A180188 Triangle read by rows: T(n,k) is the number of permutations of [n] with k circular successions (0<=k<=n-1). A circular succession in a permutation p of [n] is either a pair p(i), p(i+1), where p(i+1)=p(i)+1 or the pair p(n), p(1) if p(1)=p(n)+1.

1, 0, 2, 3, 0, 3, 8, 12, 0, 4, 45, 40, 30, 0, 5, 264, 270, 120, 60, 0, 6, 1855, 1848, 945, 280, 105, 0, 7, 14832, 14840, 7392, 2520, 560, 168, 0, 8, 133497, 133488, 66780, 22176, 5670, 1008, 252, 0, 9, 1334960, 1334970, 667440, 222600, 55440, 11340, 1680, 360, 0

Offset

1,3

Comments

For example, p=(4,1,2,5,3) has 2 circular successions: (1,2) and (3,4).

Sum of entries in row n = n! = A000142(n).

T(n,0)=nd(n-1)=A000240(n).

T(n,1)=n(n-1)d(n-2)=A180189(n).

Sum(k*T(n,k), k>=0)=n! = A000142(n) if n>=2.

Links

Alois P. Heinz, Rows n = 1..141, flattened

S. M. Tanny, Permutations and successions, J. Combinatorial Theory, Series A, 21 (1976), 196-202.

Formula

T(n,k) = n*C(n-1,k)*d(n-1-k), where d(j) = A000166(j) are the derangement numbers (see Prop. 1 of the Tanny reference).

T(n,k) = n*A008290(n-1,k), 0<=k<n, n>=1. - R. J. Mathar, Sep 08 2013

Example

T(3,2) = 3 because we have 123, 312, and 231.
The triangle starts:
1;
0,   2;
3,   0,  3;
8,  12,  0, 4;
45, 40, 30, 0, 5;

Maple

A180188 := proc (n, k) n*binomial(n-1, k)*A000166(n-1-k) end proc:
for n to 10 do seq(A180188(n, k), k = 0 .. n-1) end do; # yields sequence in triangular form

Mathematica

T[n_, k_] := n*Binomial[n-1, k]*Subfactorial[n-1-k]; Table[T[n, k], {n, 0, 10}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Feb 19 2017 *)

Crossrefs

Cf. A000142, A000166, A000240, A180189.

Sequence in context: A035549 A329970 A245255 * A316607 A194365 A216217

Adjacent sequences: A180185 A180186 A180187 * A180189 A180190 A180191

Keyword

nonn,tabl

Author

Emeric Deutsch, Sep 06 2010

Status

approved