A066357 Number of ordered (i.e., planar) trees on 2n edges with every subtree at the root having an even number of edges.

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

Offset: 0

Author

Louis Shapiro, Feb 01 2002

Keywords

• nonn
• easy

Comments

Row sums of A078990. First column of A079513.

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 one-level grid poset G[(0^n), (1^(n-1)), (1^(n-1))]. The definition of a one-level 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 2-adic valuation (checked up to n = 100). - Peter Bala, Aug 02 2016

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps, FPSAC02, Melbourne, 2002.
• Nantel Bergeron, Cesar Ceballos, Vincent Pilaud, Hopf dreams, arXiv:1807.03044 [math.CO], 2018. See p. 17.
• A. de Mier and M. Noy, A solution to the tennis ball problem, arXiv:math/0311242 [math.CO], 2003.
• J.-G. Luque and J.-Y. Thibon, Noncommutative Symmetric Functions Associated with a Code, Lazard Elimination and Witt Vectors, arXiv:math/0607254 [math.CO], 2006; Discrete Math. Theor. Comput. Sci. 9 (2007), no. 2, 59-72.
• D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), 307-344 (p. 333).
• Ran Pan, Problem 1, Project P.
• Ran Pan, Algorithmic Solution to Problem 1 (and linear extensions of general one-level grid-like posets), Project P.
• A. Regev, Enumerating triangulations by parallel diagonals, arXiv:1208.3915 [math.CO], 2012, J. Int. Seq. 15 (2012) #12.8.5

Crossrefs

Cf. A078990, A079513, A001448, A005900, A079489, A274644, A274763.

Programs

Magma

[1] cat [(&+[Binomial(4*n,k)*Binomial(3*n-k-2,n-k-1)/n: k in [0..n]]): n in [1..30]]; // G. C. Greubel, Jan 15 2019

Maple

gf := (1-sqrt(1-4*z)-sqrt(1+4*z)+sqrt(1-16*z^2))/(z*(sqrt(1-4*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 Sellers, Feb 11 2002
a := n -> `if`(n=0,1,binomial(3*n-2,n-1)*hypergeom([1-n,-4*n],[2-3*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 *)

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*A-8*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*n-k-2,n-k-1)/n))) \\ Altug Alkan, Oct 07 2015

Sage

[1] + [sum(binomial(4*n,k)*binomial(3*n-k-2,n-k-1)/n for k in (0..n)) for n in (1..30)] # G. C. Greubel, Jan 15 2019

Formula

For n>0, a(n) = Sum_{r=1..n} C(2*r-1)*a(n-r). Here C(2*r-1) is a Catalan number (A000108). - Paul Boddington, Mar 14 2003

G.f.: 2/(1+4*sqrt(x)/(sqrt(1+4*sqrt(x))-sqrt(1-4*sqrt(x)))).

D-finite with recurrence a(n)*(2*n-1)*(n+1)n = a(n-1)*(32*n^2 - 64*n + 39)*2*n - a(n-2)*(2*n-3)*(4*n-5)*(4*n-7)*16, n>1.

a(0) = 1,a(n) = (1/n)*Sum_{k=0..n} C(4*n,k)*C(3*n-k-2,n-k-1), 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 second-order operator. A combinatorial interpretation of the Barry sum would be nice. - David Callan, Apr 10 2012

a(n) ~ (3-2*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*n-2,n-1)*hypergeom([1-n,-4*n],[2-3*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