

A284843


Number of permutations on [n+2] with no circular 2successions.


0



3, 12, 55, 318, 2163, 16952, 150183, 1483290, 16153027, 192234372, 2481692343, 34538108822, 515428409715, 8210201868528, 139028223047143, 2493848817565362, 47236883312958723, 942120664413581180, 19735072617803491767, 433187081497364626542
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OFFSET

1,1


COMMENTS

Define a circular ksuccession 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 2succession, then a(n) counts d*(n+2, 2).
For example, for n = 1, the permutations in S3 that contain the substring {13} in circular 2succession are 132, 213, 321, therefore d*(3,2) consists of the permutations 123, 231, 312, and a(1) = 3.


LINKS

Table of n, a(n) for n=1..20.
Enrique Navarrete, Generalized KShift 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 2successions 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

Sequence in context: A263533 A064314 A185630 * A107318 A176281 A050147
Adjacent sequences: A284840 A284841 A284842 * A284844 A284845 A284846


KEYWORD

nonn


AUTHOR

Enrique Navarrete, Apr 03 2017


STATUS

approved



