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 A000888 a(n) = (2*n)!^2 / ((n+1)!*n!^3). 16
 1, 2, 12, 100, 980, 10584, 121968, 1472328, 18404100, 236390440, 3103161776, 41469525552, 562496897872, 7726605740000, 107289439704000, 1503840313184400, 21252802073091300, 302539888334593800, 4334635827016110000, 62464383654579522000, 904841214653480504400 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS a(n) = number of walks of 2n unit steps North, East, South, or West, starting at the origin, bounded above by y=x, below by y=-x and terminating on the ray y=x>=0. Example: a(1) counts EN, EW; a(2) counts ESNN, ESNW, ENSN, ENSW, ENEN, ENEW, EENN, EENW, EEWN, EEWW, EWEN, EWEW. - David Callan, Oct 11 2005 Bijective proof: given such a NESW walk, construct a pair (P_1, P_2) of lattice paths of upsteps U=(1,1) and downsteps D=(1,-1) as follows. To get P_1, replace each E and S with U and each W and N with D. To get P_2, replace each N and E with U and each S and W with D. For example, EENSNW -> (UUDUDD, UUUDUD). This mapping is 1-to-1 and its range is the Cartesian product of the set of Dyck n-paths and the set of nonnegative paths of length 2n. The Dyck paths are counted by the Catalan number C_n (A000108) and the nonnegative paths are counted (see for example the Callan link) by the central binomial coefficient binom(2n,n) (A000984). So this is a bijection from these NESW walks to a set of size C_n*binomial(2n,n)=a(n). - David Callan, Sep 18 2007 If A is a random matrix in USp(4) (4 X 4 complex matrices that are unitary and symplectic), then a(n)=E[(tr(A^3))^{2n}]. - Andrew V. Sutherland, Apr 01 2008 Number of walks within N^2 (the first quadrant of Z^2) starting at (0,0), ending on the vertical axis and consisting of 2 n steps taken from {(-1,-1), (-1,1), (1,-1), (1,1)}. - Manuel Kauers, Nov 18 2008 a(n) is equal to the n-th moment of the following positive function defined on x in (0,16), in Maple notation: (EllipticK(sqrt(1-x/16)) - EllipticE(sqrt(1-x/16)))/(Pi^2*sqrt(x)). This is the solution of the Hausdorff moment problem and thus it is unique. - Karol A. Penson, Feb 11 2011 The partial sum of A000888(n) divided by A013709(n) absolutely converge to 1/Pi. Thus this series is 1/Pi - Ralf Steiner, Jan 21 2016 REFERENCES E. R. Hansen, A Table of Series and Products, Prentice-Hall, Englewood Cliffs, NJ, 1975, p. 93. T. M. MacRobert, Functions of a Complex Variable, 4th ed., Macmillan & Co., London, 1958, p. 177. LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..100 M. Bousquet-Mélou and M. Mishna, Walks with small steps in the quarter plane, arXiv:0810.4387 [math.CO], 2008-2009. David Callan, Bijections for the identity 4^n = ... Kiran S. Kedlaya and Andrew V. Sutherland, Hyperelliptic curves, L-polynomials and random matrices, arXiv:0803.4462 [math.NT], 2008-2010. Ralf Steiner, 1/Pi, 2016. - Ralf Steiner, Jan 21 2016 FORMULA G.f.: 1/4*((16*x-1)*EllipticK(4*x^(1/2))+EllipticE(4*x^(1/2)))/x/Pi. - Vladeta Jovovic, Oct 12 2003 Given G.f. A(x), y = x*A(x) satisfies y = y'' * (1 - 16*x) * x/4. - Michael Somos, Sep 11 2005 a(n) = binomial(2*n,n)^2/(n+1). - Zerinvary Lajos, May 27 2006 G.f.: 2F1(1/2,1/2;2;16*x). - Paul Barry, Sep 03 2008 a(n) = 2*A125558(n) (n>=1). - Olivier Gérard, Feb 16 2011 A002894(n) = (n+1) * a(n). A001246(n) = a(n) / (n+1). A089835(n) = n! * a(n). - Michael Somos, May 12 2012 G.f. 1 + 4*x/(G(0)-4*x) where G(k) = 4*x*(2*k+1)^2 + (k+1)*(k+2) - 4*x*(k+1)*(k+2)*(2*k+3)^2/G(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Jul 30 2012 Recurrence: (n+1)*(n+2)*a(n+1)=4*(2*n+1)^2*a(n). - Vaclav Kotesovec, Sep 11 2012 EXAMPLE G.f.: 1 + 2*x + 12*x^2 + 100*x^3 + 980*x^4 + 10584*x^5 + 121968*x^6 + ... MAPLE [seq(binomial(2*n, n)^2/(n+1), n=0..17)]; # Zerinvary Lajos, May 27 2006 MATHEMATICA f[n_] := Binomial[2 n, n]^2/(n + 1); Array[f, 18, 0]  (* Robert G. Wilson v *) a[ n_] := SeriesCoefficient[ (1/8) (EllipticE[ 16 x] - (1 - 16 x) EllipticK[ 16 x]) / (Pi/2), {x, 0, n + 1}]; (* Michael Somos, Jan 23 2012 *) PROG (PARI) {a(n) = if( n<0, 0, (2*n)!^2 / n!^4 / (n+1))}; /* Michael Somos, Sep 11 2005 */ (MAGMA) [(Factorial(2*n))^2/(Factorial(n))^4/(n+1): n in [0..20]]; // Vincenzo Librandi, Aug 15 2011 CROSSREFS Cf. A000108, A002894, A089835, A125558. Sequence in context: A224403 A219534 A151392 * A079821 A303614 A124102 Adjacent sequences:  A000885 A000886 A000887 * A000889 A000890 A000891 KEYWORD nonn,easy AUTHOR STATUS approved

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Last modified December 18 23:46 EST 2018. Contains 318245 sequences. (Running on oeis4.)