A152874 Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} with k parity changes (n>=2; 1<=k <=n-1); the permutation 372185946 has 5 parity changes: 37-2-1-8-59-46.
2, 4, 2, 8, 8, 8, 24, 36, 48, 12, 72, 144, 288, 144, 72, 288, 720, 1728, 1296, 864, 144, 1152, 3456, 10368, 10368, 10368, 3456, 1152, 5760, 20160, 69120, 86400, 103680, 51840, 23040, 2880, 28800, 115200, 460800, 691200, 1036800, 691200, 460800, 115200, 28800
Offset: 2
Examples
T(4,3) = 8 because we have 1243, 1423, 4132, 4312, 2134, 2314, 3241 and 3421. Triangle starts: 2; 4, 2; 8, 8, 8; 24, 36, 48, 12; 72, 144, 288, 144, 72; ...
Links
- Alois P. Heinz, Rows n = 2..150, flattened
Programs
-
Maple
ae := proc (n, k) if `mod`(k, 2) = 0 then 2*factorial(n)^2*binomial(n-1, (1/2)*k-1)*binomial(n-1, (1/2)*k) else 2*factorial(n)^2*binomial(n-1, (1/2)*k-1/2)^2 end if end proc: ao := proc (n, k) if `mod`(k, 2) = 0 then factorial(n)*factorial(n+1)*(binomial(n, (1/2)*k)*binomial(n-1, (1/2)*k-1)+binomial(n, (1/2)*k-1)*binomial(n-1, (1/2)*k)) else 2*factorial(n)*factorial(n+1)*binomial(n, (1/2)*k-1/2)*binomial(n-1, (1/2)*k-1/2) end if end proc: T := proc (n, k) if `mod`(n, 2) = 0 then ae((1/2)*n, k) else ao((1/2)*n-1/2, k) end if end proc: for n from 2 to 10 do seq(T(n, k), k = 1 .. n-1) end do; # yields sequence in triangular form # second Maple program: b:= proc(x, y, t) option remember; `if`(x+y=0, 1, `if`(x>0, b(x-1, y, z)*x, 0)+`if`(y>0, expand(b(y-1, x, z)*y*t), 0)) end: T:= n-> (h-> (p-> seq(coeff(p, z, i), i=1..n-1))(b(h, n-h, 1)))(iquo(n, 2)): seq(T(n), n=2..12); # Alois P. Heinz, May 23 2023 -
Mathematica
b[x_, y_, t_] := b[x, y, t] = If[x + y == 0, 1, If[x > 0, b[x - 1, y, z]*x, 0] + If[y > 0, Expand[b[y - 1, x, z]*y*t], 0]]; T[n_] := Table[Coefficient[#, z, i], {i, 1, n-1}]&[b[#, n-#, 1]]&[ Quotient[n, 2]]; Table[T[n], {n, 2, 12}] // Flatten (* Jean-François Alcover, Aug 16 2023, after Alois P. Heinz *)
Formula
T(2n,k) = (n!)^2*a(n,k), where a(n,k) is the number of lattice paths from (0,0) to (n,n) with steps N=(0,1) and E=(1,0) and having k turns;
a(n,k) = 2*binomial(n-1,k/2-1)*binomial(n-1,k/2) if k even;
a(n,k) = 2*(binomial(n-1,(k-1)/2))^2 if k odd.
T(2n+1,k) = n!*(n+1)!*b(n,k), where b(n,k) is the number of lattice paths from (0,0) to (n,n+1) with steps N=(0,1) and E=(1,0) and having k turns;
b(n,k) = binomial(n,k/2)*binomial(n-1,k/2-1) + binomial(n,k/2-1)*binomial(n-1,k/2) = (binomial(n,k/2))^2*k(2n-k+1)/(n(2n-k+2)) if k even;
b(n,k) = 2*binomial(n,(k-1)/2)*binomial(n-1,(k-1)/2) if k odd.
Comments