OFFSET
0,2
COMMENTS
Radius of convergence of g.f. A(x) is r = 1/(3*4^(4/3)) where A(r) = 4.
Self-convolution cube of A078532.
FORMULA
a(n) = 9^n * binomial(4*n/3, n) / (n/3 + 1).
From Karol A. Penson, Mar 24 2024: (Start)
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);
a(n) = Integral_{x=0..6912^(1/3)} x^n*W(x), where
W(x) = h1(x) + h2(x) + h3(x), with
h1(x) = sqrt(6)*3F2([-3/4, 7/12, 11/12], [1/2, 3/4], x^3/6912)/(18*Pi*x^(1/4)),
h2(x) = sqrt(x)*3F2([-1/2, 5/6, 7/6], [3/4, 5/4], x^3/6912)/(36*Pi),
h3(x) = (5*sqrt(6)*x^(5/4)*3F2([-1/4, 13/12, 17/12], [5/4, 3/2], x^3/6912))/(5184*Pi).
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)
EXAMPLE
G.f.: A(x) = 1 + 9*x + 108*x^2 + 1458*x^3 + 21060*x^4 + 318087*x^5 + ...
where A(x) = 1 + 9*x*A(x)^(4/3).
Radius of convergence: r = 1/(3*4^(4/3)) = 0.052496710...
Related expansions:
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 + ...
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 + ...
PROG
(PARI) {a(n)=9^n*binomial(4*n/3, n)/(n/3+1)}
(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)}
for(n=0, 30, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jul 24 2012
STATUS
approved