(MAGMAMagma) [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

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T. Mansour and A. Vainshtein, <a href="httphttps://www.arXivarxiv.org/abs/math.CO/0105073">Counting occurrences of 123 in a permutation</a>, arXiv:math/0105073 [math.CO], 2001.

J. Noonan and D. Zeilberger, <a href="httphttps://arXivarxiv.org/abs/math.CO/9808080">[math/9808080] The Enumeration of Permutations With a Prescribed Number of "Forbidden" Patterns</a>, arXiv:math/9808080 [math.CO], 1998. Also <a href="http://dx.doi.org/10.1006/aama.1996.0016">doi:10.1006/aama.1996.0016</a>, Adv. in Appl. Math. 17 (1996), no. 4, 381--407. MR1422065 (97j:05003).

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G. C. Greubel, <a href="/A001089/b001089.txt">Table of n, a(n) for n = 0..1000</a>

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

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

{0}~Join~CoefficientList[Series[((x^5 - 3 x3x^4 + 5 x5x^3 - 10 x10x^2 + 6*x - 1) (1 - 4 x4x)^(1/2) - 5 x5x^5 + 7 x7x^4 - 17 x17x^3 + 20 x20x^2 - 8*x + 1)/(2 x2x^6), {x, 0, 23}], x] (* or *)

{0}~Join~CoefficientList[Series[x^5*((1 - (1 - 4 x4x)^(1/2))/(2 x2x))^11 + 3 x3x^3*( (1 - (1 - 4 x4x)^(1/2))/(2 x2x))^8, {x, 0, 23}], x] (* Michael De Vlieger, Sep 03 2016 *)

(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

Terms a(25) onward added by G. C. Greubel, Sep 19 2019

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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

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