A064352 a(n) = (3*n)!/(2*n)!.
1, 3, 30, 504, 11880, 360360, 13366080, 586051200, 29654190720, 1700755056000, 109027350432000, 7725366544896000, 599555620984320000, 50578512186237235200, 4608264443634948096000, 450974292794344230912000
Offset: 0
Links
- Harry J. Smith, Table of n, a(n) for n = 0..100
- Guillaume Chapuy, On the scaling of random Tamari intervals and Schnyder woods of random triangulations (with an asymptotic D-finite trick), arXiv:2403.18719 [math.CO], 2024.
- Karol A. Penson and Allan I. Solomon, Coherent states from combinatorial sequences, in: E. Kapuscik and A. Horzela (eds.), Quantum theory and symmetries, World Scientific, 2002, pp. 527-530; arXiv preprint, arXiv:quant-ph/0111151, 2001.
Programs
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Mathematica
Array[(3 #)!/(2 #)! &, 16, 0] (* Michael De Vlieger, Jan 13 2018 *)
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PARI
{ f3=f2=1; for (n=0, 100, if (n, f3*=3*n*(3*n - 1)*(3*n - 2); f2*=2*n*(2*n - 1)); write("b064352.txt", n, " ", f3/f2) ) } \\ Harry J. Smith, Sep 12 2009 -
Sage
[falling_factorial(3*n, n) for n in (0..15)] # Peter Luschny, Jan 13 2018
Formula
Integral representation as n-th moment of a positive function on a positive half-axis: a(n) = Integral_{x>=0} (x^n*exp(-2*x/27)*(BesselK(1/3, 2*x/27) + BesselK(2/3, 2*x/27))*(sqrt(3)/(27*Pi))).
From Carleman's criterion Sum_{n>=1} a(n)^(-1/(2*n)) = infinity the above solution of the Stieltjes moment problem is unique. - Karol A. Penson, Jan 13 2018
a(n) = n! * [x^n] 1/(1 - x)^(2*n+1). - Ilya Gutkovskiy, Jan 23 2018
Sum_{n>=1} 1/a(n) = A248760. - Amiram Eldar, Nov 15 2020
Extensions
a(15) from Harry J. Smith, Sep 12 2009