OFFSET
0,3
COMMENTS
It appears that a(n) - 1 is divisible by n*(n - 1) for n >= 2. Cf. A251568. - Peter Bala, Feb 15 2015
FORMULA
a(n) = Sum_{k=0..n} n!/k! * binomial(3*n-2*k-1, n-k) * k/(2*n-k) for n>0 with a(0)=1.
Recurrence: 2*(2*n-1)*(54*n^2 - 171*n + 116)*a(n) = (1458*n^4 - 7533*n^3 + 12474*n^2 - 6624*n - 7)*a(n-1) - (324*n^3 - 1080*n^2 + 759*n + 95)*a(n-2) + 8*(n-2)*(54*n^2 - 63*n - 1)*a(n-3). - Vaclav Kotesovec, Feb 15 2015
EXAMPLE
E.g.f.: A(x) = 1 + x + 3*x^2/2! + 25*x^3/3! + 385*x^4/4! + 8661*x^5/5! +...
such that A(x) = exp(x*G(x)) where G(x) = 1 + x*G(x)^3:
G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 +...
MATHEMATICA
Flatten[{1, Table[Sum[n!/k! * Binomial[3*n-2*k-1, n-k] * k/(2*n-k), {k, 0, n}], {n, 1, 20}]}] (* Vaclav Kotesovec, Feb 15 2015 *)
PROG
(PARI) {a(n)=local(G=1); for(i=1, n, G=1+x*G^3 +x*O(x^n)); n!*polcoeff(exp(x*G), n)}
for(n=0, 20, print1(a(n), ", "))
(PARI) {a(n) = if(n==0, 1, sum(k=1, n, n!/k! * binomial(3*n-2*k-1, n-k) * k/(2*n-k) ))}
for(n=0, 20, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Dec 05 2014
STATUS
approved