%I #15 Dec 14 2014 22:22:28
%S 1,1,3,21,261,4833,120303,3778029,143531433,6404711553,328447585179,
%T 19037277446949,1230842669484717,87829738967634849,
%U 6856701559496841159,581343578623728854397,53196439113856500195537,5225543459274294130169601,548468830470032135590262067,61258398893626609968686844597
%N E.g.f.: exp(3*x*G(x)^2) / G(x)^2 where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.
%H Paul D. Hanna, <a href="/A251573/b251573.txt">Table of n, a(n) for n = 0..200</a>
%F Let G(x) = 1 + x*G(x)^3 be the g.f. of A001764, then the e.g.f. A(x) of this sequence satisfies:
%F (1) A'(x)/A(x) = G(x)^2.
%F (2) A'(x) = exp(3*x*G(x)^2).
%F (3) A(x) = exp( Integral G(x)^2 dx ).
%F (4) A(x) = exp( Sum_{n>=1} A006013(n-1)*x^n/n ), where A006013(n-1) = binomial(3*n-2,n)/(2*n-1).
%F (5) A(x) = F(x/A(x)) where F(x) is the e.g.f. of A251583.
%F (6) A(x) = Sum_{n>=0} A251583(n)*(x/A(x))^n/n! where A251583(n) = 3^(n-1) * (n+1)^(n-3) * (n+3).
%F (7) [x^n/n!] A(x)^(n+1) = 3^(n-1) * (n+1)^(n-2) * (n+3).
%F a(n) = Sum_{k=0..n} 3^k * n!/k! * binomial(3*n-k-3, n-k) * (k-1)/(n-1) for n>1.
%F Recurrence (for n>3): 2*(n-3)*(2*n-3)*a(n) = 3*(3*n-8)*(3*n^2 - 13*n + 15)*a(n-1) - 27*(n-2)*a(n-2). - _Vaclav Kotesovec_, Dec 07 2014
%F a(n) ~ 3^(3*n-7/2) * n^(n-2) / (2^(2*n-5/2) * exp(n-1)). - _Vaclav Kotesovec_, Dec 07 2014
%e E.g.f.: A(x) = 1 + x + 3*x^2/2! + 21*x^3/3! + 261*x^4/4! + 4833*x^5/5! +...
%e such that A(x) = exp(3*x*G(x)^2) / G(x)^2
%e where G(x) = 1 + x*G(x)^3 is the g.f. of A001764:
%e G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 +...
%e The e.g.f. satisfies:
%e A(x) = 1 + x/A(x) + 5*x^2/(2!*A(x)^2) + 54*x^3/(3!*A(x)^3) + 945*x^4/(4!*A(x)^4) + 23328*x^5/(5!*A(x)^5) + 750141*x^6/(6!*A(x)^6) + 29859840*x^7/(7!*A(x)^7) +...+ 3^(n-1)*(n+1)^(n-3)*(n+3) * x^n/(n!*A(x)^n) +...
%e Note that
%e A'(x) = exp(3*x*G(x)^2) = 1 + 3*x + 21*x^2/2! + 261*x^3/3! + 4833*x^4/4! +...
%e LOGARITHMIC DERIVATIVE.
%e The logarithm of the e.g.f. begins:
%e log(A(x)) = x + 2*x^2/2 + 7*x^3/3 + 30*x^4/4 + 143*x^5/5 +...
%e and so A'(x)/A(x) = G(x)^2.
%e TABLE OF POWERS OF E.G.F.
%e Form a table of coefficients of x^k/k! in A(x)^n as follows.
%e n=1: [1, 1, 3, 21, 261, 4833, 120303, 3778029, ...];
%e n=2: [1, 2, 8, 60, 744, 13536, 330912, 10232928, ...];
%e n=3: [1, 3, 15, 123, 1557, 28179, 680427, 20771235, ...];
%e n=4: [1, 4, 24, 216, 2832, 51552, 1237248, 37404288, ...];
%e n=5: [1, 5, 35, 345, 4725, 87285, 2094975, 62949825, ...];
%e n=6: [1, 6, 48, 516, 7416, 139968, 3378528, 101278944, ...];
%e n=7: [1, 7, 63, 735, 11109, 215271, 5250987, 157613463, ...];
%e n=8: [1, 8, 80, 1008, 16032, 320064, 7921152, 238878720, ...]; ...
%e in which the main diagonal begins (see A251583):
%e [1, 2, 15, 216, 4725, 139968, 5250987, 238878720, ...]
%e and is given by the formula:
%e [x^n/n!] A(x)^(n+1) = 3^(n-1) * (n+1)^(n-2) * (n+3) for n>=0.
%t Flatten[{1,1,Table[Sum[3^k * n!/k! * Binomial[3*n-k-3, n-k] * (k-1)/(n-1),{k,0,n}],{n,2,20}]}] (* _Vaclav Kotesovec_, Dec 07 2014 *)
%t Flatten[{1,1,RecurrenceTable[{27*(n-2)*a[n-2]-3*(3*n-8)*(15-13*n+3*n^2)*a[n-1]+2*(n-3)*(2*n-3)*a[n]==0,a[2]==3,a[3]==21},a,{n,20}]}] (* _Vaclav Kotesovec_, Dec 07 2014 *)
%o (PARI) {a(n) = local(G=1);for(i=1,n,G=1+x*G^3 +x*O(x^n)); n!*polcoeff(exp(3*x*G^2)/G^2, n)}
%o for(n=0, 20, print1(a(n), ", "))
%o (PARI) {a(n) = if(n==0|n==1, 1, sum(k=0, n, 3^k * n!/k! * binomial(3*n-k-3,n-k) * (k-1)/(n-1) ))}
%o for(n=0, 20, print1(a(n), ", "))
%Y Cf. A251583, A251663, A001764, A006013.
%Y Cf. Variants: A243953, A251574, A251575, A251576, A251577, A251578, A251579, A251580.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Dec 06 2014