A145894 -id:A145894 - cp's OEIS frontend

A145893 Triangle read by rows: T(n,k) is the number of permutations p of {1,2,...,n} such that j and p(j) are of opposite parities for k values of j (0<=k<=n).

1, 1, 0, 1, 0, 1, 2, 0, 4, 0, 4, 0, 16, 0, 4, 12, 0, 72, 0, 36, 0, 36, 0, 324, 0, 324, 0, 36, 144, 0, 1728, 0, 2592, 0, 576, 0, 576, 0, 9216, 0, 20736, 0, 9216, 0, 576, 2880, 0, 57600, 0, 172800, 0, 115200, 0, 14400, 0, 14400, 0, 360000, 0, 1440000, 0, 1440000, 0, 360000, 0, 14400

Offset: 0

Emeric Deutsch, Nov 30 2008

Mirror image of A145894.
Without the 0's it is the triangle of A145891.
Sum of entries in row n = n! = A000142(n).
Columns k=0,2,4,8,10,12 give: A010551, A226282, A226283, A226284, A226285, A226286. - Alois P. Heinz, May 29 2014

			T(3,2) = 4 because we have 132, 312, 213 and 231.
Triangle starts:
   1;
   1, 0;
   1, 0,   1;
   2, 0,   4, 0;
   4, 0,  16, 0,   4;
  12, 0,  72, 0,  36, 0;
  36, 0, 324, 0, 324, 0, 36;
  ...

Alois P. Heinz, Rows n = 0..140, flattened

Cf. A000142, A145891, A145894.

T:=proc(n,k) if `mod`(n, 2) = 0 and `mod`(k, 2) = 0 then factorial((1/2)*n)^2*binomial((1/2)*n, (1/2)*k)^2 elif `mod`(n, 2) = 1 and `mod`(k, 2) = 0 then factorial((1/2)*n-1/2)*factorial((1/2)*n+1/2)*binomial((1/2)*n-1/2, (1/2)*k)*binomial((1/2)*n+1/2, (1/2)*k) else 0 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

T[n_, k_] := Which[EvenQ[n] && EvenQ[k], (n/2)!^2*Binomial[n/2, k/2]^2, OddQ[n] && EvenQ[k], (n/2-1/2)!*(n/2+1/2)!*Binomial[n/2-1/2, k/2] * Binomial[n/2+1/2, k/2], True, 0]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 20 2017, translated from Maple *)

T(2n,2k) = [n!*C(n,k)]^2; T(2n+1,2k) = n! (n+1)! C(n,k) C(n+1,k); elsewhere T(n,k)=0.

A226282 a(n) = [n/2]![(n+1)/2]!C([n/2],1)*C([(n+1)/2],1).

Original entry on oeis.org

0, 1, 4, 16, 72, 324, 1728, 9216, 57600, 360000, 2592000, 18662400, 152409600, 1244678400, 11379916800, 104044953600, 1053455155200, 10666233446400, 118513704960000, 1316818944000000, 15933509222400000, 192795461591040000, 2523867860828160000, 33039724723568640000

Offset: 1

R. H. Hardin, connection of formula with combinatoric problem via N. J. A. Sloane in the Sequence Fans Mailing List, Jun 02 2013

nonn

Number of permutations of n elements with 2 odd displacements.
Column 2 of A226288.
Column k=2 of A145893, a diagonal of A145894. - Alois P. Heinz, May 29 2014

R. H. Hardin, Table of n, a(n) for n = 1..210

Cf. A145893, A145894, A226288.

Table[Floor[n/2]!Floor[(n+1)/2]!Binomial[Floor[n/2],1]Binomial[Floor[(n+1)/2],1],{n,30}] (* Harvey P. Dale, Aug 05 2023 *)

A226283 a(n) = [n/2]![(n+1)/2]!C([n/2],2)*C([(n+1)/2],2).

Original entry on oeis.org

0, 0, 0, 4, 36, 324, 2592, 20736, 172800, 1440000, 12960000, 116640000, 1143072000, 11202105600, 119489126400, 1274550681600, 14748372172800, 170659735142400, 2133246689280000, 26665583616000000, 358503957504000000, 4819886539776000000, 69406366172774400000

Offset: 1

R. H. Hardin, connection of formula with combinatoric problem via N. J. A. Sloane in the Sequence Fans Mailing List, Jun 02 2013

nonn

Number of permutations of n elements with 4 odd displacements.
Column 3 of A226288.
Column k=4 of A145893, a diagonal of A145894. - Alois P. Heinz, May 29 2014

R. H. Hardin, Table of n, a(n) for n = 1..210

Cf. A145893, A145894, A226288.

a[n_]:=(Floor[n/2])!(Floor[(n+1)/2])!Binomial[Floor[n/2],2]Binomial[Floor[(n+1)/2],2]; Array[a,23] (* Stefano Spezia, Jul 12 2024 *)

A226284 a(n) = [n/2]![(n+1)/2]!C([n/2],3)*C([(n+1)/2],3).

