A302100 - OEIS

%I #18 Jun 18 2019 04:34:59

%S 1,1,5,41,449,6081,98133,1846377,39888353,977117825,26839621829,

%T 818332799593,27443807417569,1004188344449473,39809506543659477,

%U 1699473112658002089,77716022374143303489,3789578550994707778305,196255782523222432943109,10756748528551996006448553,622036345094017435642828161,37846075344692579622469742529

%N G.f. A(x) satisfies: A(x) = Sum_{n>=0} x^n * A(x)^n * Product_{k=0..n-1} (3*k + 1).

%H Paul D. Hanna, <a href="/A302100/b302100.txt">Table of n, a(n) for n = 0..300</a>

%F G.f. A(x) satisfies:

%F (1) A(x) = Sum_{n>=0} x^n * A(x)^n * Product_{k=0..n-1} (3*k + 1).

%F (2) A(x) = (1/x)*Series_Reversion( x/F(x) ), where F(x) = Sum_{n>=0} A007559(n)*x^n, the o.g.f. of the triple factorials.

%F (3) A(x) = 1 + x*A(x)^2 * (A(x) + 4*x*A'(x)) / (A(x) + x*A'(x)).

%F (4) A(x) = 1/(1 - x*A(x)/(1 - 3*x*A(x)/(1 - 4*x*A(x)/(1 - 6*x*A(x)/(1 - 7*x*A(x)/(1 - 9*x*A(x)/(1 - 10*x*A(x)/(1 - ...)))))))), a continued fraction.

%F a(n) ~ sqrt(2*Pi) * 3^n * n^(n - 1/6) / (Gamma(1/3) * exp(n - 1/3)). - _Vaclav Kotesovec_, Jun 18 2019

%e G.f.: A(x) = 1 + x + 5*x^2 + 41*x^3 + 449*x^4 + 6081*x^5 + 98133*x^6 + 1846377*x^7 + 39888353*x^8 + 977117825*x^9 + 26839621829*x^10 + ...

%e such that

%e A(x) = 1 + x*A(x) + 4*x^2*A(x)^2 + 28*x^3*A(x)^3 + 280*x^4*A(x)^4 + 3640*x^5*A(x)^5 + 58240*x^6*A(x)^6 + ... + x^n * A(x)^n * Product_{k=0..n-1} (3*k + 1) + ...

%o (PARI) /* Series Reversion of Triple Factorials g.f.: */

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

%o for(n=0, 30, print1(a(n), ", "))

%o (PARI) /* Differential Equation: */

%o {a(n) = my(A=1); for(i=0, n, A = 1 + x*A^2*(A + 4*x*A')/(x*A +x^2*O(x^n))'); polcoeff(A, n)}

%o for(n=0, 30, print1(a(n), ", "))

%o (PARI) /* Continued fraction: */

%o {a(n) = my(A=1, CF = 1+x +x*O(x^n)); for(i=1, n, A=CF; for(k=0, n, CF = 1/(1 - floor(3*(n-k+1)/2)*x*A*CF ) )); polcoeff(CF, n)}

%o for(n=0, 30, print1(a(n), ", "))

%Y Cf. A007559, A088368, A301363, A302535, A302565.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Apr 09 2018