A324134 Number of permutations of [n] that avoid the shuffle pattern s-k-t, where s = 12 and t = 132.

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

Offset: 0

Author

N. J. A. Sloane, Feb 16 2019

Keywords

• nonn

Examples

			From _Petros Hadjicostas_, Oct 31 2019: (Start)
In a permutation of [n] that contains the shuffle pattern s-k-t, 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)
		

Links

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

Crossrefs

Cf. A000142, A111004, A324130.

Formula

Let b(n) = A111004(n) = number of permutations avoiding a consecutive 132 pattern. Then a(n) = Sum_{i = 0..n-1} binomial(n-1,i) * (a(n-1-i) + b(i) * a(n-1-i) - b(n-1-i)) 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

Extensions

More terms from Petros Hadjicostas, Oct 30 2019