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

%I A324134 #37 Nov 01 2019 03:21:42
%S A324134 1,1,2,6,24,120,710,4800,36298,302780,2758618,27246450,289962508,
%T A324134 3308024082,40278949800,521427324542,7152011191362,103621538280688,
%U A324134 1581465201545374,25361207137790358,426374509273382756,7499269147438400178,137728268057069904088,2636572230825216681414
%N A324134 Number of permutations of [n] that avoid the shuffle pattern s-k-t, where s = 12 and t = 132.
%H A324134 Sergey Kitaev, <a href="http://dx.doi.org/10.1016/j.disc.2004.03.017">Partially Ordered Generalized Patterns</a>, Discrete Math. 298 (2005), no. 1-3, 212-229.
%F A324134 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
%e A324134 From _Petros Hadjicostas_, Oct 31 2019: (Start)
%e A324134 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.
%e A324134 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)
%Y A324134 Cf. A000142, A111004, A324130.
%K A324134 nonn
%O A324134 0,3
%A A324134 _N. J. A. Sloane_, Feb 16 2019
%E A324134 More terms from _Petros Hadjicostas_, Oct 30 2019