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 = 2, i.e., permutations with no circular 2-succession, then a(n) counts d*(n+2, 2).
For example, for n = 1, the permutations in S3 that contain the substring {13} in circular 2-succession are 132, 213, 321, therefore d*(3,2) consists of the permutations 123, 231, 312, and a(1) = 3.
LINKS
Enrique Navarrete, Generalized K-Shift Forbidden Substrings in Permutations, arXiv:1610.06217 [math.CO], 2016.
FORMULA
a(n) = (n+2) * Sum_{j = 0..n} (-1)^j * (n - j + 1) * n!/j!.
EXAMPLE
For n = 2, the permutations in S4 that contain the substrings {13, 24} in circular 2-successions are 1243, 1324, 1342, 2134, 2413, 2431, 3124, 3241, 3421, 4132, 4213, 4312, therefore d*(4,2) consists of the complementary permutations in S4, and a(2) = 12.
MATHEMATICA
Table[(n + 2) Sum[(-1)^j * (n - j + 1) * n!/j!, {j, 0, n}], {n, 20}] (* Michael De Vlieger, Apr 05 2017 *)
PROG
(PARI) a(n) = (n + 2) * sum(j=0, n, (-1)^j * (n - j + 1) * n!/j!); \\ Indranil Ghosh, Apr 06 2017
(Python)
from sympy import factorial
print([(n + 2) * sum([(-1)**j * (n - j + 1) * factorial(n)//factorial(j) for j in range(n + 1)]) for n in range(1, 50)]) # Indranil Ghosh, Apr 06 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Enrique Navarrete, Apr 03 2017
STATUS
approved