A180188 - OEIS

%I #18 Feb 19 2017 06:55:19

%S 1,0,2,3,0,3,8,12,0,4,45,40,30,0,5,264,270,120,60,0,6,1855,1848,945,

%T 280,105,0,7,14832,14840,7392,2520,560,168,0,8,133497,133488,66780,

%U 22176,5670,1008,252,0,9,1334960,1334970,667440,222600,55440,11340,1680,360,0

%N 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.

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

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

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

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

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

%H Alois P. Heinz, <a href="/A180188/b180188.txt">Rows n = 1..141, flattened</a>

%H S. M. Tanny, <a href="http://dx.doi.org/10.1016/0097-3165(76)90063-7">Permutations and successions</a>, J. Combinatorial Theory, Series A, 21 (1976), 196-202.

%F 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).

%F T(n,k) = n*A008290(n-1,k), 0<=k<n, n>=1. - _R. J. Mathar_, Sep 08 2013

%e T(3,2) = 3 because we have 123, 312, and 231.

%e The triangle starts:

%e 1;

%e 0,   2;

%e 3,   0,  3;

%e 8,  12,  0, 4;

%e 45, 40, 30, 0, 5;

%p A180188 := proc (n, k) n*binomial(n-1, k)*A000166(n-1-k) end proc:

%p for n to 10 do seq(A180188(n, k), k = 0 .. n-1) end do; # yields sequence in triangular form

%t 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 *)

%Y Cf. A000142, A000166, A000240, A180189.

%K nonn,tabl

%O 1,3

%A _Emeric Deutsch_, Sep 06 2010