A396568 Number of bracelets (turnover necklaces) of length 4 using exactly 2 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.

4, 24, 144, 760, 4512, 26220, 163020, 1032484, 6782244, 45419980, 311775420, 2174371512, 15421367820, 110807959560, 806060157124, 5924773164244, 43967355888600, 329024223763512, 2481154828428784, 18839416123775160, 143951380037336340, 1106253702164150544, 8546422021727402760, 66346825443130687180

Offset

3,1

Comments

These sequences arise from a bijection between bracelet colorings and unicyclic graphs, where each unlabeled rooted tree of size n represents a distinct color. Hence the total number of available colors is N = A000081(n).

Let a(n) be the number of bracelets of length 4 using exactly 2 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 2 colors, the possible color distributions are (3,1) and (2,2). The (3,1) pattern gives 2 distinct bracelets (the single bead can be adjacent or opposite), and the (2,2) pattern gives 2 distinct bracelets (adjacent pairs or opposite pairs). Hence there are 2 + 2 = 4 distinct bracelets per color set. Therefore a(n) = 4 * binomial(N,2). Offset 3,1 because n >= 3.

The parameter m=4 (bracelet length) is fixed; the present sequence varies r=2 (exactly 2 colors) and n (via A000081(n)). See the 3D array defined in A396539 for the general case.

This sequence is the r=2 slice (m=4) of the same 3D array as A396539 (r=4) and A396566 (r=3). Not to be confused with A214308 (varying length) or A056357 (unlabeled 2-color bracelets).

Links

Table of n, a(n) for n=3..26.

Formula

a(n) = 4 * binomial(A000081(n), 2) = 2 * A000081(n) * (A000081(n) - 1) for n >= 3.

Example

For n=3, N = A000081(3) = 2 distinct colors, say X and Y. The only unordered pair of colors is {X, Y}. The 4 distinct bracelets of length 4 using exactly these two colors are:
  X X X Y (single Y adjacent)
  X X Y X (single Y opposite)
  X X Y Y (adjacent pairs)
  X Y X Y (opposite pairs)
All rotations and reflections are considered identical. Hence a(3) = 4.
For n=4, N = A000081(4) = 4 colors (A, B, C, D). Choosing any unordered pair of distinct colors, say A and B, there are again 4 bracelets as listed above (with X,Y replaced by A,B). Since there are C(4,2) = 6 such color pairs, the total number is 6 * 4 = 24, matching a(4) = 4 * C(4,2) = 24.
For n=5, N = A000081(5) = 9 colors. The number of unordered color pairs is C(9,2) = 36, and each gives 4 bracelets, so a(5) = 36 * 4 = 144.

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 4 * comb(A000081(n), 2)
print([a(n) for n in range(3, 27)])

Crossrefs

Cf. A000081 (number of colors), A396539, A396566, A214308, A056357.

Sequence in context: A370234 A114169 A121102 * A307526 A067411 A045915

Adjacent sequences: A396564 A396566 A396567 * A396569 A396570 A396571

Keyword

nonn

Author

Chen Wei-Shi, May 29 2026

Status

approved