344 (p. 333).


FORMULA

For n>0, a(n) = Sum_{r=1..n} C(2*r1)*a(nr). Here C(2*r1) is a Catalan number (A000108).  Paul Boddington, Mar 14 2003
G.f.: 2/(1+4*sqrt(x)/(sqrt(1+4*sqrt(x))sqrt(14*sqrt(x)))).
Dfinite with recurrence a(n)*(2*n1)*(n+1)n = a(n1)*(32*n^2  64*n + 39)*2*n  a(n2)*(2*n3)*(4*n5)*(4*n7)*16, n>1.
a(0) = 1,a(n) = (1/n)*Sum_{k=0..n} C(4*n,k)*C(3*nk2,nk1), n>1.  Paul Barry, Apr 09 2007
a(n) = ((2^(4*n))/Gamma(1/2)) * ((6*(2*n+1)*Gamma(2*n+1/2)/Gamma(2*n+3))2*Gamma(n+1/2)/Gamma(n+2)).  David Dickson (dcmd(AT)unimelb.edu.au), Nov 10 2009
Convolution of A079489 with itself: (1, 6, 53, 554, ...) = (1, 3, 22, 211, ...)*(1, 3, 22, 211, ...).
Proof. Working with Dyck paths, we must show that Dyck paths of size (semilength) 2n, all of whose components (constituent primitive Dyck paths) have even size, are equinumerous with ordered pairs of nonempty Dyck paths of total size 2n in each of which the first component is of odd size and all other components (if any) are of even size. Given a Dyck path P of the former class, use the first return decomposition to write P (uniquely) as the concatenation of U A_1 A_2 ... A_j O E D Q where U denotes upstep, D denotes downstep, A_1,...,A_j are all primitive Dyck paths of even size with j>=0, O is a primitive Dyck path of odd size, E is a Dyck path of even size, and Q is a Dyck path in which all components are of even size. Then P > (O A_1 A_2 ... A_j, U E D Q) is the desired bijection. QED  David Callan, Apr 11 2012
a(n) = C(2*n+1) + 2*C(2*n)  2^(2*n+1)*C(n), where C(n) is the Catalan number A000108. This formula can be obtained by manipulating generating functions. The equivalence of this formula and the Barry (Apr 09 2007) sum can be established by the WZ method with a secondorder operator. A combinatorial interpretation of the Barry sum would be nice.  David Callan, Apr 10 2012
a(n) ~ (32*sqrt(2)) * 2^(4*n) / (n^(3/2) * sqrt(2*Pi)).  Vaclav Kotesovec, Mar 21 2014
exp( Sum_{n >= 1} binomial(4*n,2*n)*x^n/n ) = 1 + 6*x + 53*x^2 + 554*x^3 + ... is an o.g.f. for this sequence omitting the initial term. See A001448.  Peter Bala, Oct 02 2015
a(n) = binomial(3*n2,n1)*hypergeom([1n,4*n],[23*n],1)/n for n>=1.  Peter Luschny, Oct 15 2015
a(n) = 3*(2*n+1) /(2*n+2) /(4*n+1) *binomial(4*n+2,2*n+1) 4^n /(2*n+1) *binomial(2*n+2,n+1) [Merlini et al F_n formula]  R. J. Mathar, Oct 01 2021


MAPLE

gf := (1sqrt(14*z)sqrt(1+4*z)+sqrt(116*z^2))/(z*(sqrt(14*z)sqrt(1+4*z))):s := series(gf, z, 80): for i from 0 to 50 by 2 do printf(`%d, `, coeff(s, z, i)) od: # James A. Sellers, Feb 11 2002
a := n > `if`(n=0, 1, binomial(3*n2, n1)*hypergeom([1n, 4*n], [23*n], 1)/n): seq(simplify(a(n)), n=0..20); # Peter Luschny, Oct 15 2015


MATHEMATICA

CoefficientList[Series[2/(1 + 4 Sqrt[x]/(Sqrt[1 + 4 Sqrt[x]]  Sqrt[1  4 Sqrt[x]])), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 21 2014 *)


PROG

(PARI) a(n)=local(A); if(n<1, n==0, A=sqrt(1+4*x+O(x^(2*n+2))); A=subst(A, x, x); polcoeff(((2*A8*x)/A^2)^2, 2*n))
(PARI) vector (100, n, n; if(n<1, 1, sum(k=0, n, binomial(4*n, k)*binomial(3*nk2, nk1)/n))) \\ Altug Alkan, Oct 07 2015
(Magma) [1] cat [(&+[Binomial(4*n, k)*Binomial(3*nk2, nk1)/n: k in [0..n]]): n in [1..30]]; // G. C. Greubel, Jan 15 2019
(Sage) [1] + [sum(binomial(4*n, k)*binomial(3*nk2, nk1)/n for k in (0..n)) for n in (1..30)] # G. C. Greubel, Jan 15 2019


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



