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 A324134 Number of permutations of [n] that avoid the shuffle pattern s-k-t, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 LINKS Sergey Kitaev, Partially Ordered Generalized Patterns, Discrete Math. 298 (2005), no. 1-3, 212-229. 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 EXAMPLE 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  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

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Last modified March 25 23:47 EDT 2023. Contains 361529 sequences. (Running on oeis4.)