OFFSET
1,1
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.
LINKS
Enrique Navarrete, Generalized K-Shift Forbidden Substrings in Permutations, arXiv:1610.06217 [math.CO], 2016.
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
EXAMPLE
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.
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Enrique Navarrete, Apr 03 2017
STATUS
approved