using just two colours, how many different bracelets are there for

- 1 bead
- 2 beads
- 3 beads
- 4 beads
- 5 beads
- 6 beads?

if a bracelet can be rotated to another then it

__isn't__different

if it can be reflected to another (or turned over) then it

__isn't__different

1 bead (2 options)

2 beads (3 options)

3 beads (4 options)

4 beads (6 options)

for 5 beads there are 8 options:

5 colours the same = 2

4 of 1 colour and 1 of the other = 2

3 of 1 colour and 2 of the other = 4

for 6 beads there are 13 options:

or, as a list:

this can be developed to consider 'friends'

a bracelet is a 'friend' of another if it is a rotation of it

e.g.

rBrBBB

BrBBBr

rBBBrB

BBBrBr

BBrBrB

BrBrBB

are all friends

how many 'friends' do various bracelets have (including themselves...)?

- quite often this is the number of beads
- when isn't it the number of beads?

for example, there are only 2 friends for this bracelet

one way to think about this is to consider cutting the bracelet somewhere and see where else you can cut it to obtain the same string

there are only 2 possible strings for the example above: BrBrBr and rBrBrB

this can develop into the 'bracelet' (necklace) argument for Fermat's 'little' theorem

when the number of beads is prime then the number of 'friends' for all cases (other than the single colour ones) is the prime number

same colour cases

3 'friends'

again, 3 'friends'

Solomon W Golomb Combinatorial proof of Fermat's "Little" Theorem. The American Mathematical Monthly, Vol. 63, No. 10 (Dec., 1956), p. 718

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