

A324134


Number of permutations of [n] that avoid the shuffle pattern skt, where s = 12 and t = 132.


1



1, 1, 2, 6, 24, 120, 710, 4800, 36298, 302780, 2758618, 27246450, 289962508, 3308024082, 40278949800, 521427324542, 7152011191362, 103621538280688, 1581465201545374, 25361207137790358, 426374509273382756, 7499269147438400178, 137728268057069904088, 2636572230825216681414
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OFFSET

0,3


LINKS

Table of n, a(n) for n=0..23.
Sergey Kitaev, Partially Ordered Generalized Patterns, Discrete Math. 298 (2005), no. 13, 212229.


FORMULA

Let b(n) = A111004(n) = number of permutations avoiding a consecutive 132 pattern. Then a(n) = Sum_{i = 0..n1} binomial(n1,i) * (a(n1i) + b(i) * a(n1i)  b(n1i)) for n >= 1 with a(0) = b(0) = 1. [See the recurrence for C_n on p. 220 of Kitaev (2005).]  Petros Hadjicostas, Oct 30 2019


EXAMPLE

From Petros Hadjicostas, Oct 31 2019: (Start)
In a permutation of [n] that contains the shuffle pattern skt, where s = 12 and t = 132, k should be greater than the numbers in pattern s and the numbers in pattern t. (The numbers in each of the patterns s and t should be contiguous.) Clearly, for n = 0..5, all permutations of [n] avoid this shuffle pattern (since we need at least six numbers to get this pattern). Hence, a(n) = n! for n = 0..5.
For n = 6, k should be equal to 6, and for the pattern s = 12 we have the 10 choices 12, 13, 14, 15, 23, 24, 25, 34, 35, and 45. The corresponding permutations of [6] that contain this shuffle pattern are 126354, 136254, 146253, 156243, 236154, 246153, 256143, 346152, 356142, and 456132. Thus, a(6) = 6!  10 = 710. (End)


CROSSREFS

Cf. A000142, A111004, A324130.
Sequence in context: A242573 A223034 A177530 * A324135 A177531 A121987
Adjacent sequences: A324131 A324132 A324133 * A324135 A324136 A324137


KEYWORD

nonn


AUTHOR

N. J. A. Sloane, Feb 16 2019


EXTENSIONS

More terms from Petros Hadjicostas, Oct 30 2019


STATUS

approved



