OFFSET
0,2
COMMENTS
A deck has 4 suits of n cards each. The deck is shuffled and dealt into 4 hands of n cards each. A match occurs for every card in the i-th hand of suit i. a(n) is the number of ways of achieving no matches. The probability of no matches is a(n)/((4n)!/n!^4).
LINKS
Jeremy Tan, Table of n, a(n) for n = 0..200
Shalosh B. Ekhad, Christoph Koutschan and Doron Zeilberger, There are EXACTLY 1493804444499093354916284290188948031229880469556 Ways to Derange a Standard Deck of Cards (ignoring suits) [and many other such useful facts], arXiv:2101.10147 [math.CO], 2021.
Shalosh B. Ekhad, Terms, recurrences and asymptotics for multiset derangements.
S. Even and J. Gillis, Derangements and Laguerre polynomials, Mathematical Proceedings of the Cambridge Philosophical Society, Volume 79, Issue 1, January 1976, pp. 135-143.
B. H. Margolius, The Dinner-Diner Matching Problem, Mathematics Magazine, 76 (2003), pp. 107-118.
FORMULA
a(n) = Integral_{x=0..oo} exp(-x)*L_n(x)^4 dx, where L_n(x) is the Laguerre polynomial of degree n (Even and Gillis).
D-finite with recurrence n^3*(2*n-1)*(5*n-6)*(10*n-13)*a(n) = (8300*n^6 - 37350*n^5 + 66698*n^4 - 60393*n^3 + 29297*n^2 - 7263*n + 738)*a(n-1) - (n-1)*(16300*n^5 - 81500*n^4 + 151553*n^3 - 123364*n^2 + 39501*n - 4338)*a(n-2) + 162*(n-2)^3*(n-1)*(5*n-1)*(10*n-3)*a(n-3) (Ekhad).
a(n) = [(w*x*y*z)^n] ((x+y+z)*(w+y+z)*(w+x+z)*(w+x+y))^n.
a(n) ~ 3^(4*n + 3) / (32 * Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Mar 29 2024
EXAMPLE
There are a(13) = 20342533966643026042641 bridge deals where North, South, East and West are void in clubs, diamonds, hearts and spades, respectively.
MATHEMATICA
Table[Integrate[Exp[-x] LaguerreL[n, x]^4, {x, 0, Infinity}], {n, 0, 16}]
(* or *)
rec = n^3(2n-1)(5n-6)(10n-13) a[n] == (8300n^6-37350n^5+66698n^4-60393n^3+29297n^2-7263n+738) a[n-1] - (n-1)(16300n^5-81500n^4+151553n^3-123364n^2+39501n-4338) a[n-2] + 162(n-2)^3(n-1)(5n-1)(10n-3) a[n-3];
RecurrenceTable[{rec, a[0] == 1, a[1] == 9, a[2] == 297}, a, {n, 0, 16}]
PROG
(Python)
def A371252(n):
l = [1, 9, 297]
for k in range(3, n+1):
m1 = (((((8300*k-37350)*k+66698)*k-60393)*k+29297)*k-7263)*k+738
m2 = (k-1)*(((((16300*k-81500)*k+151553)*k-123364)*k+39501)*k-4338)
m3 = 162*(k-2)**3*(k-1)*(5*k-1)*(10*k-3)
r = (m1*l[-1] - m2*l[-2] + m3*l[-3]) // (k**3*(2*k-1)*(5*k-6)*(10*k-13))
l.append(r)
return l[n]
CROSSREFS
KEYWORD
nonn
AUTHOR
Jeremy Tan, Mar 16 2024
STATUS
approved