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 A210474 The number of different lattice paths from (0,0) to (2n,0) using steps of S={(i,i) or (i,-i): i=1,2,...,n} with j flaws(j=1,2,...,n-1), where the j flaws is the sum of lengths of down steps below the x-axis. (For down steps that are partly above and partly below the x-axis we just count the part below the x-axis.) This number is independent of the number of flaws. 1
 1, 0, 4, 24, 156, 1072, 7668, 56520, 426380, 3276384, 25556196, 201828728, 1610647932, 12968268432, 105219588308, 859440482856, 7061361325164, 58320764249280, 483922589498820, 4032178320794328, 33723925620989532, 283021269941508336, 2382598282012764084, 20114924440891152264 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 REFERENCES JiSun Huh and SeungKyung Park, The Chung-Feller Theorem on Generalized Lattice Paths, submitted LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..200 FORMULA a(n) = (1/(n-1))*Sum_{k=1..n-1} binomial(n-1, k)*(binomial(n, k-1)+binomial(n-1, k-2))*4^(n-k) for n > 1. a(n) = Sum_{k=1..n-1} C_{k+1}*C(n-2,k-1)*2^k where C_n is n-th Catalan number. G.f.: (1-x)*(1+3*x-sqrt(9*x^2-10*x+1))/(8*x). a(n) ~ 3^(2*n-1)/(sqrt(Pi/2)*n^(3/2)). - Vaclav Kotesovec, Jan 28 2013 Conjecture:  (n+1)*a(n) +2*(-5*n+3)*a(n-1) +9*(n-3)*a(n-2)=0. - R. J. Mathar, Jun 14 2016 MAPLE gf := (1-x)*(1+3*x-sqrt(9*x^2-10*x+1))/(8*x): s := series(gf, x, 100): for i from 0 to 50 do printf(‘%d, ’, coeff(s, x, i)) od: MATHEMATICA CoefficientList[Series[(1-x)*(1+3*x-Sqrt[9*x^2-10*x+1])/(8*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Jan 28 2013 *) PROG (PARI) x = 'x + O('x^66); Vec((1-x)*(1+3*x-sqrt(9*x^2-10*x+1))/(8*x)) /* Joerg Arndt, Jan 23 2013 */ CROSSREFS Sequence in context: A178879 A091166 A078108 * A117337 A272865 A084130 Adjacent sequences:  A210471 A210472 A210473 * A210475 A210476 A210477 KEYWORD nonn AUTHOR Jisun Huh, Jan 23 2013 STATUS approved

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Last modified May 25 17:19 EDT 2020. Contains 334595 sequences. (Running on oeis4.)