cp's OEIS Frontend

A214668 G.f. A(x) satisfies A(x) = 1 + 9*x*A(x)^(4/3).

%I A214668 #27 Sep 02 2025 04:06:27
%S A214668 1,9,108,1458,21060,318087,4960116,79227720,1289516436,21308126895,
%T A214668 356506456680,6027199821864,102804351279084,1766931074710515,
%U A214668 30570993847594800,532022685332573016,9306598678048938420,163549467160708850910,2886019647490699098588
%N A214668 G.f. A(x) satisfies A(x) = 1 + 9*x*A(x)^(4/3).
%C A214668 Radius of convergence of g.f. A(x) is r = 1/(3*4^(4/3)) where A(r) = 4.
%C A214668 Self-convolution cube of A078532.
%F A214668 a(n) = 9^n * binomial(4*n/3, n) / (n/3 + 1).
%F A214668 From _Karol A. Penson_, Mar 24 2024: (Start)
%F A214668 G.f. = 4F3([1/4, 1/2, 3/4, 1], [1/3, 2/3, 2], 6912*z^3) + 9*z*3F2([7/12, 5/6, 13/12], [2/3, 7/3], 6912*z^3) + 108*z^2*3F2([11/12, 7/6, 17/12], [4/3, 8/3], 6912*z^3);
%F A214668 a(n) = Integral_{x=0..6912^(1/3)} x^n*W(x), where
%F A214668  W(x) = h1(x) + h2(x) + h3(x), with
%F A214668  h1(x) = sqrt(6)*3F2([-3/4, 7/12, 11/12], [1/2, 3/4], x^3/6912)/(18*Pi*x^(1/4)),
%F A214668  h2(x) = sqrt(x)*3F2([-1/2, 5/6, 7/6], [3/4, 5/4], x^3/6912)/(36*Pi),
%F A214668  h3(x) = (5*sqrt(6)*x^(5/4)*3F2([-1/4, 13/12, 17/12], [5/4, 3/2], x^3/6912))/(5184*Pi).
%F A214668 This integral representation is unique as W(x) is the solution of the Hausdorff power moment problem on x = (0, 6912^(1/3)). Using only the definition of a(n), W(x) can be proven to be positive. W(x) is singular at x = 0, with singularity x^(-1/4), and for x > 0 is first monotonically decreasing up to a local minimum at x around x = 2, then it is monotonically increasing up to a local maximum at x around x = 10.8, and then finally is monotonically decreasing up to zero at x = 6912^(1/3). For x -> 6912^(1/3), W'(x) tends to -infinity. (End)
%F A214668 G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^(5/3)). - _Seiichi Manyama_, Jun 18 2025
%F A214668 D-finite with recurrence (n-1)*(n-2)*(n+3)*a(n) - 216*(4*n-9)*(4*n-3)*(2*n-3)*a(n-3) = 0. - _R. J. Mathar_, Jul 30 2025
%F A214668 a(n) ~ 2^(8*n/3 + 1/2) * 3^(n+1) / (sqrt(Pi) * n^(3/2)). - _Amiram Eldar_, Sep 02 2025
%e A214668 G.f.: A(x) = 1 + 9*x + 108*x^2 + 1458*x^3 + 21060*x^4 + 318087*x^5 + ...
%e A214668 where A(x) = 1 + 9*x*A(x)^(4/3).
%e A214668 Radius of convergence: r = 1/(3*4^(4/3)) = 0.052496710...
%e A214668 Related expansions:
%e A214668 A(x)^(4/3) = 1 + 12*x + 162*x^2 + 2340*x^3 + 35343*x^4 + 551124*x^5 + ... + a(n+1)/9*x^n + ...
%e A214668 A(x)^(1/3) = 1 + 3*x + 27*x^2 + 315*x^3 + 4158*x^4 + 59049*x^5 + 880308*x^6 + 13586859*x^7 + 215233605*x^8 + ... + A078532(n)*x^n + ...
%t A214668 a[n_] := 9^n * Binomial[4*n/3, n] / (n/3 + 1); Array[a, 20, 0] (* _Amiram Eldar_, Sep 02 2025 *)
%o A214668 (PARI) {a(n)=9^n*binomial(4*n/3, n)/(n/3+1)}
%o A214668 (PARI) {a(n)=local(A=1+x);for(i=1,n,A =1+9*x*(A+x*O(x^n))^(4/3));polcoeff(A,n)}
%o A214668 for(n=0,30,print1(a(n),", "))
%Y A214668 Cf. A078532, A214377, A214553.
%K A214668 nonn,changed
%O A214668 0,2
%A A214668 _Paul D. Hanna_, Jul 24 2012