%I #40 Nov 25 2024 04:58:28
%S 1,4,28,220,1816,15424,133456,1169872,10354528,92331904,828204928,
%T 7464652672,67547774464,613295870464,5584367987968,50974595472640,
%U 466307503244800,4273832891668480,39237007284226048,360768875975526400,3321625537178669056,30619908430235828224,282578914501599305728
%N a(n) = Sum_{k=0..n} C(n + 2*k, 3*k) * C(3*k, 2*k).
%C Compare to: Sum_{k=0..n} (-1)^k * C(n+2*k,3*k) * C(3*k,2*k) = (-2)^n for n>=0.
%H Michael De Vlieger, <a href="/A243116/b243116.txt">Table of n, a(n) for n = 0..1027</a>
%H Hacène Belbachir and Abdelghani Mehdaoui, <a href="https://lrecits.usthb.dz/6.5.pdf">Diagonal sums in Pascal pyramid (1, 2, r)</a>, Les Annales RECITS (2019) Vol. 6, 45-52.
%F G.f.: Sum_{n>=0} C(3*n, n) * x^n / (1-x)^(3*n+1). - _Paul D. Hanna_, Aug 30 2014
%F G.f.: 1/(1-x) / ( 3 / G(x/(1-x)^3) - 2 ), where G(x) = 1 + x*G(x)^3 is the g.f. of A001764. - _Paul D. Hanna_, Aug 30 2014
%F G.f. satisfies: A(x) = 1 + (4-3*x)*A(x) - (4 - 39*x + 12*x^2 - 4*x^3)*A(x)^3. - _Paul D. Hanna_, Sep 05 2014
%F a(n) = Sum_{k=0..n} A109955(n,k) * A005809(k).
%F a(n) = -(-2)^n + 2*Sum_{k=0..[n/2]} C(n+4*k, 6*k) * C(6*k, 4*k).
%F Recurrence: 2*n*(2*n-1)*(3*n-4)*a(n) = (3*n-2)*(39*n^2 - 65*n + 18)*a(n-1) - 2*(n-1)*(18*n^2 - 33*n + 10)*a(n-2) + 4*(n-2)*(n-1)*(3*n-1)*a(n-3). - _Vaclav Kotesovec_, Aug 21 2014
%F From _Peter Bala_, Mar 11 2022: (Start)
%F a(n) = Sum_{k = 0..floor(n/4)} (-1)^k*binomial(n,k)*binomial(4*n-4*k,3*n).
%F a(n) = [x^n] ( (1 + x)^4 - x^4 )^n. Cf. A122868(n) = [x^n] ( (1 + x)^3 - x^3 )^n.
%F It follows that the Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k. (End)
%F a(n) = [x^n] (1 + x + (1 + x)^3)^n. - _Seiichi Manyama_, Nov 25 2024
%e G.f.: A(x) = 1 + 4*x + 28*x^2 + 220*x^3 + 1816*x^4 + 15424*x^5 +...
%e where
%e A(x) = 1/(1-x) + 3*x/(1-x)^4 + 15*x^2/(1-x)^7 + 84*x^3/(1-x)^10 + 495*x^4/(1-x)^13 + 3003*x^5/(1-x)^16 + 18564*x^6/(1-x)^19 + 116280*x^7/(1-x)^22 + 735471*x^8/(1-x)^25 +...+ C(3*n, n)*x^n/(1-x)^(3*n+1) +...
%e ILLUSTRATION OF TERMS.
%e The sequence A005809(k) = C(3*k, 2*k) begins:
%e [1, 3, 15, 84, 495, 3003, 18564, 116280, 735471, 4686825, ...];
%e the triangle A109955(n,k) = C(n + 2*k, 3*k) begins:
%e 1;
%e 1, 1;
%e 1, 4, 1;
%e 1, 10, 7, 1;
%e 1, 20, 28, 10, 1;
%e 1, 35, 84, 55, 13, 1;
%e 1, 56, 210, 220, 91, 16, 1;
%e 1, 84, 462, 715, 455, 136, 19, 1; ...
%e where a(n) = Sum_{k=0..n} A109955(n,k) * A005809(k):
%e a(1) = 1*1 + 1*3 = 4;
%e a(2) = 1*1 + 4*3 + 1*15 = 28;
%e a(3) = 1*1 + 10*3 + 7*15 + 1*84 = 220;
%e a(4) = 1*1 + 20*3 + 28*15 + 10*84 + 1*495 = 1816; ...
%e compare to: Sum_{k=0..n} (-1)^k * A109955(n,k) * A005809(k) = (-2)^n.
%t Table[Sum[Binomial[n + 2*k, 3*k] * Binomial[3*k, 2*k],{k,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Aug 21 2014 *)
%o (PARI) {a(n)=sum(k=0,n, binomial(n+2*k,3*k) * binomial(3*k,2*k))}
%o for(n=0,30,print1(a(n),", "))
%o (PARI) {a(n)=-(-2)^n + 2*sum(k=0, n\2, binomial(n+4*k, 6*k) * binomial(6*k, 4*k))}
%o for(n=0, 30, print1(a(n), ", "))
%o (PARI) {a(n)=local(A=1); A=sum(m=0, n, binomial(3*m, m) * x^m/(1-x +x*O(x^n))^(3*m+1)); polcoeff(A, n)}
%o for(n=0, 30, print1(a(n), ", "))
%Y Cf. A109955, A005809.
%K nonn,changed
%O 0,2
%A _Paul D. Hanna_, Aug 20 2014