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 A001089 Number of permutations of [n] containing exactly 2 increasing subsequences of length 3. 6
 0, 0, 0, 0, 3, 24, 133, 635, 2807, 11864, 48756, 196707, 783750, 3095708, 12152855, 47500635, 185082495, 719559600, 2793121080, 10830450780, 41965864794, 162539516448, 629399492330, 2437072038302, 9437097796918 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 LINKS G. C. Greubel, Table of n, a(n) for n = 0..1000 M. Fulmek, Enumeration of permutations containing a prescribed number of occurrences of a pattern of length three, arXiv:math/0112092 [math.CO], 2001-2002. Also doi:10.1016/S0196-8858(02)00501-8Adv. Appl. Math., 30, 2003, 607-632. T. Mansour and A. Vainshtein, Counting occurrences of 123 in a permutation, arXiv:math/0105073 [math.CO], 2001. Toufik Mansour, Sherry H. F. Yan and Laura L. M. Yang, Counting occurrences of 231 in an involution, Discrete Mathematics 306 (2006), pages 564-572. J. Noonan and D. Zeilberger, The Enumeration of Permutations With a Prescribed Number of "Forbidden" Patterns, arXiv:math/9808080 [math.CO], 1998. Also doi:10.1006/aama.1996.0016, Adv. in Appl. Math. 17 (1996), no. 4, 381--407. MR1422065 (97j:05003). FORMULA Noonan and Zeilberger conjectured that a(n) = ((59*n^2+117*n+100)/(2*n*(2*n-1)*(n+5)))*binomial(2*n,n-4). This was proved by Fulmek. G.f.: ((x^5 -3*x^4 +5*x^3 -10*x^2 +6*x -1)*(1-4*x)^(1/2) - 5*x^5 +7*x^4 -17*x^3 +20*x^2 -8*x +1)/(2*x^6). - Mark van Hoeij, Oct 25 2011 G.f.: x^5*C(x)^11 + 3*x^3*C(x)^8, where C(x) is g.f. for the Catalan numbers (A000108). - Michael D. Weiner, Sep 02 2016 Conjecture: -(n+5)*(n-4)*(59*n^2-n+42)*a(n) +2*(n-1)*(2*n-3)*(59*n^2 +117*n+100)*a(n-1) = 0. - R. J. Mathar, Jan 04 2017 EXAMPLE For n=4, there are 4! = 24 permutations of 1234. The identity permutation 1234 has four increasing subsequences of length 3 (123, 124, 134, and 234), and the permutation 2314 has only one increasing subsequence of length 3 (234). Only the permutations 1243, 1324, and 2134 have exactly two increasing subsequences of length 3, and since there are three of them, a(4) = 3. - Michael B. Porter, Sep 03 2016 MAPLE seq(`if`(n=0, 0, (100+117*n+59*n^2)*binomial(2*n, n-4)/(2*n*(2*n-1)*(n+5))), n = 0..30); # G. C. Greubel, Sep 19 2019 MATHEMATICA {0}~Join~CoefficientList[Series[((x^5-3x^4+5x^3-10x^2+6*x-1)(1-4x)^(1/2) - 5x^5+7x^4-17x^3+20x^2-8*x+1)/(2x^6), {x, 0, 23}], x] (* or *) {0}~Join~CoefficientList[Series[x^5*((1-(1-4x)^(1/2))/(2x))^11 +3x^3*( (1-(1-4x)^(1/2))/(2x))^8, {x, 0, 23}], x] (* Michael De Vlieger, Sep 03 2016 *) PROG (PARI) a(n) = (100+117*n+59*n^2)*binomial(2*n, n-4)/(2*n*(2*n-1)*(n+5)) \\ G. C. Greubel, Sep 19 2019 (MAGMA) [0, 0, 0, 0] cat [(100+117*n+59*n^2)*Binomial(2*n, n-4)/(2*n*(2*n-1)*(n+5)): n in [4..30]]; // G. C. Greubel, Sep 19 2019 (Sage) [0, 0, 0, 0]+[(100+117*n+59*n^2)*binomial(2*n, n-4)/(2*n*(2*n-1)*(n+5)) for n in (4..30)] # G. C. Greubel, Sep 19 2019 (GAP) Concatenation([0, 0, 0, 0], List([4..30], n-> (100+117*n+59*n^2)* Binomial(2*n, n-4)/(2*n*(2*n-1)*(n+5)))); # G. C. Greubel, Sep 19 2019 CROSSREFS Cf. A003517, A084249, A138159. Leading column of A229158. Sequence in context: A326789 A305543 A183900 * A069515 A206949 A215636 Adjacent sequences:  A001086 A001087 A001088 * A001090 A001091 A001092 KEYWORD nonn AUTHOR John Thomas Noonan [ noonan(AT)euclid.math.temple.edu ] EXTENSIONS Terms a(25) onward added by G. C. Greubel, Sep 19 2019 STATUS approved

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Last modified November 20 05:07 EST 2019. Contains 329323 sequences. (Running on oeis4.)