A396566 - OEIS

%I #6 Jun 02 2026 21:42:54

%S 24,504,6840,103776,1481430,23148840,370145514,6239664480,

%T 108190392360,1946102171640,35845601561076,677067443453190,

%U 13040933356636620,255861630224692890,5098732402525355154,103074299201812858200,2110069526732687732856,43695406785298898577360,914231452074242154776820

%N Number of bracelets (turnover necklaces) of length 4 using exactly 3 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.

%C 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.

%C This sequence is the r=3 slice (m=4) of the same 3D array as A396539 (r=4). Not to be confused with A214310 (varying length) or A214307 (representatives).

%F a(n) = 6 * binomial(A000081(n), 3) for n >= 4.

%e 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:

%e (1) A B A C

%e (2) A B C A

%e All rotations and reflections are considered identical. For n=5, N = 9 colors, there are 504 distinct bracelets.

%o (Python)

%o from math import comb

%o from functools import lru_cache

%o @lru_cache(maxsize=None)

%o def A000081(n):

%o     if n <= 1:

%o         return 1

%o     total = 0

%o     for j in range(1, n):

%o         s = 0

%o         for d in range(1, j + 1):

%o             if j % d == 0:

%o                 s += d * A000081(d)

%o         total += s * A000081(n - j)

%o     return total // (n - 1)

%o def a(n):

%o     return 6 * comb(A000081(n), 3)

%o print([a(n) for n in range(4, 23)])

%Y 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), A396539 (r=4 slice of the same 3D array).

%K nonn

%O 4,1

%A _Chen Wei-Shi_, May 29 2026