A323809 Squares visited by a knight on a spirally numbered board, moving always to the lowest available unvisited square, or one step back if no unvisited square is available.

0, 9, 2, 5, 8, 3, 6, 1, 4, 7, 10, 13, 28, 31, 14, 11, 26, 23, 44, 19, 22, 43, 40, 17, 34, 37, 18, 15, 32, 29, 52, 25, 46, 21, 42, 69, 20, 39, 16, 33, 12, 27, 24, 45, 74, 41, 68, 103, 36, 61, 94, 57, 54, 85, 50, 47, 76, 113, 72, 107, 150, 67, 102, 63, 66, 35, 38, 65, 62

Offset: 0

Author

Daniël Karssen, Jan 28 2019

Keywords

• nonn
• walk
• changed

Comments

This is an infinite extension of A316328, with which it coincides for the first 2016 terms. - N. J. A. Sloane, Jan 29 2019

From M. F. Hasler, Nov 04 2019: (Start)

At move 99999, the least yet unvisited square has number 66048, which is near the border of the visited region. This suggests that the knight will eventually visit every square. Can this be proved or disproved through a counterexample?

More formally, let us call "isolated" a set of unvisited squares which is connected through knight moves, but which cannot be extended to a larger such set by adding a further square. Can there exist at some moment a finite isolated set which the knight cannot reach? (Without the last condition, the square a(2016) would clearly satisfy the condition just before the knight reaches it.)

Such subsets have a good chance of preserving this property forever. It should be possible to prove that an isolated subset sufficiently far (2 knight moves?) from any other unvisited square (or from the infinite connected subset of unvisited squares) remains so forever. (This condition of distance could replace the time-dependent condition "reachable by the knight".)

If such (forever) isolated sets do exist, with what frequency will they occur? Could they have a nonzero asymptotic density? Will this (if so, how) depend on the way the knight measures "lowest available" (cf. variants where the taxicab or Euclidean or sup norm distance from the origin is used)? (End)

Examples

			The board is numbered following a square spiral:
  16--15--14--13--12   :
   |               |   :
  17   4---3---2  11  28
   |   |       |   |   |
  18   5   0---1  10  27
   |   |           |   |
  19   6---7---8---9  26
   |                   |
  20--21--22--23--24--25
.
From _M. F. Hasler_, Nov 06 2019: (Start)
At move 2015, the knight lands on a(2015) = 2083, from where no unvisited squares can be reached. So the knight moves back to a(2016) = a(2014) = 2466, from where it goes on to the unvisited square a(2017) = 2667.
Similarly, at moves 2985, 3120, 3378, 7493, 8785, 9738, 10985, 11861, 11936, 12160, 18499, 18730, 19947 and 22251, the knight get "trapped" and has to move to the previous square on the next move.
On move 23044, the same happens on square 25808, and the knight must move back to square a(23045) = a(23043) = 27111. However, there is still no unvisited square in reach, so the knight has to make another step back to a(23046) = a(23042) = 28446, before it can move on to a(23047) = 29123. (End)
		

Links

• Daniël Karssen, Table of n, a(n) for n = 0..99999
• M. F. Hasler, Knight tours, OEIS wiki, Nov. 2019.
• Daniël Karssen, Figure showing the first 1e5 steps of the sequence

Crossrefs

The sequences involved in this set of related sequences are A316588, A316328, A316334, A316667, A323808, A323809, A323810 and A323811.

Cf. A326924 & A326922 (using L2-norm), A328908 & A328928 (L1-norm), A328909 & A328929 (sup norm); A326916 & A326918 (digits on spiral), A326413 and A328698 (variants with other tie breaker).

Programs

PARI

Nmax=1e5 /* number of terms to compute */; {local( K=[[(-1)^(i\2)<<(i>4),(-1)^i<<(i<5)]|i<-[1..8]], pos(x,y)=if(y>=abs(x),4*y^2-y-x,-x>=abs(y),4*x^2-x-y,-y>=abs(x),(4*y-3)*y+x,(4*x-3)*x+y), coords(n, m=sqrtint(n), k=m\/2)=if(m<=n-=4*k^2, [n-3*k, -k], n>=0, [-k, k-n], n>=-m, [-k-n, k], [k, 3*k+n]), U=0, Umin=0, t(x, p=pos(x[1],x[2]))=if(pt(x+K), K))[1], back=0); my(A=List(0)); for(n=1, Nmax, back||U+=1<<(A[n]-Umin); while(bittest(U,0), U>>=1; Umin++); listput(A, nxt(A[n])); if(A[n+1] != oo, back=0, A[n+1]=A[n+1-back+=2])); print("Index of the last term: ", #A-1); A323809(n)=A[n+1];}

Formula

a(n) = A323808(n+1) - 1. - M. F. Hasler, Nov 06 2019

Extensions

Edited by M. F. Hasler, Nov 02 2019