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A125305
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Number of 132-segmented permutations of length n.
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4
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1, 1, 2, 6, 18, 57, 190, 654, 2306, 8290, 30272, 111973, 418666, 1579803, 6008464, 23009240, 88645034, 343334976, 1336105472, 5221667740, 20485272152, 80645855014, 318489386884, 1261428593649, 5009356014722, 19941674099985
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OFFSET
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0,3
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COMMENTS
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From [Callan 2006] Theorem 3, the number of permutations on [n] that avoid a nonconsecutive 132 pattern is Sum_{k=0..n/3} binomial(n-2k, k) C_{n-2k}. - Michael Somos, Sep 07 2018
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LINKS
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FORMULA
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a(n) = sum(binomial(n-2*k,k)*catalan(n-2*k),k=0..floor(n/3)); generating function = C(x + x^3), where C(x) is the generating function for the Catalan numbers.
G.f.: A(x)=1/(1-z/(1-z/(1-z/(...)))) where z=x+x^3 (continued fraction, a special case of the previous formula). - Joerg Arndt, Mar 18 2011
Recurrence: (n+1)*a(n) = 2*(2*n-1)*a(n-1) - (n+1)*a(n-2) + 8*(n-2)*a(n-3) + 2*(2*n-7)*a(n-5). - Vaclav Kotesovec, Mar 21 2014
a(n) ~ sqrt(48 - 4*(27+3*sqrt(273))^(2/3) + 9*(27+3*sqrt(273))^(1/3)) / ((27+3*sqrt(273))^(1/6) * sqrt(3*Pi)) * (16 + (118+6*sqrt(273))^(2/3) + 4*(118+6*sqrt(273))^(1/3))^n / ((118+6*sqrt(273))^(n/3) * n^(3/2) * 3^n). - Vaclav Kotesovec, Mar 21 2014
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EXAMPLE
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a(4)=18 because of the 24 permutations of {1,2,3,4} only 1243, 1342, 1423, 1432, 2143, 2413 are not 132-segmented; i.e., those permutations have more occurrences of the pattern (1-3-2) than of the pattern (132).
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MAPLE
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a := n->sum(binomial(n-2*k, k)*catalan(n-2*k), k=0..floor(n/3)); seq(a(n), n=0..25);
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MATHEMATICA
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Table[Sum[Binomial[n - 2 k, k] Binomial[2 n - 4 * k, n - 2 k]/(n - 2 k + 1), {k, 0, Floor[n/3]}], {n, 0, 40}] (* Vaclav Kotesovec, Mar 21 2014 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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