A293653 - OEIS

%I #58 Jul 16 2023 16:54:55

%S 1,2,6,30,180,1440,12960,142560,1710720,23950080,359251200,6107270400,

%T 109930867200,2198617344000,46170964224000,1061932177152000,

%U 25486372251648000,662645678542848000,17891433320656896000,518851566299049984000,15565546988971499520000,498097503647087984640000

%N Young urn sequence (number of possible evolutions in n steps of the "Young" Pólya urn).

%H Robert G. Wilson v, <a href="/A293653/b293653.txt">Table of n, a(n) for n = 0..422</a>

%H Cyril Banderier, Philippe Marchal, and Michael Wallner, <a href="http://lipn.univ-paris13.fr/~banderier/Papers/urns.pdf">Periodic Pólya urns and an application to Young tableaux</a>, Leibniz International Proceedings in Informatics (LIPIcs), 29th International Conference on Probabilistic, Combinatorial and Asymptotic Methods for the Analysis of Algorithms (AofA 2018), 1-12, also <a href="https://arxiv.org/abs/1806.03133">arXiv:1806.03133</a> [cs.DM], 2018.

%H Cyril Banderier, Philippe Marchal, Michael Wallner, <a href="https://arxiv.org/abs/1912.01035">Periodic Pólya Urns, the Density Method, and Asymptotics of Young Tableaux</a>, arXiv:1912.01035 [math.PR], 2019.

%F a(n+2) = (3/2)*a(n+1) + (9*n^2+21*n+12)/4*a(n), a(0) = 1, a(1) = 2.

%F Also, a(2n+k) has a hypergeometric expression (for k=0,1, see Maple code below) (proved).

%e We based the following explanations on Figure 1 from the Banderier et al. reference:

%e We have an urn with black and white balls in it.

%e At odd-numbered steps, we apply rule M1: we choose a ball, check its color, and add to the urn a ball of the same color.

%e At even-numbered steps, we apply rule M2: we choose a ball and check its color;

%e   if it is black, we add 1 white ball and 1 black ball;

%e   if it is white, we add 2 white balls.

%e At step 0, we start with the urn containing 1 black ball and 1 white ball (in short, 1b/1w). Accordingly, a(0)=1.

%e At step 1, we apply rule M1. This leads to 2 compositions: 2b/1w and 1b/2w. Accordingly, a(1) = 1 + 1 = 2.

%e At step 2, we apply rule M2. This leads to 3 compositions: 3b/2w, 2b/3w, and 1b/4w, each having multiplicity two. Accordingly, a(2) = 2 + 2 + 2 = 6. This leads to 4 compositions at step 3: 4b/2w (with multiplicity 6), and 3b/3w, 2b/4w, and 1b/5w (each of these last three with multiplicity 8). Accordingly, a(3) = 6 + 8 + 8 + 8 = 30.

%p a:=proc(n) if n mod 2= 0 then 3^n /GAMMA(2/3) * GAMMA(n/2+2/3) * GAMMA(n/2+1);

%p else 3^n /GAMMA(2/3) * GAMMA(n/2+7/6) * GAMMA(n/2+1/2);  fi; end:

%p seq(a(n), n=0..30);

%t f[n_] := FullSimplify[ 3^n/Gamma[2/3]*If[OddQ@ n, Gamma[n/2 + 7/6] Gamma[n/2 + 1/2], Gamma[n/2 + 2/3]Gamma[n/2 + 1]]]; Array[f, 20, 0] (* _Robert G. Wilson v_, Feb 07 2018 *)

%o (GAP) a:=[1,2];; for n in [3..25] do a[n]:=(3/2)*a[n-1]+(9*(n-3)^2+21*n-51)/4*a[n-2]; od; a; # _Muniru A Asiru_, Dec 19 2018

%Y Cf. A123144.

%K nonn

%O 0,2

%A _Cyril Banderier_, Feb 06 2018