(In reply to
My take by Tristan)
Nice approach and nice writeup. However, something isn't right in your counts for the 4bead necklaces since (n^4nn*(n1)/2)/4 = (1622/2)/4=13/4 when n=2.
From http://mathworld.wolfram.com/Necklace.html comes the formula (my notation, however)
N(b,n)=(1/b)(p(b1)n^(b/b1)+...+p(bD)n^(b/bD)
for the number N(b,n) of bbead necklaces with beads of n possible colors, where two necklaces are "the same" only if one can be rotated to coincide with the other (no flipping), and where
D=number of divisors (including 1 and b) of b,
b1,...,bD=the divisors of b,
p(k)=number of whole numbers j such that 1 <= j <= k and gcd(j,k)=1 (Euler's totient function).
For 3 beads, n colors this gives
N(3,n)=(1/3)(n^3+2n)
and for 4 beads, n colors
N(4,n)=(1/4)(n^4+n^2+2n).
This N(4,n) agrees with Charlie's numbers.
Edited on September 11, 2004, 1:16 pm

Posted by Richard
on 20040910 23:11:18 