A094061 Number of n-moves paths of a king starting and ending at the origin of an infinite chessboard.

1, 0, 8, 24, 216, 1200, 8840, 58800, 423640, 3000480, 21824208, 158964960, 1171230984, 8668531872, 64574844048, 483114856224, 3630440899800, 27379154692032, 207172490054816, 1572194644061184, 11962847247681616, 91242602561647680, 697438669619791008

Offset: 0

Author

Matthijs Coster, Apr 29 2004

Keywords

• nonn
• easy

Comments

The chessboard here is the full four-quadrant board Z X Z.

This is an analog of A054474 for walks on a square grid where the steps can be made diagonally as well.

a(n) is the constant term in the expansion of ((x + 1/x) * (y + 1/y) + x^2 + 1/x^2 + y^2 + 1/y^2)^n. - Seiichi Manyama, Nov 03 2019

References

• D. Joyner, "Adventures in Group Theory: Rubik's Cube, Merlin's Machine and Other Mathematical Toys", Johns Hopkins University Press, 2002, pp. 79

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000
• Peter Bala, A note on A094061

Crossrefs

Row 2 of A327751.

Cf. A002426, A098070, A126869, A253974, A254129, A254459, A329024.

Programs

Maple

a:=array(0..30):a[0]:=1:a[1]:=0:a[2]:=8:a[3]:=24:for n from 3 to 29 do a[n+1]:= (n*(5*n+1)*a[n]+2*(15*n^2+6*n-5)*a[n-1]-8*(5*n^2-23*n+21)*a[n-2]-64*(n-2)^2*a[n-3])/(n+1)^2: print(n+1,a[n+1]) od:
# second Maple program
a:= proc(n) option remember; `if`(n<3, (n-1)*(9*n-2)/2,
      ((n-1)*(3*n-1)*(3*n-4) *a(n-1)
      +(108*n^3-396*n^2+452*n-152) *a(n-2)
      +32*(3*n-2)*(n-2)^2 *a(n-3))/ (n^2*(3*n-5)))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Nov 02 2012

Mathematica

a[n_]:=Module[{f=(x+x^-1+y+y^-1+x y+x^-1y+x^-1y^-1+x y^-1)^n,s}, s=Series[f,{x,0,0},{y,0,0}]; SeriesCoefficient[s,{0,0}]]; Table[a[n], {n,1,22}] (* Armin Vollmer (Armin.Vollmer(AT)kabelleipzig.de), May 01 2006 *)
CoefficientList[Series[1/(1+4*x)*LegendreP[-1/2,1-32*x*(1+x)/(1+4*x)^2], {x, 0, 20}], x] (* Vaclav Kotesovec, Aug 16 2013 *)

Maxima

a[0]:1$
a[1]:0$
a[2]:8$
a[3]:24$
a[n]:=((n-1)*(3*n-1)*(3*n-4) *a[n-1]
      +(108*n^3-396*n^2+452*n-152) *a[n-2]
      +32*(3*n-2)*(n-2)^2 *a[n-3])/ (n^2*(3*n-5))$
A094061(n):=a[n]$
makelist(A094061(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

PARI

{a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n, k)*polcoef((1+x+1/x)^k, 0)^2)} \\ Seiichi Manyama, Oct 29 2019

PARI

{a(n) = polcoef(polcoef(((x+1/x)*(y+1/y)+x^2+1/x^2+y^2+1/y^2)^n, 0), 0)} \\ Seiichi Manyama, Nov 03 2019

Formula

D-finite with recurrence (n+1)^2*a(n+1) = n*(5*n+1)*a(n) + 2*(15*n^2+6*n-5)*a(n-1) - 8*(5*n^2-23*n+21) *a(n-2) - 64*(n-2)^2*a(n-3).

G.f.: (2/(Pi*(1+4*x))) * EllipticK(4*sqrt(x*(1+x))/(1+4*x)) = 1/(1+4*x) * hypergeom([1/2,1/2], [1], 16*(x*(1+x))/(1+4*x)^2). - Sergey Perepechko, Jan 15 2011

a(n) ~ 2^(3*n+1)/(3*Pi*n). - Vaclav Kotesovec, Aug 16 2013

a(n) = (1/Pi^2) * Integral_{y = 0..Pi} Integral_{x = 0..Pi} (2*cos(x) + 2*cos(y) + 4*cos(x)*cos(y))^n dx dy. - Peter Bala, Feb 14 2017

a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * A002426(k)^2. - Seiichi Manyama, Oct 29 2019

From Peter Bala, Feb 08 2022: (Start)

The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.

Conjecture: the stronger congruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for all primes p >= 5 and positive integers n and k. (End)

a(n) = Sum_{j = 0..n} Sum_{k = 0..j} binomial(2*j,j)^2*binomial(j,k)* binomial(n+j-k,2*j)*(-4)^(n-j-k). - Peter Bala, Mar 19 2022

Extensions

More terms from and entry improved by Sergey Perepechko, Sep 06 2004