A076731 Table T(n,k) giving number of ways of obtaining exactly 0 correct answers on an (n,k)-matching problem (1 <= k <= n).
0, 1, 1, 2, 3, 2, 3, 7, 11, 9, 4, 13, 32, 53, 44, 5, 21, 71, 181, 309, 265, 6, 31, 134, 465, 1214, 2119, 1854, 7, 43, 227, 1001, 3539, 9403, 16687, 14833, 8, 57, 356, 1909, 8544, 30637, 82508, 148329, 133496, 9, 73, 527, 3333, 18089, 81901, 296967, 808393
Offset: 1
Examples
0; 1,1; 2,3,2; 3,7,11,9; ... Formatted as a square array: 0 1 2 3 4 5 6 7 8 1 3 7 13 21 31 43 57 which equals A002061 2 11 32 71 134 227 356 which equals A094792 9 53 181 465 1001 1909 which equals A094793 44 309 1214 3539 8544 which equals A094794 265 2119 9403 30637 which equals A023043 1854 16687 82508 which equals A023044 14833 148329 which equals A023045 Columns give A000255 A000153 A000261 A001909 A001910 Formatted as a triangular array (mirror image of A086764): 0 1 1 2 3 2 3 7 11 9 4 13 32 53 44 5 21 71 181 309 265 6 31 134 465 1214 2119 1854 7 43 227 1001 3539 9403 16687 14833 8 57 356 1909 8544 30637 82508 148329 133496
Links
- D. Hanson, K. Seyffarth and J. H. Weston, Matchings, Derangements, Rencontres, Mathematics Magazine, Vol. 56, No. 4, September 1983.
- StackExchange, How many injective functions f:[1,...,m]->[1,...,n] have no fixed point?
Programs
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Maple
A076731 := proc(n,k): (1/(n-k)!)*A061312(n-1,k-1) end: A061312:=proc(n,k): add(((-1)^j)*binomial(k+1,j)*(n+1-j)!, j=0..k+1) end: for n from 1 to 7 do seq(A076731(n,k), k=1..n) od; seq(seq(A076731(n,k), k=1..n), n=1..9); # Johannes W. Meijer, Jul 27 2011
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Mathematica
t[n_,k_] := k!(n - k)! SeriesCoefficient[Exp[z(1-u+u^2z)/(1-z u)]/(1-z u), {z,0,n}, {u,0,k}]; Table[t[n,k], {n,9}, {k,n}] //TableForm (* David Bevan, Apr 29 2013 *) t[n_, k_] := Pochhammer[n-k+1, k]*Hypergeometric1F1[-k, -n, -1]; Table[t[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Nov 29 2013 *)
Formula
T(n,k) = F(n,k)*Sum{((-1)^j)*C(k, j)*(n-j)! (j=0 to k)}, where F(n,k) = 1/(n-k)! for 1 <= k <= n.
From Johannes W. Meijer, Jul 27 2011: (Start)
T(n,k) = (n-1)*T(n-1,k-1) + (k-1)*T(n-2,k-2) with T(n,1) = (n-1) and T(n,n) = A000166(n) [Hanson et al.]
T(n,k) = (1/(n-k)!)*A061312(n-1,k-1)
sum(T(n,k), k=1..n) = A193464(n); row sums. (End)
T(n,k) = k!(n-k)![z^n*u^k]J(z,u) where J(z,u) = exp(z(1-u+z*u^2)/(1-z*u))/(1-z*u) is the exponential generating function of labeled digraphs consisting just of directed paths and oriented cycles (of length at least 2), z marking the vertices and u the edges; [z^n*u^k]J(z,u) is the coefficient of z^n*u^k in J(z,u). - David Bevan, Apr 29 2013
Extensions
Additional comments from Zerinvary Lajos, Mar 30 2006
Comments