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This sequence uses the same board numbering as A326413, and like that sequence, if the next step encounters two or more squares with the same square number, it then chooses the square which is the closest to the original 0-squared origin. But if two or more squares are found which also have the same distance to the origin, then the square which was first drawn in the spiral numbering is chosen, i.e., the smallest standard spiral numbered square as in A316667.

The sequence is finite. After 1069 steps a square with the number 9 (standard spiral number = 473) is visited, after which all neighboring squares have been visited.

Sequence A326916(n) gives the number of the square visited at step n, i.e., its rank in the spiral, starting with 0, as illustrated, e.g., in A326413. The digit on that square, i.e., a(n) can be obtained through A007376, cf. formula. - M. F. Hasler, Oct 21 2019

Differs from A326924 (where only the 2-norm is considered) from a(73) = 175 on.

The sequence is also finite, when the knight lands on square number a(1092366) there is no unvisited square within reach.

The 1-norm or taxicab distance from the origin of the square a(n) is given in A328928(n).

It appears that this knight's tour would also completely fills the board, if we consider the infinite extension where the knight is allowed to move back on its last step(s) when there's no unvisited square available: no isolated sets of unvisited squares as defined in A323809, seem to occur. Is there a proof or disproof for this? - M. F. Hasler, Nov 04 2019

Take the standard counterclockwise square spiral starting at 0, as in A304586, but only write one digit at a time in the cells of the spiral: 0,1,2,3,4,5,6,7,8,9,1,0,1,1,1,2,...

Place a chess knight at cell 0. Move it to the lowest-numbered cell it can attack, and if there is a tie, move it to the cell closest (in Euclidean distance) to the start, and if there is still a tie, move to the left(*).

No cell can be visited more than once.

Inspired by the Trapped Knight video and A316667.

Just as for A316667, the sequence is finite. After a while, the knight has no unvisited squares it can reach, and the sequence ends with a(1217) = 4.

(*)Moving to the left means choose the point with the lowest x-coordinate. This leads to an unambiguous choice of tied squares only for the 'move left' case.

Board is numbered with the square spiral:

.

17--16--15--14--13

| |

18 5---4---3 12 .

| | | |

19 6 1---2 11 .

| | |

20 7---8---9--10 .

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21--22--23--24--25--26

A (1,2) leaper is a chess knight.

a(2)-a(5) were computed by Daniël Karssen.

A (1,2) leaper is a chess knight.

a(2)-a(5) were computed by Daniël Karssen.

This is an infinite extension of A316667 with which it agrees for the first 2016 terms. - N. J. A. Sloane, Jan 28 2019

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)

A variant of Angelini's "Kneil's Knumberphile Knight", inspired by Sloane's "The Trapped Knight", cf. A316328 and links:

Consider an infinite chess board with squares numbered along the infinite square spiral starting with 0 at the origin (as in A174344, A274923 and A296030). The squares are filled with successive digits of the integers: 0, 1, 2, ..., 9, 1, 0, 1, 1, ... (= A007376 starting with 0). The knight moves at each step to the yet unvisited square with the lowest digit on it, and in case of a tie, the one closest to the origin, first by Euclidean distance, then by appearance on the spiral (i.e., number of the square). This sequence lists the number of the square visited in the n-th move, if the knight starts at the origin, viz a(0) = 0.

It turns out that following these rules, the knight gets trapped at the 1070th move, when he can't reach any unvisited square.

See A326918 for the sequence of visited digits, given as A007376(a(n)).

Many squares, e.g., 2: (1,1), 4: (-1,1), 5: (-1,0), 6: (-1,-1), 7: (0,-1), 8: (1,-1), 9: (2,-1), ..., will never be visited, even in the infinite extension of the sequence where the knight can move back if it gets trapped, in order to resume with a new unvisited square, as in A323809. - M. F. Hasler, Nov 08 2019

This is a variation of A316667. The same knight move rules apply, but the knight will not move to a square which has seven or eight visited neighbors unless no other square is available. If the only unvisited squares available to move to have seven or eight neighbors it will choose the one with the lowest number of neighbors first, and if a tie still exists it will choose the one with the smallest spiral number.

The sequence is finite. After 3935788 steps the square with spiral number 3352743 is reached after which all surrounding squares have been visited. This is the longest possible path using the given knight-leap rules for the eight possible values of visited neighbor count. A329520 gives the other path lengths.

The sequences matches the values of A316667 for the first 332 terms, but on the 332nd step the knight sees that square 193 has seven visited neighbors and thus chooses square 393 instead. Along its path the knight encounters 38812 squares where it would have chosen a square with seven or eight neighbors if only the lowest spiral number was considered; 21897 squares had seven neighbors, 16915 squares had eight neighbors. Of those squares thirteen times a square with seven neighbors was actually chosen due to no other square with a lower neighbor count being available. The first such encounter is after 380990 steps while on the square with number 295910. On this first encounter a square with eight neighbors is also available, and has a lower board number than the square with seven neighbors. Thus if the rules were changed to always select the lowest board number regardless of the number of neighbors in such cases then the knight would choose the eight-neighbor square and thus be trapped after 380990 steps.

Due to the knight's avoidance of trapping or potentially trapping squares numerous squares which are inside the knight's overall path are never visited; the first such example is square 193 mentioned above. This is in contrast to standard knight's tours which typically cover all internal squares.