A198951 - OEIS

%I #47 Feb 21 2020 07:00:19

%S 1,1,1,2,6,16,39,99,271,763,2146,6062,17359,50337,147057,431874,

%T 1275273,3786649,11298031,33846202,101762937,306997821,929038518,

%U 2819426688,8578433304,26163061776,79970186791,244938841096,751646959402,2310683396056,7115199919151

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

%C 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

%H Alois P. Heinz, <a href="/A198951/b198951.txt">Table of n, a(n) for n = 0..600</a>

%F G.f. satisfies:

%F (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) ).

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

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

%F (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) ).

%F 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

%F 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

%F 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

%F From _Peter Bala_, Jun 21 2015: (Start)

%F 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.

%F 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.

%F 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)

%e G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 6*x^4 + 16*x^5 + 39*x^6 + 99*x^7 + ...

%e Related expansions:

%e A(x)^3 = 1 + 3*x + 6*x^2 + 13*x^3 + 36*x^4 + 105*x^5 + 292*x^6 + ...

%e A(x)^4 = 1 + 4*x + 10*x^2 + 24*x^3 + 67*x^4 + 200*x^5 + 582*x^6 + ...

%e The logarithm of the g.f. equals the series:

%e 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 +

%e (1 + 3^2*x^2*A(x)^2 + 3^2*x^4*A(x)^4 + x^6*A(x)^6)*x^3/3 +

%e (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 +

%e (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 + ...

%e more explicitly,

%e 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 + ...

%p a:= n-> coeff(series(RootOf(A=(1+x*A)*(1+x^3*A^3), A), x, n+1), x, n):

%p seq(a(n), n=0..30);  # _Alois P. Heinz_, May 16 2012

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

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

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

%o (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)}

%o (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)}

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

%K nonn,easy

%O 0,4

%A _Paul D. Hanna_, Oct 31 2011