A353108 a(n) is the number of cycles of n numbers arranged so that every integer in 1..n*(n-1)+1 occurs as the sum of up to n adjacent numbers. Both a solution and its reverse are counted unless they are identical.
1, 1, 2, 4, 2, 10, 0, 12, 8, 12, 0, 36, 0, 40, 0, 0, 12, 102, 0, 84, 0, 0, 0
Offset: 1
Examples
For n = 1, the only solution consists of the single number { 1 }, and a "cycle" consisting of { 1 } is the same whether read forward or backward, so a(1) = 1.
For n = 2, the only solution (starting at 1) consists of the two numbers { 1, 2 }; arranging these around a circle as
1
/ \
\ /
2
gives the same cycle, i.e., { 1, 2 } whether read clockwise or counterclockwise from 1, so a(2) = 1.
For n = 3, the two cycles (starting at 1) are { 1, 2, 4 } and { 1, 4, 2 }, so a(3) = 2.
For n = 8, the twelve solutions are
{ 1, 2, 10, 19, 4, 7, 9, 5 },
{ 1, 3, 5, 11, 2, 12, 17, 6 },
{ 1, 3, 8, 2, 16, 7, 15, 5 },
{ 1, 4, 2, 10, 18, 3, 11, 8 },
{ 1, 4, 22, 7, 3, 6, 2, 12 },
{ 1, 6, 12, 4, 21, 3, 2, 8 },
and the same six cycles read in the opposite direction from 1 (e.g.,
{ 1, 2, 10, 19, 4, 7, 9, 5 }
read in reverse order starting at 1 is
{ 1, 5, 9, 7, 4, 19, 10, 2 }
each of which counts as a separate solution), so a(8) = 12.
Links
- Donald Bell, Puzzle #171: Can you work out which numbers are on the bracelet?, New Scientist, 8 June 2022.
Crossrefs
Cf. A058241.
Formula
a(n) = 2 * A058241(n) for n > 2.
Extensions
a(12)-a(23) computed from A058241 by Max Alekseyev, Jun 10 2023
Comments