OFFSET
4,1
COMMENTS
Let a(n) be the number of bracelets of length 4 using exactly 3 colors, where the colors are chosen from a set of size N = A000081(n) (unlabeled rooted trees with n nodes). For a fixed set of 3 colors, the color distribution must be (2,1,1). There are 3 choices for the color that appears twice, and for each such choice, there are 2 distinct bracelets (the two occurrences can be adjacent or opposite). Hence there are 3 * 2 = 6 distinct bracelets per color set. Therefore a(n) = 6 * binomial(N,3). Offset 4,1 because n >= 4.
FORMULA
a(n) = 6 * binomial(A000081(n), 3) for n >= 4.
EXAMPLE
For n=4, N = A000081(4) = 4 colors (A, B, C, D). Choosing colors A, B, C with A appearing twice, the 2 bracelets are:
(1) A B A C
(2) A B C A
All rotations and reflections are considered identical. For n=5, N = 9 colors, there are 504 distinct bracelets.
PROG
(Python)
from math import comb
from functools import lru_cache
@lru_cache(maxsize=None)
def A000081(n):
if n <= 1:
return 1
total = 0
for j in range(1, n):
s = 0
for d in range(1, j + 1):
if j % d == 0:
s += d * A000081(d)
total += s * A000081(n - j)
return total // (n - 1)
def a(n):
return 6 * comb(A000081(n), 3)
print([a(n) for n in range(4, 23)])
CROSSREFS
KEYWORD
nonn
AUTHOR
Chen Wei-Shi, May 29 2026
STATUS
approved
