

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
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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 1to1 and its range is the Cartesian product of the set of Dyck npaths 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 nth moment of the following positive function defined on x in (0,16), in Maple notation: (EllipticK(sqrt(1x/16))  EllipticE(sqrt(1x/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, PrenticeHall, 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. BousquetMélou and M. Mishna, Walks with small steps in the quarter plane, arXiv:0810.4387 [math.CO], 20082009.
David Callan, Bijections for the identity 4^n = ...
Kiran S. Kedlaya and Andrew V. Sutherland, Hyperelliptic curves, Lpolynomials and random matrices, arXiv:0803.4462 [math.NT], 20082010.
Ralf Steiner, 1/Pi, 2016.  Ralf Steiner, Jan 21 2016


FORMULA

G.f.: 1/4*((16*x1)*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, 1step).  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

N. J. A. Sloane


STATUS

approved



