O.g.f. satisfies: A(x) = 1 + x*A(x)^4 = 1/(1 - x*A(x)^3).
a(n) = binomial(4*n,n-1)/n, n >= 1, a(0) = 1. From the Lagrange series of the o.g.f. A(x) with its above given implicit equation.
Integral representation as n-th Hausdorff power moment of a positive function on the interval [0, 256/27]:
a(n) = Integral_{x=0..256/27}(x^n((3/256) * sqrt(2) * sqrt(3) * ((2/27) * 3^(3/4) * 27^(1/4) * 256^(/4) * hypergeom([-1/12, 1/4, 7/12], [1/2, 3/4], (27/256)*x)/(sqrt(Pi) * x^(3/4)) - (2/27) * sqrt(2) * sqrt(27) * sqrt(256) * hypergeom([1/6, 1/2, 5/6], [3/4, 5/4], (27/256)*x)/ (sqrt(Pi) * sqrt(x)) - (1/81) * 3^(1/4) * 27^(3/4) * 256^(1/4) * hypergeom([5/12, 3/4, 13/12], [5/4, 3/2], (27/256)*x/(sqrt(Pi)*x^(1/4)))/sqrt(Pi))).
This representation is unique as it represents the solution of the Hausdorff moment problem.
O.g.f.: hypergeom([1/4, 1/2, 3/4], [2/3, 4/3], (256/27)*x);
E.g.f.: hypergeom([1/4, 1/2, 3/4], [2/3, 1, 4/3], (256/27)*x). (End)
a(n) = upper left term in M^n, M = the production matrix:
1, 1
3, 3, 1
6, 6, 3, 1
...
(where 1, 3, 6, 10, ...) is the triangular series. -
Gary W. Adamson, Jul 08 2011
O.g.f. satisfies g = 1+x*g^4. If h is the series reversion of x*g, so h(x*g)=x, then (x-h(x))/x^2 is the o.g.f. of
A006013. -
Mark van Hoeij, Nov 10 2011
a(n) = Sum_{i=0..n-1} Sum_{j=0..n-1-i} Sum_{k=0..n-1-i-j} a(i)*a(j)*a(k)*a(n-1-i-j-k) for n>=1; and a(0) = 1. -
Robert FERREOL, Apr 02 2015
a(n) ~ 2^(8*n+1/2) / (sqrt(Pi) * n^(3/2) * 3^(3*n+3/2)). -
Vaclav Kotesovec, Jun 03 2015
D-finite with recurrence: a(n+1) = a(n)*4*(4*n + 3)*(4*n + 2)*(4*n + 1)/((3*n + 2)*(3*n + 3)*(3*n + 4)). -
Chai Wah Wu, Feb 19 2016
E.g.f.: F([1/4, 1/2, 3/4], [2/3, 1, 4/3], 256*x/27), where F is the generalized hypergeometric function. -
Stefano Spezia, Dec 27 2019
a(n) = hypergeom([1 - n, -3*n], [2], 1). Row sums of
A173020. -
Peter Bala, Aug 31 2023
G.f.: t*exp(4*t*hypergeom([1, 1, 5/4, 3/2, 7/4], [4/3, 5/3, 2, 2], (256*t)/27))+1. -
Karol A. Penson, Dec 20 2023
a(n) = Integral_{x=0..256/27} x^(n)*W(x)dx, n>=0, where W(x) = x^(-3/4) * sqrt(4*R(x) - 3^(3/4)*x^(1/4)/sqrt(R(x)))/(2*3^(1/4)*Pi), with R(x) = ((i + sqrt(3))*(4*sqrt(256 - 27*x) -12*i*sqrt(3*x))^(1/3))/16 - ((i - sqrt(3))*(4*sqrt(256 - 27*x) + 12*i* sqrt(3*x))^(1/3))/16, where i is the imaginary unit.
The elementary function W(x) is positive on the interval x = (0, 256/27) and is equal to the combination of hypergeometric functions in my formula from 2010; see above.
(Pi*W(x))^6 satisfies an algebraic equation of order 6, with integer polynomials as coefficients. (End)
G.f.: (Sum_{n >= 0} binomial(4*n+1, n)*x^n) / (Sum_{n >= 0} binomial(4*n, n)*x^n). -
Peter Bala, Dec 14 2024
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