A396765 Number of distinct types of connected cyclic Alexander quandles of order 2n-1.

1, 1, 2, 3, 2, 3, 5, 2, 4, 5, 2, 3, 4, 3, 5, 7, 2, 4, 8, 4, 7, 7, 4, 3, 6, 4, 5, 4, 3, 3, 11, 2, 4, 7, 2, 7, 11, 4, 5, 7, 4, 3, 4, 4, 7, 5, 4, 6, 11, 4, 8, 7, 4, 3, 11, 6, 9, 4, 4, 8, 6, 6, 6, 11, 4, 7, 5, 6, 7, 7, 2, 9, 4, 4, 5, 11, 8, 8, 11, 4, 5, 9, 4, 3, 10, 3, 5, 8, 2, 3, 17, 8, 6, 8, 3, 7, 13, 4, 8, 11

Offset

1,3

Comments

For m >= 1, consider operations on Z/mZ of the form x*y = t*x + (1-t)*y mod m. The count is over odd orders m = 2n-1 and residues t modulo m satisfying gcd(t,m) = gcd(1-t,m) = 1. For each such t, the operation gives a connected cyclic Alexander quandle. Its type is the multiplicative order of t modulo m. This sequence counts the number of distinct types obtained in this way. For even m there are no admissible residues t modulo m: if gcd(t,m) = 1 and m is even, then t is odd, so 1-t is even and gcd(1-t,m) > 1.

References

W. Edwin Clark, Mohamed Elhamdadi, Masahico Saito and Timothy Yeatman, Quandle colorings of knots and applications, Journal of Knot Theory and Its Ramifications 23 (2014), no. 6, 1450035.

Links

Table of n, a(n) for n=1..100.

W. E. Clark, M. Elhamdadi, M. Saito and T. Yeatman, Quandle Colorings of Knots and Applications, arXiv preprint arXiv:1312.3307 [math.GT], 2013-2014.

Formula

a(n) = #{ ord_{2n-1}(t) : 0 <= t < 2n-1, gcd(t,2n-1) = gcd(1-t,2n-1) = 1 }.

Example

For n = 3, the order is 2n-1 = 5. The admissible residues are t = 2, 3, 4. Their multiplicative orders modulo 5 are 4, 4, 2 respectively, so the distinct types are 2 and 4. Hence a(3) = 2.
For n = 4, the order is 2n-1 = 7. The admissible residues are t = 2, 3, 4, 5, 6. Their multiplicative orders modulo 7 are 3, 6, 3, 6, 2 respectively, so the distinct types are 2, 3 and 6. Hence a(4) = 3.

Prog

(Python)
from math import gcd
def order_mod(t, n):
    if n == 1:
        return 1
    k = 1
    x = t % n
    while x != 1 % n:
        x = (x*t) % n
        k += 1
    return k
def b(m):
    if m == 1:
        return 1
    types = set()
    for t in range(m):
        if gcd(t, m) == 1 and gcd(1 - t, m) == 1:
            types.add(order_mod(t, m))
    return len(types)
def a(n):
    return b(2*n - 1)
print([a(n) for n in range(1, 101)])

Crossrefs

Cf. A181769, A181771, A193024, A193067, A225744.

Sequence in context: A174063 A327277 A124459 * A283360 A256366 A046147

Adjacent sequences: A396762 A396763 A396764 * A396767 A396768 A396770

Keyword

nonn

Author

Neil Panse, Jun 05 2026

Status

approved