OFFSET
0,3
COMMENTS
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.
It is conjectured that the sequence contains all nonnegative integers.
LINKS
Scott R. Shannon, Table of n, a(n) for n = 0..1000
Scott R. Shannon, Image of the first 1000 terms on the square spiral. The 1 bits are black while the 0 bits are light gray. The bits that comprise each term are connected by a red line.
EXAMPLE
The spiral begins:
.
.
0---1---0---0---0---0---1 1
| | |
0 0---0---0---0---0 0 1
| | | | |
0 1 0---0---1 0 0 0
| | | | | | |
1 1 0 0---1 1 0 1
| | | | | |
0 1 0---0---1---0 0 0
| | | |
0 0---0---0---0---0---0 1
| |
0---1---0---0---1---0---0---1
.
a(0) = 0 by definition.
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.
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.
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Scott R. Shannon, May 30 2026
STATUS
approved
