A324138 Number of permutations of [n] that avoid the shuffle pattern s-k-t, where s = 123 and t = 132.

1, 1, 2, 6, 24, 120, 720, 5020, 39755, 351518, 3425572, 36419844, 419026188, 5182797757, 68535001302, 964404124479, 14383519018582, 226579159065496, 3758349089828472, 65466833442028670, 1194655878120996337, 22788580047064423474, 453513206778006345040

Offset

0,3

Links

Table of n, a(n) for n=0..22.

Sergey Kitaev, Partially Ordered Generalized Patterns, Discrete Math. 298 (2005), no. 1-3, 212-229.

Formula

Let b(n) = A049774(n) = number of permutations of [n] that avoid consecutive pattern s = 123 and c(n) = A111004(n) = number of permutations of [n] that avoid consecutive pattern t = 132. Then a(n) = Sum_{i = 0..n-1} binomial(n-1,i) * (b(i)*a(n-1-i) + c(i)*a(n-1-i) - b(i)*c(n-1-i)) for n >= 1 with a(0) = b(0) = c(0) = 1. [This follows from the recurrence for C_n on p. 220 in Kitaev (2005).] - Petros Hadjicostas, Nov 01 2019

Example

From Petros Hadjicostas, Nov 01 2019: (Start)
In a permutation of [n] that contains the shuffle pattern s-k-t, where s = 123 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..6, all permutations of [n] avoid this shuffle pattern (since we need at least seven numbers to get this pattern). Hence, a(n) = n! for n = 0..6.
For n = 7, k should be equal to 7, and for the pattern s = 123 we have binomial(6,3) = 20 choices: 123, 124, 125, ..., 456. The corresponding permutations of [7] that contain this shuffle pattern are 1237465, 1247365, 1257364, ..., 4567132. Thus, a(7) = 7! - 20 = 5020. (End)

Crossrefs

Cf. A000142, A049774, A111004.

Sequence in context: A177535 A263929 A324139 * A324137 A177552 A177547

Adjacent sequences: A324135 A324136 A324137 * A324139 A324140 A324141

Keyword

nonn

Author

N. J. A. Sloane, Feb 16 2019

Extensions

More terms from Petros Hadjicostas, Nov 01 2019

Status

approved