

A226697


Central symmetric closed knight's tour on an 8x8 board, attributed to Euler. (n,m) position after (a(n,m)1)th move.


1



37, 62, 43, 56, 35, 60, 41, 50, 44, 55, 36, 61, 42, 49, 34, 59, 63, 38, 53, 46, 57, 40, 51, 48, 54, 45, 64, 39, 52, 47, 58, 33, 1, 26, 15, 20, 7, 32, 13, 22, 16, 19, 8, 25, 14, 21, 6, 31, 27, 2, 17, 10, 29, 4, 23, 12, 18, 9, 28, 3, 24, 11, 30, 5
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OFFSET

1,1


COMMENTS

Each row length is 8, and there are 8 rows.
This is an 8x8 matrix with entries a(n,m) = a(8*(n1) + m), n, m =1, 2, ..., 8, with the given length 64 sequence {a(n)}. The knight starts (move No. 0) at the square with position (n,m) = (5,1) because a(5,1) = 1. The square position (n,m) of the knight after move No. k is found from a(n,m) = k + 1, for k = 0, 1, ..., 63. See A226698 for the inverse sequence:
A226698(a(n)1) = n, n= 1, 2, ..., 64.
This is a Hamiltonian path which is reentrant because with move No. 64 one can close the path at position (5,1) to obtain a Hamiltonian circuit.
In the Gardner reference this path is attributed to L. Euler (1759). The lower half of the board is covered first. The closed path has central symmetry. Each symmetric pair has an absolute difference of 32: a(n,m)  a(9n,9m) = 32, n,m =1, 2, ..., 8.
For a figure with this central symmetric Hamiltonian cycle see also Diagram 4 on p. 24 of the ElkiesStanley reference.


REFERENCES

Martin Gardner, Mathematical Magic Show, The Math. Assoc. of Am., Washington DC, 1989, Ch. 14, Knights of the Square Table,Fig. 86, p. 191. German Translation: Mathematische Hexereien, Ullstein, 1977, Abb. 86, S. 186.


LINKS

Table of n, a(n) for n=1..64.
N. D. Elkies and R. P. Stanley, The mathematical knight, Math. Intelligencer, 25 (No. 1) (2003), 2234.


EXAMPLE

The board as an 8x8 matrix a(n,m)
n\m 1 2 3 4 5 6 7 8
1: 37 62 43 56 35 60 41 50
2: 44 55 36 61 42 49 34 59
3: 63 38 53 46 57 40 51 48
4: 54 45 64 39 52 47 58 33
5: 1 26 15 20 7 32 13 22
6: 16 19 8 25 14 21 6 31
7: 27 2 17 10 29 4 23 12
8: 18 9 28 3 24 11 30 5
(5,1) marks the start position of the knight because a(n,m) = 1 (move 0). Move 1 leads to square (7,2) because a(7,2) = 2, etc. Move 63 leads to a(4,3) = 64, the end of the reentrant Hamiltonian path on the board.
Central symmetry: a(4,8) = 32 + 1 = 33, a(4,7) = 32 + 26 = 58, etc. such that the upper half can be found from the lower half.


CROSSREFS

Cf. A226698 (inverse).
Sequence in context: A103946 A107142 A158018 * A117475 A145486 A319673
Adjacent sequences: A226694 A226695 A226696 * A226698 A226699 A226700


KEYWORD

nonn,fini,full,tabf


AUTHOR

Wolfdieter Lang, Jun 25 2013


STATUS

approved



