A161120
Number of cycles with entries of opposite parities in all fixed-point-free involutions of {1,2,...,2n}.
Original entry on oeis.org
0, 1, 4, 27, 240, 2625, 34020, 509355, 8648640, 164189025, 3445942500, 79222218075, 1979900722800, 53443570205025, 1549547301802500, 48028060502296875, 1584712538529120000, 55458748565165570625
Offset: 0
a(2)=4 because in the 3 fixed-point-free involutions of {1,2,3,4}, namely (12)(34), (13)(24), (14)(23), we have a total of 4 cycles with entries of opposite parities.
A161121
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 the same parity (0 <= k <= 2*floor(n/2)).
Original entry on oeis.org
1, 1, 2, 0, 1, 6, 0, 9, 24, 0, 72, 0, 9, 120, 0, 600, 0, 225, 720, 0, 5400, 0, 4050, 0, 225, 5040, 0, 52920, 0, 66150, 0, 11025, 40320, 0, 564480, 0, 1058400, 0, 352800, 0, 11025, 362880, 0, 6531840, 0, 17146080, 0, 9525600, 0, 893025, 3628800, 0, 81648000, 0
Offset: 0
T(3,2)=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;
1;
2, 0, 1;
6, 0, 9;
24, 0, 72, 0, 9;
120, 0, 600, 0, 225;
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T := proc (n, k) if n < k then 0 elif `mod`(k, 2) = 0 then binomial(n, k)^2*factorial(n-k)*(product(2*j-1, j = 1 .. (1/2)*k))^2 else 0 end if end proc: for n from 0 to 10 do seq(T(n, k), k = 0 .. 2*floor((1/2)*n)) end do; # yields sequence in triangular form
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T[n_, k_] := If[EvenQ[k], (n-k)! Binomial[n, k]^2 ((k-1)!!)^2, 0];
Table[T[n, k], {n, 0, 10}, {k, 0, 2 Quotient[n, 2]}] // Flatten (* Jean-François Alcover, Feb 01 2023 *)
A161122
Number of cycles with entries of the same parity in all fixed-point-free involutions of {1,2,...,2n}.
Original entry on oeis.org
0, 0, 2, 18, 180, 2100, 28350, 436590, 7567560, 145945800, 3101348250, 72020198250, 1814908995900, 49332526343100, 1438865351673750, 44826189802143750, 1485668004871050000, 52196469237802890000, 1937793920453432291250, 75801938653031321981250, 3116301922402398792562500
Offset: 0
a(2)=2 because in the 3 permutations (12)(34), (13)(24), (14)(23) we have a total of 2 cycles with entries of the same parity.
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DoubleFactorial:=func< n | &*[n..2 by -2] >; [ n*(n-1)*DoubleFactorial(2*n-3): n in [0..22]]; // Vincenzo Librandi, Jul 21 2017
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seq(n*(n-1)*(product(2*j-1, j = 1 .. n-1)), n = 0 .. 18);
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Table[n (n - 1) (2 n -3)!!, {n, 0, 20}] (* Vincenzo Librandi, Jul 21 2017 *)
Showing 1-3 of 3 results.
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