A161122 -id:A161122 - cp's OEIS frontend

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

Emeric Deutsch, Jun 02 2009

T(n,k) is the number of basis elements in the order-n Brauer algebra that have propagation number k. - John M. Campbell, Dec 08 2021

			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;

Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50)

Cf. A000142, A001147, A001818, A161120, A161121, A161122.

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

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

T(n,k) = k!*binomial(n,k)^2*(n-k-1)!!^2 if n-k is even; T(n,k) = 0 if n-k is odd.
Sum of row n = (2n-1)!! = A001147(n).
T(n,n) = n! = A000142(n).
T(2n,0) = A001818(n).
Sum_{k>=0} k*T(n,k) = n^2*(2n-3)!! = A161120(n).

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

Emeric Deutsch, Jun 02 2009

nonn

			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.

Cf. A161119, A161121, A161122.

seq(n^2*(product(2*j-1, j = 1 .. n-1)), n = 0 .. 18);

a(n)=n^2*(2n-3)!!
a(n)=Sum(k*A161119(n,k), k=0..n)
E.g.f.: x(1-x)/(1-2x)^(3/2). [From Paul Barry, Sep 13 2010]
E.g.f.: x/2*U(0)  where U(k)= 1 + (2*k+1)/(1 - x/(x + (k+1)/U(k+1))) ; (continued fraction, 3-step). - Sergei N. Gladkovskii, Sep 25 2012
D-finite with recurrence a(n) +(-2*n-1)*a(n-1) +a(n-2) +(-2*n+7)*a(n-3)=0. - R. J. Mathar, Jul 26 2022

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

Emeric Deutsch, Jun 02 2009

Row n contains 1 + 2*floor(n/2) terms.
Sum of row n = (2n-1)!! (A001147).
a(n,0) = n! (A000142).
a(2n,2n) = A001818(n).
Sum_{k>=0} k*T(n,k) = n*(n-1)*(2n-3)!! = A161122(n).

			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;

Robert Israel, Table of n, a(n) for n = 0..10081 (rows 0 to 141, flattened)

Cf. A000142, A001147, A001818, A161119, A161120, A161122.

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

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 *)

T(n,k) = (n-k)!*binomial(n,k)^2*((k-1)!!)^2 if k is even; T(n,k) = 0 if k is odd.

cp's OEIS Frontend

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).

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A161120 Number of cycles with entries of opposite parities in all fixed-point-free involutions of {1,2,...,2n}.

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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)).

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