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G.f.: A(x) = Sum_{n>=0} (3n)!/(n!)^3 * x^(3n)/(1-x-x^2)^(3n+1).
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%I #25 Dec 08 2021 11:53:47

%S 1,1,2,9,29,92,343,1281,4720,17899,68933,266364,1037423,4072439,

%T 16065148,63658521,253356763,1012049086,4055596343,16299779331,

%U 65683233938,265310551667,1073968967929,4355988107100,17699727361051

%N G.f.: A(x) = Sum_{n>=0} (3n)!/(n!)^3 * x^(3n)/(1-x-x^2)^(3n+1).

%C Limit_{n->oo} a(n+1)/a(n) = (Fibonacci(3)*sqrt(5) + Lucas(3))/2 = sqrt(5) + 2.

%H Iain Fox, <a href="/A181545/b181545.txt">Table of n, a(n) for n = 0..1600</a>

%H C. Banderier, P. Hitczenko, <a href="http://doi.org/10.1016/j.dam.2011.12.011">Enumeration and asymptotics of restricted compositions having the same number of parts</a>, Disc. Appl. Math. 160 (18) (2012) 2542-2554. Proposition 3.1.

%H Steffen Eger, <a href="http://arxiv.org/abs/1511.00622">On the Number of Many-to-Many Alignments of N Sequences</a>, arXiv:1511.00622 [math.CO], 2015.

%H Edyta Hetmaniok, Barbara Smoleń, Roman Wituła, <a href="http://ceur-ws.org/Vol-1853/p07.pdf">The Stirling triangles</a>, Proceedings of the Symposium for Young Scientists in Technology, Engineering and Mathematics (SYSTEM 2017), Kaunas, Lithuania, April 28, 2017, p. 35-41.

%F a(n) = Sum_{k=0..floor(n/2)} binomial(n-k,k)^3.

%F G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} binomial(n,k)^3 * x^k.

%F G.f.: A(x) = G( x^3/(1-x-x^2)^3 )/(1-x-x^2) where G(x) satisfies:

%F * G(x^3) = G( x*(1+3*x+9*x^2)/(1+6*x)^3 )/(1+6*x)

%F and G(x) is the g.f. of A006480.

%F Recurrence: (n-3)*n^2*a(n) = (n-3)*(3*n^2 - 3*n + 1)*a(n-1) - (n-1)*a(n-2) + 2*(n-2)*(11*n^2 - 44*n + 34)*a(n-3) + (n-3)*a(n-4) + (n-1)*(3*n^2 - 21*n + 37)*a(n-5) + (n-4)^2*(n-1)*a(n-6). - _Vaclav Kotesovec_, Jul 31 2014

%F a(n) ~ sqrt((9+4*sqrt(5))/12) * (2+sqrt(5))^n / (Pi*n). - _Vaclav Kotesovec_, Jul 31 2014

%F Equivalently, a(n) ~ phi^(3*n + 3) / (2*sqrt(3)*Pi*n), where phi = A001622 is the golden ratio. - _Vaclav Kotesovec_, Dec 08 2021

%e G.f. A(x) = 1 + x + 2*x^2 + 9*x^3 + 29*x^4 + 92*x^5 + 343*x^6 + ...

%e which equals the series:

%e A(x) = 1/(1-x-x^2) + 3!/1!^3*x^3/(1-x-x^2)^4 + 6!/2!^3*x^6/(1-x-x^2)^7 + 9!/3!^3*x^9/(1-x-x^2)^10 + 12!/4!^3*x^12/(1-x-x^2)^13 + ...

%e The g.f. also equals the series:

%e A(x) = 1 +

%e x*(1 + x) +

%e x^2*(1 + 2^3*x + x^2) +

%e x^3*(1 + 3^3*x + 3^3*x^2 + x^3) +

%e x^4*(1 + 4^3*x + 6^3*x^2 + 4^3*x^3 + x^4) +

%e x^5*(1 + 5^3*x + 10^3*x^2 + 10^3*x^3 + 5^3*x^4 + x^5) + ...

%e The terms begin:

%e a(0) = a(1) = 1^3;

%e a(2) = 1^3 + 1^3 = 2;

%e a(3) = 1^3 + 2^3 = 9;

%e a(4) = 1^3 + 3^3 + 1^3 = 29;

%e a(5) = 1^3 + 4^3 + 3^3 = 92;

%e a(6) = 1^3 + 5^3 + 6^3 + 1^3 = 343;

%e a(7) = 1^3 + 6^3 + 10^3 + 4^3 = 1281; ...

%t Table[Sum[Binomial[n-k,k]^3,{k,0,Floor[n/2]}],{n,0,20}] (* _Vaclav Kotesovec_, Jul 31 2014 *)

%o (PARI) {a(n)=sum(k=0,n\2,binomial(n-k,k)^3)}

%o (PARI) {a(n)=polcoeff(sum(m=0, n+1, x^m*sum(k=0, m, binomial(m, k)^3*x^k + x*O(x^n))), n)}

%o (PARI) {a(n)=polcoeff(sum(m=0,n,x^(3*m)/(1-x-x^2+x*O(x^n))^(3*m+1)*(3*m)!/(m!)^3),n)}

%Y Cf. A006480, A217421, A181546, A051286.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Oct 29 2010