A004987 a(n) = (3^n/n!)*Product_{k=0..n-1} (3*k + 1). 3-central binomial coefficients.
1, 3, 18, 126, 945, 7371, 58968, 480168, 3961386, 33011550, 277297020, 2344420260, 19927572210, 170150808870, 1458435504600, 12542545339560, 108179453553705, 935434098376155, 8107095519260010, 70403724246205350, 612512400941986545, 5337608065351597035, 46582761297613937760
Offset: 0
Examples
G.f.: 1 + 3*x + 18*x^2 + 126*x^3 + 945*x^4 + 7371*x^5 + 58968*x^6 + 480168*x^7 + ...
Links
- Gheorghe Coserea, Table of n, a(n) for n = 0..200
- A. Bostan, S. Boukraa, J.-M. Maillard, and J.-A. Weil, Diagonals of rational functions and selected differential Galois groups, arXiv preprint arXiv:1507.03227 [math-ph], 2015.
- A. Straub, V. H. Moll, and T. Amdeberhan, The p-adic valuation of k-central binomial coefficients, Acta Arith. 140 (2009) 31-41, eq (1.10).
Crossrefs
Programs
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GAP
List([0..25], n-> 3^n*Product([0..n-1], k-> 3*k+1)/Factorial(n) ); # G. C. Greubel, Aug 22 2019
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Magma
[1] cat [3^n*&*[3*k+1: k in [0..n-1]]/Factorial(n): n in [1..25]]; // G. C. Greubel, Aug 22 2019
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Maple
a:= n-> (3^n/n!)*mul(3*k+1, k=0..n-1); seq(a(n), n=0..25); # G. C. Greubel, Aug 22 2019
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Mathematica
Table[(-9)^n Binomial[-1/3, n], {n, 0, 25}] (* Jean-François Alcover, Sep 28 2016, after Peter Luschny *) -
PARI
a(n) = prod(k=0, n-1, 3*k + 1)*3^n/n! \\ Michel Marcus, Jun 30 2013
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PARI
my(x='x, y='y); R = (1 - 9*x*y) / (1 - 2*x - 3*y + 3*y^2 + 9*x^2*y); diag(n, expr, var) = { my(a = vector(n)); for (i = 1, #var, expr = taylor(expr, var[#var - i + 1], n)); for (k = 1, n, a[k] = expr; for (i = 1, #var, a[k] = polcoeff(a[k], k-1))); return(a); }; diag(20, R, [x,y]) \\ Gheorghe Coserea, Jul 01 2016 -
PARI
Vec((1-9*x+O(x^25))^(-1/3)) \\ yields the same as: apply( {A004987(n)=prod(k=0, n-1, 9*k+3)\n!}, [0..24]) \\ M. F. Hasler, Nov 12 2024 -
Sage
[9^n*rising_factorial(1/3, n)/factorial(n) for n in (0..25)] # G. C. Greubel, Aug 22 2019
Formula
G.f.: (1 - 9*x)^(-1/3).
a(n) = (3^n/n!)*A007559(n), n >= 1, a(0) := 1.
a(n) ~ Gamma(1/3)^-1*n^(-2/3)*3^(2*n)*{1 - 1/9*n^-1 + ...}.
Representation as n-th moment of a positive function on (0, 9): a(n) = Integral_{x=0..9} ( x^n*(1/(Pi*sqrt(3)*6*(x/9)^(2/3)*(1-x/9)^(1/3))) ), n >= 0. This function is the solution of the Hausdorff moment problem on (0, 9) with moments equal to a(n). As a consequence this representation is unique. - Karol A. Penson, Jan 30 2003
D-finite with recurrence: n*a(n) + 3*(2-3*n)*a(n-1)=0. - R. J. Mathar, Jun 07 2013
0 = a(n) * (81*a(n+1) - 15*a(n+2)) + a(n+1) * (-3*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Jan 27 2014
G.f. A(x)=:y satisfies 0 = y'' * y - 4 * y' * y'. - Michael Somos, Jan 27 2014
a(n) = (-9)^n*binomial(-1/3, n). - Peter Luschny, Mar 23 2014
E.g.f.: is the hypergeometric function of type 1F1, in Maple notation hypergeom([1/3], [1], 9*x). - Karol A. Penson, Dec 19 2015
Sum_{n>=0} 1/a(n) = (sqrt(3)*Pi + 3*(12 + log(3)))/32 = 1.3980385924595932... - Ilya Gutkovskiy, Jul 01 2016
Binomial transform of A216316. - Peter Bala, Jul 02 2023
From Peter Bala, Mar 31 2024: (Start)
a(n) = (9^n)*Sum_{k = 0..2*n} (-1)^k*binomial(-1/3, k)* binomial(-1/3, 2*n - k).
(9^n)*a(n) = Sum_{k = 0..2*n} (-1)^k*a(k)*a(2*n-k).
Sum_{k = 0..n} a(k)*a(n-k) = A004988(n).
Sum_{k = 0..2*n} a(k)*a(2*n-k) = 18^n/(2*n)! * Product_{k = 1..n} (6*k - 1)*(3*k - 2). (End)
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^3). - Seiichi Manyama, Jun 20 2025
Extensions
More terms from Ralf Stephan, Mar 13 2004
More terms from Benoit Cloitre, Jun 05 2004
Comments