A145892 Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k adjacent pairs of the form (even,even) (0<=k<=floor(n/2)-1).
1, 1, 2, 6, 12, 12, 72, 48, 144, 432, 144, 1440, 2880, 720, 2880, 17280, 17280, 2880, 43200, 172800, 129600, 17280, 86400, 864000, 1728000, 864000, 86400, 1814400, 12096000, 18144000, 7257600, 604800, 3628800, 54432000, 181440000, 181440000, 54432000, 3628800
Offset: 0
Examples
T(4,1) = 12 because we have 1243, 1423, 1324, 1342, 3124, 3142, 2413, 4213, 2431, 4231, 3241 and 3421. Triangle starts: 1; 1; 2; 6; 12, 12; 72, 48; 144, 432, 144; 1440, 2880, 720; ...
Programs
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Maple
T:=proc(n,k) if `mod`(n, 2) = 0 then factorial((1/2)*n)^2*binomial((1/2)*n-1, k)*binomial((1/2)*n+1, k+1) else factorial((1/2)*n-1/2)*factorial((1/2)*n+1/2)*binomial((1/2)*n-3/2, k)*binomial((1/2)* n+3/2, k+2) end if end proc: 1; 1; for n from 2 to 12 do seq(T(n, k), k = 0 .. floor((1/2)*n)-1) end do; # yields sequence in triangular form
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Mathematica
T[n_,k_]:=If[EvenQ[n],((n/2)!)^2Binomial[n/2-1,k]Binomial[n/2+1,k+1], ((n-1)/2)!((n+1)/2)!Binomial[(n-3)/2,k]Binomial[(n+3)/2,k+2]]; Join[{1,1},Flatten[Table[T[n,k],{n,0,12},{k,0,Floor[n/2]-1}]]] (* Stefano Spezia, Jul 12 2024 *)
Formula
T(2n,k) = (n!)^2*C(n-1,k)*C(n+1,k+1); T(2n+1,k) = n!(n+1)! * C(n-1,k) * C(n+2,k+2).
Comments