A124753 a(3n+k) = (k+1)*binomial(4n+k, n)/(3n+k+1), where k is n reduced mod 3.

1, 1, 1, 1, 2, 3, 4, 9, 15, 22, 52, 91, 140, 340, 612, 969, 2394, 4389, 7084, 17710, 32890, 53820, 135720, 254475, 420732, 1068012, 2017356, 3362260, 8579560, 16301164, 27343888, 70068713, 133767543, 225568798, 580034052, 1111731933, 1882933364, 4855986044, 9338434700

Offset: 0

Author

Paul Barry, Nov 06 2006

Keywords

• nonn
• easy

Comments

Row sums of Riordan array (1,x(1-x^3))^(-1). Also row sums of A124752.

a(n) is the number of ordered trees (A000108) with n vertices in which every non-leaf non-root vertex has exactly two children that are leaves. For example, a(4) counts the 2 trees

\ /

| and \|/ . - David Callan, Aug 22 2014

Links

• J.-B. Priez and A. Virmaux, Non-commutative Frobenius characteristic of generalized parking functions: Application to enumeration, arXiv:1411.4161 [math.CO], 2014-2015.

Crossrefs

Cf. A084080, A002293, A069271 (trisection), A006632 (trisection).

Cf. A047749, A118968.

Programs

Maple

A124753 := proc(n)
    local k,np;
    k := modp(n,3) ;
    np := floor(n/3) ;
    (k+1)*binomial(np+n,np)/(n+1) ;
end proc:
seq(A124753(n),n=0..40) ; # R. J. Mathar, Oct 30 2014

Mathematica

a[n_] := Module[{q, k}, {q, k} = QuotientRemainder[n, 3]; (k+1)*Binomial[4q + k, q]/(3q + k + 1)];
Table[a[n], {n, 0, 34}] (* Jean-François Alcover, Nov 20 2017 *)

PARI

{a(n)=local(A=1+x); for(i=1,n,A=1+x*A*exp(sum(m=1,n\3,3*polcoeff(log(A+x*O(x^n)),3*m)*x^(3*m))+x*O(x^n))); polcoeff(A,n)} \\ Paul D. Hanna, Jun 04 2012

PARI

apr(n, p, r) = r*binomial(n*p+r, n)/(n*p+r);
a(n) = apr(n\3, 4, n%3+1); \\ Seiichi Manyama, Jul 20 2025

Formula

a(3n) = A002293(n), a(3n+1) = A069271(n), a(3n+2) = A006632(n+1).

a(n) = ((mod(n,3)+1)*C(4*floor(n/3)+mod(n,3), floor(n/3))/ (3*floor(n/3) + 1 + mod(n, 3))). - Paul Barry, Dec 14 2006

G.f. satisfies: A(x) = 1 + x*A(x)^2*A(w*x)*A(w^2*x), where w = exp(2*Pi*I/3). - Paul D. Hanna, Jun 04 2012

G.f. satisfies: A(x) = 1 + x*A(x)*G(x^3) where G(x) = 1 + x*G(x)^4 is the g.f. of A002293. - Paul D. Hanna, Jun 04 2012

Conjecture: +8019*n*(n-1)*(n+1)*a(n) +17496*n*(n-1)*(n-3)*a(n-1) +2592*(3*n-5)*(n-1)*(3*n-16)*a(n-2) +216*(-224*n^3+48*n^2+3926*n-6331)*a(n-3) +576*(-288*n^3+2448*n^2-6558*n+5443)*a(n-4) +768*(-288*n^3+3600*n^2-14878*n+20375)*a(n-5) -8192*(4*n-23)*(2*n-11)*(4*n-21)*a(n-6)=0. - R. J. Mathar, Oct 30 2014

a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/3)} a(3*k) * a(n-1-3*k). - Seiichi Manyama, Jul 07 2025