A254129 Number of 2n-move closed knight paths on an unbounded chessboard from a given square to the same square.
1, 8, 168, 5840, 261800, 13180608, 702273264, 38641656768, 2171652448680, 123938999632448, 7158206751686848, 417418594698260064, 24535017440445455216, 1451786144317963971200, 86396682439552099487040, 5166936574734171792925440, 310340697572034456203934120
Offset: 0
Examples
a(1) = 8. For illustration, let's assume we're on a usual 8 X 8 chessboard, with the knight initially at D4. Then there are 8 paths bringing it back to D4 in 2 moves: D4-E6-D4; D4-F5-D4; D4-F3-D4; D4-E2-D4; D4-C2-D4; D4-B3-D4; D4-B5-D4; D4-C6-D4.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..550 (first 85 terms from David A. Corneth)
- Shalosh B. Ekhad and Doron Zeilberger, In How Many Ways Can the Chess Pieces Walk n Steps, Staying on the Board?, May 19 2011; Local copy, pdf file only, no active links
- Mohamud Mohammed and Doron Zeilberger, Maple program SMAZ; Local copy
- Doron Zeilberger, Input file inSMAZ6; Local copy
- Doron Zeilberger, Output from Maple program SMAZ; Local copy
Programs
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Maple
G:= cos(x+2*y)+cos(x-2*y)+cos(2*x+y)+cos(2*x-y): F:= 1: a[0]:= 1: for n from 1 to 20 do F:= combine(F*G^2,trig); a[n]:= 4^n*remove(has,F,cos); od: seq(a[n],n=0..20); # Robert Israel, Jan 26 2015 # second Maple program: b:= proc(n, x, y) option remember; `if`(max(x, y)>2*n or x+y>3*n, 0, `if`(n=0, 1, add(b(n-1, abs(x+l[1]), abs(y+l[2])), l=[[1, 2], [2, 1], [-1, 2], [-2, 1], [1, -2], [2, -1], [-1, -2], [-2, -1]]))) end: a:= n-> b(2*n, 0$2): seq(a(n), n=0..25); # Alois P. Heinz, Jan 29 2015
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Mathematica
b[n_, x_, y_] := b[n, x, y] = If[Max[x, y] > 2n || x+y > 3n, 0, If[n == 0, 1, Sum[b[n-1, Abs[x+l[[1]]], Abs[y+l[[2]]]], {l, {{1, 2}, {2, 1}, {-1, 2}, {-2, 1}, {1, -2}, {2, -1}, {-1, -2}, {-2, -1}}}]]]; a[n_] := b[2n, 0, 0]; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Apr 30 2019, after Alois P. Heinz *) -
PARI
a(n)={my(l=listcreate(),v=vector(2*n+1));m=matrix(1,1);m[1,1]=1;listput(l,m);v[1]=1;for(i=2,2*n+1, m=matrix(4*i-3,4*i-3);for(j=1,#l[i-1],for(k=1,#l[i-1],m[j+2-2,k+2-1]+=l[i-1][j,k];m[j+2-2,k+2+1]+=l[i-1][j,k];m[j+2-1,k+2-2]+=l[i-1][j,k];m[j+2-1,k+2+2]+=l[i-1][j,k];m[j+2+1,k+2-2]+=l[i-1][j,k];m[j+2+1,k+2+2]+=l[i-1][j,k];m[j+2+2,k+2-1]+=l[i-1][j,k];m[j+2+2,k+2+1]+=l[i-1][j,k]));v[i]=m[2*i-1,2*i-1];listput(l,m););listput(l,v);v[#v]} \\ David A. Corneth, Jan 26 2015 -
PARI
{a(n) = polcoef(polcoef((x^2*y+x*y^2+y^2/x+y/x^2+1/(x^2*y)+1/(x*y^2)+x/y^2+x^2/y)^(2*n), 0), 0)} \\ Seiichi Manyama, Nov 02 2019
Formula
From Robert Israel, Jan 26 2015: (Start)
a(n) = 4^n * (2*Pi)^(-2) * int_0^(2*Pi) int_0^(2*Pi) ds dt (cos(s+2*t)+cos(s-2*t)+cos(2*s+t)+cos(2*s-t))^(2*n).
G.f.: (2*Pi)^(-2) * int_0^(2*Pi) int_0^(2*Pi) ds dt 1/(1 - 4*x*(cos(s+2*t)+cos(s-2*t)+cos(2*s+t)+cos(2*s-t))^2).
(End)
a(n) ~ 64^n / (5*Pi*n). - Vaclav Kotesovec, Jan 28 2015
Recurrence (conjectured): 3*n^2*(3*n-2)*(3*n-1)*(4*n-3)*(4*n-1)*(58625*n^6 - 574525*n^5 + 2317575*n^4 - 4929815*n^3 + 5836090*n^2 - 3647730*n + 940788)*a(n) = 4*(563444875*n^12 - 7212094400*n^11 + 40894216825*n^10 - 135653664390*n^9 + 292742658975*n^8 - 432166599360*n^7 + 446527351283*n^6 - 324481592710*n^5 + 164046706898*n^4 - 56035458036*n^3 + 12203976528*n^2 - 1507156200*n + 78246000)*a(n-1) - 64*(n-1)*(2*n-3)^2*(167726125*n^9 - 1643716025*n^8 + 6735239425*n^7 - 15048594215*n^6 + 20072439970*n^5 - 16473493280*n^4 + 8273936628*n^3 - 2437948332*n^2 + 377982648*n - 22556880)*a(n-2) + 165888*(n-2)*(n-1)*(2*n - 5)^2*(2*n - 3)^2*(58625*n^6 - 222775*n^5 + 324325*n^4 - 232265*n^3 + 86220*n^2 - 15570*n + 1008)*a(n-3). - Vaclav Kotesovec, Jan 30 2015
a(n) = the constant term in the expansion of (x^2*y + x*y^2 + y^2/x + y/x^2 + 1/(x^2*y) + 1/(y^2*x) + x/y^2 + x^2/y)^(2*n). - Peter Bala, Feb 14 2017
The conjectured recurrence of Vaclav Kotesovec is true. Running the input file inSMAZ6 (see Links) on the Maple program SMAZ gives the recurrence followed by the certificate shown in the output file oSMAZ6. - Doron Zeilberger, Mar 29 2019
Comments