A284845 Number of permutations on [n+4] with no circular 4-successions.

90, 468, 2982, 22320, 191106, 1838220, 19599822, 229257288, 2917290090, 40107565764, 592302134070, 9349254600288, 157059054215442, 2797498002296700, 52657059745734366, 1044337677676754040, 21765735199891598202, 475573090189331643828, 10870086948032475194310

Offset: 1

Author

Enrique Navarrete, Apr 03 2017

Keywords

• nonn

Comments

Define a circular k-succession in a permutation p on [n] as either a pair p(i),p(i+1) if p(i+1)=p(i)+k, or as the pair p(n),p(1) if p(1)=p(n)+k. If we let d*(n,k) be the number of permutations on [n] that avoid substrings (j,j+k), 1<=j<=n, k=4, i.e., permutations with no circular 4-succession, then a(n) counts d*(n+4,4).

For example, a(1)=90 since there are 90 permutations in S5 with no circular 4-succession, i.e., permutations that avoid the substring {15} such as 15234 or 53241.

Examples

			a(2)=468 since there are 468 permutations in S6 with no circular 4-succession, i.e., permutations that avoid substrings {15,26} such as 261345 or 653142.
		

Links

• Enrique Navarrete, Generalized K-Shift Forbidden Substrings in Permutations, arXiv:1610.06217 [math.CO], 2016.

Programs

Maple

A284845 := proc(n)
    local j;
    add( (-1)^j*binomial(n,j)*(n-j+3)!,j=0..n) ;
    %*(n+4) ;
end proc:
seq(A284845(n),n=1..20) ; # R. J. Mathar, Jul 15 2017

Mathematica

Table[(n + 4) Sum[(-1)^j Binomial[n, j] * (n - j + 3)!, {j, 0, n}], {n,0, 20}] (* or *) Table[(4+n) (3+n)! Hypergeometric1F1[-n,-3-n,-1],{n, 0, 20}] (* Indranil Ghosh, Apr 07 2017 *)

Formula

a(n) = (n+4)* Sum_{j=0..n} (-1)^j*binomial(n,j)*(n-j+3)!.

Conjecture: a(n) = (n+4)*A277609(n+3). - R. J. Mathar, Jul 15 2017