A161119 Triangle read by rows: T(n,k) is the number of fixed-point-free involutions of {1,2,...,2n} having k cycles with entries of opposite parities (0 <= k <= n).
1, 0, 1, 1, 0, 2, 0, 9, 0, 6, 9, 0, 72, 0, 24, 0, 225, 0, 600, 0, 120, 225, 0, 4050, 0, 5400, 0, 720, 0, 11025, 0, 66150, 0, 52920, 0, 5040, 11025, 0, 352800, 0, 1058400, 0, 564480, 0, 40320, 0, 893025, 0, 9525600, 0, 17146080, 0, 6531840, 0, 362880, 893025, 0, 44651250, 0, 238140000, 0, 285768000, 0, 81648000, 0, 3628800
Offset: 0
Examples
T(3,1)=9 because we have (12)(35)(46), (14)(26)(35), (16)(24)(35), (23)(15)(46), (25)(13)(46), (34)(15)(26), (36)(15)(24), (45)(13)(26), (56)(13)(24). Triangle starts: 1; 0, 1; 1, 0, 2; 0, 9, 0, 6; 9, 0, 72, 0, 24;
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50)
Programs
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Maple
T := proc (n, k) if `mod`(n-k, 2) = 1 then 0 else binomial(n, k)^2*factorial(k)*(product(2*j-1, j = 1 .. (1/2)*n-(1/2)*k))^2 end if end proc: for n from 0 to 10 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form
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PARI
dfo(n) = if (n<0, (-1)^n/dfo(-n), (2*n)! / n! / 2^n); \\ A001147 T(n,k) = if ((n-k)%2, 0, k!*binomial(n,k)^2*dfo((n-k)/2)^2); row(n) = vector(n+1, k, T(n,k-1)) \\ Michel Marcus, Dec 09 2021
Comments