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A059056 Penrice Christmas gift numbers, Card-matching numbers (Dinner-Diner matching numbers): Triangle T(n,k) = number of ways to get k matches for a deck with n cards, 2 of each kind. 22
1, 0, 0, 1, 1, 0, 4, 0, 1, 10, 24, 27, 16, 12, 0, 1, 297, 672, 736, 480, 246, 64, 24, 0, 1, 13756, 30480, 32365, 21760, 10300, 3568, 970, 160, 40, 0, 1, 925705, 2016480, 2116836, 1418720, 677655, 243360, 67920, 14688, 2655, 320, 60, 0, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,7
COMMENTS
This is a triangle of card matching numbers. A deck has n kinds of cards, 2 of each kind. The deck is shuffled and dealt in to n hands with 2 cards each. A match occurs for every card in the j-th hand of kind j. Triangle T(n,k) is the number of ways of achieving exactly k matches (k=0..2n). The probability of exactly k matches is T(n,k)/((2n)!/2^n).
Rows are of length 1,3,5,7,... = A005408(n). [Edited by M. F. Hasler, Sep 21 2015]
Analogous to A008290. - Zerinvary Lajos, Jun 10 2005
REFERENCES
F. N. David and D. E. Barton, Combinatorial Chance, Hafner, NY, 1962, Ch. 7 and Ch. 12.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 174-178.
R. P. Stanley, Enumerative Combinatorics Volume I, Cambridge University Press, 1997, p. 71.
LINKS
F. F. Knudsen and I. Skau, On the Asymptotic Solution of a Card-Matching Problem, Mathematics Magazine 69 (1996), 190-197.
B. H. Margolius, The Dinner-Diner Matching Problem, Mathematics Magazine, 76 (2003), 107-118.
S. G. Penrice, Derangements, permanents and Christmas presents, The American Mathematical Monthly 98(1991), 617-620.
FORMULA
G.f.: sum(coeff(R(x, n, k), x, j)*(t-1)^j*(n*k-j)!, j=0..n*k) where n is the number of kinds of cards, k is the number of cards of each kind (here k is 2) and R(x, n, k) is the rook polynomial given by R(x, n, k)=(k!^2*sum(x^j/((k-j)!^2*j!))^n (see Stanley or Riordan). coeff(R(x, n, k), x, j) indicates the j-th coefficient on x of the rook polynomial.
EXAMPLE
There are 4 ways of matching exactly 2 cards when there are 2 different kinds of cards, 2 of each in each of the two decks so T(2,2)=4.
Triangle begins:
1
"0", 0, 1
1, '0', "4", 0, 1
10, 24, 27, '16', "12", 0, 1
297, 672, 736, 480, 246, '64', "24", 0, 1
13756, 30480, 32365, 21760, 10300, 3568, 970, '160', "40", 0, 1
925705, 2016480, 2116836, 1418720, 677655, 243360, 67920, 14688, 2655, '320', "60", 0, 1
Diagonal " ": T(n,2n-2) = 0, 4, 12, 24, 40, 60, 84, 112, 144, ... equals A046092
Diagonal ' ': T(n,2n-3) = 0, 16, 64, 160, 320, 560, 896, 1344, ... equals A102860
MAPLE
p := (x, k)->k!^2*sum(x^j/((k-j)!^2*j!), j=0..k); R := (x, n, k)->p(x, k)^n; f := (t, n, k)->sum(coeff(R(x, n, k), x, j)*(t-1)^j*(n*k-j)!, j=0..n*k);
for n from 0 to 7 do seq(coeff(f(t, n, 2), t, m)/2^n, m=0..2*n); od;
MATHEMATICA
p[x_, k_] := k!^2*Sum[ x^j/((k-j)!^2*j!), {j, 0, k}];
R[x_, n_, k_] := p[x, k]^n;
f[t_, n_, k_] := Sum[ Coefficient[ R[x, n, k], x, j]*(t-1)^j*(n*k-j)!, {j, 0, n*k}];
Table[ Coefficient[ f[t, n, 2]/2^n, t, m], {n, 0, 6}, {m, 0, 2*n}] // Flatten
(* Jean-François Alcover, Sep 17 2012, translated from Maple *)
CROSSREFS
Sequence in context: A342372 A289222 A121301 * A344393 A127153 A178979
KEYWORD
nonn,tabf,nice
AUTHOR
Barbara Haas Margolius (margolius(AT)math.csuohio.edu)
EXTENSIONS
Additional comments from Zerinvary Lajos, Jun 18 2007
Edited by M. F. Hasler, Sep 21 2015
STATUS
approved

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Last modified April 27 19:34 EDT 2024. Contains 372020 sequences. (Running on oeis4.)