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%I #135 Oct 02 2023 20:15:55
%S 1,2,12,100,980,10584,121968,1472328,18404100,236390440,3103161776,
%T 41469525552,562496897872,7726605740000,107289439704000,
%U 1503840313184400,21252802073091300,302539888334593800,4334635827016110000,62464383654579522000,904841214653480504400
%N a(n) = (2*n)!^2 / ((n+1)!*n!^3).
%C a(n) is the 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
%C Bijective proof: given such an 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 binomial(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
%C 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
%C 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
%C 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
%C The partial sums of a(n)/A013709(n) absolutely converge to 1/Pi. - _Ralf Steiner_, Jan 21 2016
%D E. R. Hansen, A Table of Series and Products, Prentice-Hall, Englewood Cliffs, NJ, 1975, p. 93.
%D T. M. MacRobert, Functions of a Complex Variable, 4th ed., Macmillan & Co., London, 1958, p. 177.
%H Vincenzo Librandi, <a href="/A000888/b000888.txt">Table of n, a(n) for n = 0..100</a>
%H Marco S. Bianchi, <a href="https://arxiv.org/abs/2306.06239">Protected and uniformly transcendental</a>, arXiv:2306.06239 [hep-th], 2023.
%H M. Bousquet-Mélou and M. Mishna, <a href="https://arxiv.org/abs/0810.4387">Walks with small steps in the quarter plane</a>, arXiv:0810.4387 [math.CO], 2008-2009.
%H David Callan, <a href="http://www.stat.wisc.edu/~callan/notes/">Bijections for the identity 4^n = ... </a>.
%H Kiran S. Kedlaya and Andrew V. Sutherland, <a href="https://arxiv.org/abs/0803.4462">Hyperelliptic curves, L-polynomials and random matrices</a>, arXiv:0803.4462 [math.NT], 2008-2010.
%H Helmut Prodinger, <a href="https://arxiv.org/abs/1911.07604">Two New Identities Involving the Catalan Numbers: A classical approach</a>, arXiv:1911.07604 [math.CO], 2019.
%H Ralf Steiner, <a href="https://www.researchgate.net/publication/340005810_Beispiele_zur_modifizierten_Wallis-Lambert-Reihe">Beispiele zur modifizierten Wallis-Lambert-Reihe</a>, 2016.
%F G.f.: 1/4*((16*x-1)*EllipticK(4*x^(1/2)) + EllipticE(4*x^(1/2)))/x/Pi. - _Vladeta Jovovic_, Oct 12 2003
%F Given G.f. A(x), y = x*A(x) satisfies y = y'' * (1 - 16*x) * x/4. - _Michael Somos_, Sep 11 2005
%F a(n) = binomial(2*n,n)^2/(n+1). - _Zerinvary Lajos_, May 27 2006
%F G.f.: 2F1(1/2,1/2;2;16*x). - _Paul Barry_, Sep 03 2008
%F a(n) = 2*A125558(n) (n >= 1). - _Olivier Gérard_, Feb 16 2011
%F A002894(n) = (n+1) * a(n). A001246(n) = a(n) / (n+1). A089835(n) = n! * a(n). - _Michael Somos_, May 12 2012
%F 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). - _Sergei N. Gladkovskii_, Jul 30 2012
%F D-finite with recurrence: (n+1)*(n+2)*a(n+1) = 4*(2*n+1)^2*a(n). - _Vaclav Kotesovec_, Sep 11 2012
%F a(n) = C(n)*binomial(2*n,n) = Sum_{k=0..2*n} binomial(2*n,k)*C(k)*C(2*n-k) where C(k) are Catalan numbers (A000108), see Prodinger. - _Michel Marcus_, Nov 19 2019
%F Sum_{n>=0} a(n)/16^n = 4/Pi (A088538). - Amiram Eldar, May 06 2023
%e G.f.: 1 + 2*x + 12*x^2 + 100*x^3 + 980*x^4 + 10584*x^5 + 121968*x^6 + ...
%p [seq(binomial(2*n,n)^2/(n+1),n=0..17)]; # _Zerinvary Lajos_, May 27 2006
%t f[n_] := Binomial[2 n, n]^2/(n + 1); Array[f, 18, 0] (* _Robert G. Wilson v_ *)
%t 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 *)
%o (PARI) {a(n) = if( n<0, 0, (2*n)!^2 / n!^4 / (n+1))}; /* _Michael Somos_, Sep 11 2005 */
%o (Magma) [(Factorial(2*n))^2/(Factorial(n))^4/(n+1): n in [0..20]]; // _Vincenzo Librandi_, Aug 15 2011
%Y Cf. A000108, A002894, A088538, A089835, A125558.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
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