A198951 G.f. satisfies: A(x) = (1 + x*A(x))*(1 + x^3*A(x)^3).

1, 1, 1, 2, 6, 16, 39, 99, 271, 763, 2146, 6062, 17359, 50337, 147057, 431874, 1275273, 3786649, 11298031, 33846202, 101762937, 306997821, 929038518, 2819426688, 8578433304, 26163061776, 79970186791, 244938841096, 751646959402, 2310683396056, 7115199919151

Offset: 0

Author

Paul D. Hanna, Oct 31 2011

Keywords

• nonn
• easy

Comments

a(n) is also the number of rooted labeled trees on n nodes such that each node has 0, 1, 3, or 4 children. - Patrick Devlin, Mar 04 2012

Examples

			G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 6*x^4 + 16*x^5 + 39*x^6 + 99*x^7 + ...
Related expansions:
A(x)^3 = 1 + 3*x + 6*x^2 + 13*x^3 + 36*x^4 + 105*x^5 + 292*x^6 + ...
A(x)^4 = 1 + 4*x + 10*x^2 + 24*x^3 + 67*x^4 + 200*x^5 + 582*x^6 + ...
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + x^2*A(x)^2)*x + (1 + 2^2*x^2*A(x)^2 + x^4*A(x)^4)*x^2/2 +
(1 + 3^2*x^2*A(x)^2 + 3^2*x^4*A(x)^4 + x^6*A(x)^6)*x^3/3 +
(1 + 4^2*x^2*A(x)^2 + 6^2*x^4*A(x)^4 + 4^2*x^6*A(x)^6 + x^8*A(x)^8)*x^4/4 +
(1 + 5^2*x^2*A(x)^2 + 10^2*x^4*A(x)^4 + 10^2*x^6*A(x)^6 + 5^2*x^8*A(x)^8 + x^10*A(x)^10)*x^5/5 + ...
more explicitly,
log(A(x)) = x + x^2/2 + 4*x^3/3 + 17*x^4/4 + 51*x^5/5 + 136*x^6/6 + 393*x^7/7 + 1233*x^8/8 + ...
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..600

Crossrefs

Cf. A198953, A007863, A036765, A186241, A200731, A228960.

Programs

Maple

a:= n-> coeff(series(RootOf(A=(1+x*A)*(1+x^3*A^3), A), x, n+1), x, n):
seq(a(n), n=0..30);  # Alois P. Heinz, May 16 2012

Mathematica

InverseSeries[ Series[ x/((1 + x)*(1 + x^3)), {x, 0, 31}], x] // CoefficientList[#, x]& // Rest (* Jean-François Alcover, Sep 10 2013 *)

PARI

{a(n)=local(A=1/x*serreverse(x/(1+x+x^3+x^4+x*O(x^n)))); polcoeff(A, n)}

PARI

{a(n)=polcoeff((1+x+x^3+x^4+x*O(x^n))^(n+1)/(n+1), n)}

PARI

{a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n,sum(j=0, m, binomial(m, j)^2*x^(2*j)*(A+x*O(x^n))^(2*j))*x^m/m))); polcoeff(A, n)}

PARI

{a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, (1-x^2*A^2)^(2*m+1)*sum(j=0, n\2, binomial(m+j, j)^2*x^(2*j)*(A^2+x*O(x^n))^j)*x^m/m))); polcoeff(A, n, x)}

Formula

G.f. satisfies:

(1) A(x) = exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} C(n,k)^2 * x^(2*k) * A(x)^(2*k) ).

(2) A(x) = (1/x)*Series_Reversion(x/((1+x)*(1+x^3))).

(3) a(n) = [x^n] (1 + x + x^3 + x^4)^(n+1) / (n+1).

(4) A(x) = exp( Sum_{n>=1} x^n/n * (1-x^2*A(x)^2)^(2*n+1)*Sum_{k>=0} C(n+k,k)^2 * x^(2*k) * A(x)^(2*k) ).

D-finite with recurrence: 3*(n+1)*(3*n+2)*(3*n+4)*(119*n^3 - 210*n^2 + 73*n - 6)*a(n) = 2*(6664*n^6 - 1764*n^5 - 11585*n^4 + 426*n^3 + 4129*n^2 - 102*n - 288)*a(n-1) - 18*(n-1)*(1190*n^5 - 910*n^4 - 1937*n^3 + 895*n^2 + 606*n - 216)*a(n-2) + 162*(n-2)*(n-1)*(2*n-3)*(119*n^3 + 147*n^2 + 10*n - 24)*a(n-3). - Vaclav Kotesovec, Sep 09 2013

a(n) ~ c*d^n/n^(3/2), where d = 1/81*((2144134+520506*sqrt(17))^(2/3)+112*(2144134+520506*sqrt(17))^(1/3)-2036)/(2144134+520506*sqrt(17))^(1/3) = 3.23407602060970245... is the root of the equation -324 + 180*d - 112*d^2 + 27*d^3 = 0 and c = 0.6286981954423757284622435... - Vaclav Kotesovec, Sep 09 2013

A(1/d) = 370/243 + (3*sqrt(17)/509 - 3070/123687)*(2144134+520506*sqrt(17))^(1/3) + (141*sqrt(17)/2072648 - 129529/503653464)*(2144134+520506*sqrt(17))^(2/3) = 2.053716618436594614948796... - Vaclav Kotesovec, Sep 10 2013

From Peter Bala, Jun 21 2015: (Start)

a(n) = 1/(n + 1)*Sum_{k = 0..floor(n/3)} binomial(n + 1,k)* binomial(n + 1,n - 3*k). Applying Maple's sumrecursion command to this formula gives the above recurrence of Kotesovec.

More generally, the coefficient of x^n in A(x)^r equals r/(n + r)*Sum_{k = 0..floor(n/3)} binomial(n + r,k)*binomial(n + r,n - 3*k) by the Lagrange-Bürmann formula.

O.g.f. A(x) = exp(Sum_{n >= 1} A228960(n)*x^n/n), where A228960(n) = Sum_{k = 0..floor(n/3)} binomial(n,k)*binomial(n,3*k). Cf. A036765, A186241 and A200731. (End)