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A324132
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Number of permutations of [n] that avoid the shuffle pattern s-k-t, where s = 1 and t = 1234.
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1
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1, 1, 2, 6, 24, 120, 715, 4946, 38963, 344350, 3374300, 36309360, 425630742, 5398849407, 73675463140, 1076309968529, 16759420085633, 277094358530356, 4848092369455556, 89489787445905318, 1738016295746132796, 35427829798218756153, 756269443056959088424, 16871986369868627423375
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OFFSET
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0,3
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LINKS
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FORMULA
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Let b(n) = A117158(n) = number of permutations avoiding a consecutive 1234 pattern. Then a(n) = 2*a(n-1) - b(n-1) + Sum_{i = 1..n-1} binomial(n-1,i) * b(i) * a(n-1-i) for n >= 1 with a(0) = b(0) = 1. [See the recurrence for C_n on p. 220 of Kitaev (2005).]
E.g.f.: If A(x) is the e.g.f. of (a(n): n >= 0) and B(x) is the e.g.f. of (b(n): n >= 0) (i.e., B(x) = 2/(exp(-x) + cos(x) - sin(x))), then A'(x) = (1 + B(x)) * A(x) - B(x) with A(0) = B(0) = 1. [Theorem 16, p. 219, in Kitaev (2005)] (End)
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EXAMPLE
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In a permutation of [n] that contains the shuffle pattern s-k-t, where s = 1 and t = 1234, 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, the permutations of [n] that contain this shuffle pattern should have k = 6 and the last four numbers in these permutations (with pattern t) should be one of the choices 1234, 1235, 1245, 1345, and 2345. The corresponding permutations that contain this shuffle pattern are 561234, 461235, 361245, 261345, and 162345. Hence a(6) = 6! - 5 = 715. (End)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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