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A396539
Number of bracelets (turnover necklaces) of length 4 using exactly 4 colors, where the colors are chosen from a palette of size N = A000081(n) (the number of unlabeled rooted trees with n nodes), and each color corresponds to a distinct tree shape.
2
3, 378, 14535, 583740, 20740020, 818890215, 33128023503, 1434342872340, 64413854851335, 3036649176072765, 147728685433584465, 7431492259342213440, 383693601452301792795, 20304027597295616632395, 1096964233375116271929753, 60410210457682687411896075, 3383028314439620964543868575
OFFSET
4,1
COMMENTS
Let a(n) be the number of bracelets of length 4 using exactly 4 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 4 colors, there are 24 linear arrangements and the dihedral group D4 has 8 elements, giving 24/8 = 3 distinct bracelets per color set. Hence a(n) = 3 * binomial(N,4).
This sequence is the r=4 slice (m=4) of the 3D array A338859(4,n) = A321791(4, A000081(n)). Not to be confused with A214312 (varying length) or A214309 (representatives).
FORMULA
a(n) = 3 * binomial(A000081(n), 4) for n >= 4.
EXAMPLE
For n=4, N = A000081(4) = 4 colors (A, B, C, D). The 3 bracelets are:
(1) A B C D
(2) A B D C
(3) A C B D
All rotations and reflections are considered identical.
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(k):
return 3 * comb(A000081(k), 4)
print([a(k) for k in range(4, 21)])
CROSSREFS
Cf. A000081 (number of unlabeled rooted trees with n nodes, gives total colors N), A321791 (bracelets of length n using up to k colors), A338859 (unicyclic graphs with m cycle nodes and trees of size k).
Sequence in context: A068988 A136025 A370448 * A157577 A062604 A395836
KEYWORD
nonn
AUTHOR
Chen Wei-Shi, May 28 2026
STATUS
approved