A396603 - OEIS

%I #12 Jun 02 2026 12:55:09

%S 0,1,32,2,64,3,1024,16,8,68,4,6,5,48,10,128,12,18,4096,34,40,256,512,

%T 130,66,42,17,9,96,33,24,14,11,258,136,520,260,8192,2048,72,138,

%U 131072,100,528,2080,20,514,160,36,65,104,73,19,67,21,26,168,74,162,1028,288

%N Lexicographically earliest sequence of distinct nonegative integers whose binary values, when written on a square spiral, one bit per square, high order bits first, contain no pair of 1 bits that are a knight's move apart.

%C After a(0) = 0 when determining the next term in the sequence one must not only ensure that it contains no 1 bits that are a knight's move apart from any previous 1 bit, including the earlier 1 bits in the term itself, but also that the next square after the end of the term is not a knight's move apart from any previous 1 bit as the next term must begin with a 1 bit.

%C It is conjectured that the sequence contains all nonnegative integers.

%H Scott R. Shannon, <a href="/A396603/b396603.txt">Table of n, a(n) for n = 0..1000</a>

%H Scott R. Shannon, <a href="/A396603/a396603.png">Image of the first 1000 terms on the square spiral</a>. The 1 bits are black while the 0 bits are light gray. The bits that comprise each term are connected by a red line.

%e The spiral begins:

%e .

%e                                 .

%e     0---1---0---0---0---0---1   1

%e     |                       |   |

%e     0   0---0---0---0---0   0   1

%e     |   |               |   |   |

%e     0   1   0---0---1   0   0   0

%e     |   |   |       |   |   |   |

%e     1   1   0   0---1   1   0   1

%e     |   |   |           |   |   |

%e     0   1   0---0---1---0   0   0

%e     |   |                   |   |

%e     0   0---0---0---0---0---0   1

%e     |                           |

%e     0---1---0---0---1---0---0---1

%e .

%e a(0) = 0 by definition.

%e a(1) = 1 as there are no 1 other bits on the spiral, and the third square, the next square after the end of a(1) = 1, can contain a 1 bit as it is not a knight's move away from any existing 1 bit.

%e a(2) = 32. The 5th square cannot contain a 1 bit as it is a knight's move away from the 2nd square which contains a 1 bit, so a(2) cannot be 2 or 3. Likewise the 6th, 7th and 8th square cannot contain a 1 bit, leaving 32 as the smallest number that contains no 1 bits in these squares while allowing a 1 bit in the 9th square to start the fourth term.

%Y Cf. A396604 (start at 1), A007088, A396607, A396608, A394166.

%K nonn,base

%O 0,3

%A _Scott R. Shannon_, May 30 2026