cp's OEIS Frontend

A330266 Number of ways to shuffle a deck of 4n cards, with 4 cards in each of n ranks, so that adjacent cards have different ranks.

%I A330266 #33 Oct 29 2023 20:39:33
%S A330266 1,0,1152,15095808,751480602624,93995798935633920,
%T A330266 25111340235557122867200,12742555660097789273088983040,
%U A330266 11259023892340311657074592904642560,16205462460428776872054787528078739374080,36051066700209244649349258741114804984663244800,118807003903158552156678227915553602167323425243136000
%N A330266 Number of ways to shuffle a deck of 4n cards, with 4 cards in each of n ranks, so that adjacent cards have different ranks.
%H A330266 James Enouen, <a href="https://arxiv.org/abs/1911.07426">What is the Perfect Shuffle?</a>, arXiv:1911.07426 [math.CO], 2019.
%F A330266 a(n) = Integral_{x=0..oo} (x^4 - 12x^3 + 36x^2 - 24x)^n*exp(-x) dx.
%F A330266 a(n) = 24^n * A321633(n).
%F A330266 Conjecture: Limit_{n->oo} a(n)/(4n)! = 1/e^3. The conjecture is based on the observation of the asymptotic behavior of A007060 and A193624; it seems that it can be generalized in the following way. Let b(n) be the number of ways to shuffle a deck of k*n cards, with k cards in each of n ranks, so that adjacent cards have different ranks. Then, lim_{n->oo} b(n)/(kn)! = 1/e^(k-1); maybe we could prove it with the help of rook polynomials theory or in some other way. - _Sergey Kirgizov_, Sep 29 2023
%e A330266 a(13) = 3668033946384704437729512814619767610579526911188666362431432294400 is the number of ways to shuffle a standard 52-card deck of playing cards so that no two cards of the same rank are adjacent.
%t A330266 Table[Integrate[(x^4 - 12x^3 + 36x^2 - 24x)^n *Exp[-x],{x,0,Infinity}],{n,0,10}] (* _Stefano Spezia_, Dec 09 2019 *)
%Y A330266 Cf. A007060 (2n cards), A193624 (3n cards).
%K A330266 nonn
%O A330266 0,3
%A A330266 _David Radcliffe_, Dec 07 2019