A064352 - OEIS

%I #42 Apr 02 2024 16:06:53

%S 1,3,30,504,11880,360360,13366080,586051200,29654190720,1700755056000,

%T 109027350432000,7725366544896000,599555620984320000,

%U 50578512186237235200,4608264443634948096000,450974292794344230912000

%N a(n) = (3*n)!/(2*n)!.

%H Harry J. Smith, <a href="/A064352/b064352.txt">Table of n, a(n) for n = 0..100</a>

%H Guillaume Chapuy, <a href="https://arxiv.org/abs/2403.18719">On the scaling of random Tamari intervals and Schnyder woods of random triangulations (with an asymptotic D-finite trick)</a>, arXiv:2403.18719 [math.CO], 2024.

%H Karol A. Penson and Allan I. Solomon, <a href="https://doi.org/10.1142/9789812777850_0066">Coherent states from combinatorial sequences</a>, in: E. Kapuscik and A. Horzela (eds.), Quantum theory and symmetries, World Scientific, 2002, pp. 527-530; <a href="https://arxiv.org/abs/quant-ph/0111151">arXiv preprint</a>, arXiv:quant-ph/0111151, 2001.

%F 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))).

%F 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

%F a(n) = n! * [x^n] 1/(1 - x)^(2*n+1). - _Ilya Gutkovskiy_, Jan 23 2018

%F Sum_{n>=1} 1/a(n) = A248760. - _Amiram Eldar_, Nov 15 2020

%t Array[(3 #)!/(2 #)! &, 16, 0] (* _Michael De Vlieger_, Jan 13 2018 *)

%o (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

%o (Sage)

%o [falling_factorial(3*n, n) for n in (0..15)] # _Peter Luschny_, Jan 13 2018

%Y Cf. A001813, A166338, A248760.

%K nonn

%O 0,2

%A _Karol A. Penson_, Sep 19 2001

%E a(15) from _Harry J. Smith_, Sep 12 2009