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

%I A124753 #61 Jul 20 2025 15:02:18
%S A124753 1,1,1,1,2,3,4,9,15,22,52,91,140,340,612,969,2394,4389,7084,17710,
%T A124753 32890,53820,135720,254475,420732,1068012,2017356,3362260,8579560,
%U A124753 16301164,27343888,70068713,133767543,225568798,580034052,1111731933,1882933364,4855986044,9338434700
%N A124753 a(3n+k) = (k+1)*binomial(4n+k, n)/(3n+k+1), where k is n reduced mod 3.
%C A124753 Row sums of Riordan array (1,x(1-x^3))^(-1). Also row sums of A124752.
%C A124753 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
%C A124753   \ /
%C A124753    |   and  \|/ . - _David Callan_, Aug 22 2014
%H A124753 J.-B. Priez and A. Virmaux, <a href="http://arxiv.org/abs/1411.4161">Non-commutative Frobenius characteristic of generalized parking functions: Application to enumeration</a>, arXiv:1411.4161 [math.CO], 2014-2015.
%F A124753 a(3n) = A002293(n), a(3n+1) = A069271(n), a(3n+2) = A006632(n+1).
%F A124753 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
%F A124753 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
%F A124753 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
%F A124753 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
%F A124753 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
%p A124753 A124753 := proc(n)
%p A124753     local k,np;
%p A124753     k := modp(n,3) ;
%p A124753     np := floor(n/3) ;
%p A124753     (k+1)*binomial(np+n,np)/(n+1) ;
%p A124753 end proc:
%p A124753 seq(A124753(n),n=0..40) ; # _R. J. Mathar_, Oct 30 2014
%t A124753 a[n_] := Module[{q, k}, {q, k} = QuotientRemainder[n, 3]; (k+1)*Binomial[4q + k, q]/(3q + k + 1)];
%t A124753 Table[a[n], {n, 0, 34}] (* _Jean-François Alcover_, Nov 20 2017 *)
%o A124753 (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
%o A124753 (PARI) apr(n, p, r) = r*binomial(n*p+r, n)/(n*p+r);
%o A124753 a(n) = apr(n\3, 4, n%3+1); \\ _Seiichi Manyama_, Jul 20 2025
%Y A124753 Cf. A084080, A002293, A069271 (trisection), A006632 (trisection).
%Y A124753 Cf. A047749, A118968.
%K A124753 nonn,easy
%O A124753 0,5
%A A124753 _Paul Barry_, Nov 06 2006