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.

1, 0, 1152, 15095808, 751480602624, 93995798935633920, 25111340235557122867200, 12742555660097789273088983040, 11259023892340311657074592904642560, 16205462460428776872054787528078739374080, 36051066700209244649349258741114804984663244800, 118807003903158552156678227915553602167323425243136000

Offset: 0

Author

David Radcliffe, Dec 07 2019

Keywords

• nonn

Examples

			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.
		

Links

• James Enouen, What is the Perfect Shuffle?, arXiv:1911.07426 [math.CO], 2019.

Crossrefs

Cf. A007060 (2n cards), A193624 (3n cards).

Programs

Mathematica

Table[Integrate[(x^4 - 12x^3 + 36x^2 - 24x)^n *Exp[-x],{x,0,Infinity}],{n,0,10}] (* Stefano Spezia, Dec 09 2019 *)

Formula

a(n) = Integral_{x=0..oo} (x^4 - 12x^3 + 36x^2 - 24x)^n*exp(-x) dx.

a(n) = 24^n * A321633(n).

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