Original entry on oeis.org

0, 0, 0, 0, 0, 36, 576, 9216, 115200, 1440000, 17280000, 207360000, 2540160000, 31116960000, 398297088000, 5098202726400, 68825736806400, 929147446886400, 13273534955520000, 189621927936000000, 2868031660032000000, 43378978857984000000, 694063661727744000000

Offset: 1

R. H. Hardin, connection of formula with combinatoric problem via N. J. A. Sloane in the Sequence Fans Mailing List, Jun 02 2013

nonn

Number of permutations of n elements with 6 odd displacements.
Column 4 of A226288.
Column k=6 of A145893, a diagonal of A145894. - Alois P. Heinz, May 29 2014

R. H. Hardin, Table of n, a(n) for n = 1..210

Cf. A145893, A145894, A226288.

Table[Floor[n/2]!Floor[(n+1)/2]!Binomial[Floor[n/2],3]Binomial[Floor[ (n+1)/2], 3],{n,30}] (* Harvey P. Dale, Dec 12 2018 *)

A226285 a(n) = [n/2]![(n+1)/2]!C([n/2],4)*C([(n+1)/2],4).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 576, 14400, 360000, 6480000, 116640000, 1905120000, 31116960000, 497871360000, 7965941760000, 129048256512000, 2090581755494400, 34843029258240000, 580717154304000000, 10038110810112000000, 173515915431936000000, 3123286477774848000000

Offset: 1

R. H. Hardin, connection of formula with combinatoric problem via N. J. A. Sloane in the Sequence Fans Mailing List, Jun 02 2013

nonn

Number of permutations of n elements with 8 odd displacements.
Column 5 of A226288.
Column k=8 of A145893, a diagonal of A145894. - Alois P. Heinz, May 29 2014

R. H. Hardin, Table of n, a(n) for n = 1..210

Cf. A145893, A145894, A226288.

a[n_]:= Floor[n/2]!*Floor[(n+1)/2]!*Binomial[Floor[n/2],4]*Binomial[Floor[(n+1)/2],4]; Array[a,21] (* Stefano Spezia, Jul 12 2024 *)

A226286 [n/2]![(n+1)/2]!C([n/2],5)*C([(n+1)/2],5).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 14400, 518400, 18662400, 457228800, 11202105600, 238978252800, 5098202726400, 103238605209600, 2090581755494400, 41811635109888000, 836232702197760000, 16864026160988160000

Offset: 1

R. H. Hardin, connection of formula with combinatoric problem via N. J. A. Sloane in the Sequence Fans Mailing List, Jun 02 2013

nonn

Number of permutations of n elements with 10 odd displacements.
Column 6 of A226288.
Column k=10 of A145893, a diagonal of A145894. - Alois P. Heinz, May 29 2014

R. H. Hardin, Table of n, a(n) for n = 1..210

A226287 a(n) = [n/2]![(n+1)/2]!C([n/2],6)*C([(n+1)/2],6).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 518400, 25401600, 1244678400, 39829708800, 1274550681600, 34412868403200, 929147446886400, 23228686172160000, 580717154304000000, 14053355134156800000, 340091194246594560000

Offset: 1

R. H. Hardin, connection of formula with combinatoric problem via N. J. A. Sloane in the Sequence Fans Mailing List, Jun 02 2013

nonn

Number of permutations of n elements with 12 odd displacements.
Column 7 of A226288.
Column k=12 of A145893, a diagonal of A145894. - Alois P. Heinz, May 29 2014

R. H. Hardin, Table of n, a(n) for n = 1..210

cp's OEIS Frontend

A145893 Triangle read by rows: T(n,k) is the number of permutations p of {1,2,...,n} such that j and p(j) are of opposite parities for k values of j (0<=k<=n).

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A226282 a(n) = [n/2]!*[(n+1)/2]!*C([n/2],1)*C([(n+1)/2],1).

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A226283 a(n) = [n/2]!*[(n+1)/2]!*C([n/2],2)*C([(n+1)/2],2).

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A226284 a(n) = [n/2]!*[(n+1)/2]!*C([n/2],3)*C([(n+1)/2],3).

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A226285 a(n) = [n/2]!*[(n+1)/2]!*C([n/2],4)*C([(n+1)/2],4).

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A226286 [n/2]!*[(n+1)/2]!*C([n/2],5)*C([(n+1)/2],5).

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A226287 a(n) = [n/2]!*[(n+1)/2]!*C([n/2],6)*C([(n+1)/2],6).

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A226282 a(n) = [n/2]![(n+1)/2]!C([n/2],1)*C([(n+1)/2],1).

A226283 a(n) = [n/2]![(n+1)/2]!C([n/2],2)*C([(n+1)/2],2).

A226284 a(n) = [n/2]![(n+1)/2]!C([n/2],3)*C([(n+1)/2],3).

A226285 a(n) = [n/2]![(n+1)/2]!C([n/2],4)*C([(n+1)/2],4).

A226286 [n/2]![(n+1)/2]!C([n/2],5)*C([(n+1)/2],5).

A226287 a(n) = [n/2]![(n+1)/2]!C([n/2],6)*C([(n+1)/2],6).