OFFSET
1,1
COMMENTS
The regular hexagon with side length n has circumradius n, so a(n) >= 6 for every n.
If a polygon with integer side lengths and circumradius n has k sides, then scaling all side lengths by a positive integer m gives a polygon with integer side lengths and circumradius m*n having the same number of sides.
The sequence is unbounded. This follows from an explicit family of cyclic polygons with arbitrarily many integer side lengths and integer circumradius.
If n has at least two distinct prime divisors congruent to 1 modulo 4, then a(n) >= 8. This follows from the Gaussian 8-gon construction and scaling.
Exact computation shows that a(n) is even for 1 <= n <= 400.
Conjecture: a(n) is even for every n.
For every prime p, a(p) = 6.
For every prime r == 1 (mod 3), choose positive integers s_r, u_r satisfying s_r^2 + 3*u_r^2 = 4*r^2 with s_r as small as possible, and put x_r = arcsin(s_r/(2*r)). For a semiprime radius n = p*q, the symbols x_p and x_q below denote the corresponding values.
For primes p <= q, a(p*q) is always in {6, 8, 18}. It is 18 if p and q are odd, p == 1 (mod 3), q == 1 (mod 3), and x_p + x_q < Pi/6. If this case does not hold, it is 8 if p < q, p == 1 (mod 4), and q == 1 (mod 4). In all remaining cases it is 6.
The condition x_p + x_q < Pi/6 is equivalent to 3*(u_p*s_q + u_q*s_p) < 3*u_p*u_q - s_p*s_q.
In particular, a(2*p) = a(3*p) = 6 for every prime p.
Proofs of these results are given in the linked document; the finite parts of the semiprime classification are accompanied by exact Maple certificates.
LINKS
Sean A. Irvine, Table of n, a(n) for n = 1..1000
John H. Conway, Charles Radin and Lorenzo Sadun, On Angles Whose Squared Trigonometric Functions are Rational, arXiv:math-ph/9812019, 1998.
Felix Huber, Maple program
FORMULA
a(n) >= 6.
a(m*n) >= a(n) for all positive integers m and n.
a((4*(k - 1)^2 + 1)^(k - 1)) >= 2*k for all k >= 2.
limsup_{n -> infinity} a(n)*log(log(n))/log(n) >= 1.
a(n) >= 8 if n has at least two distinct prime divisors congruent to 1 modulo 4.
For every prime r == 1 (mod 3), choose positive integers s_r, u_r satisfying s_r^2 + 3*u_r^2 = 4*r^2 with s_r as small as possible. Then:
For primes p <= q, a(p*q) = 18 if p and q are odd, p == 1 (mod 3), q == 1 (mod 3), and 3*(u_p*s_q + u_q*s_p) < 3*u_p*u_q - s_p*s_q.
For primes p < q, a(p*q) = 8 if p == 1 (mod 4), q == 1 (mod 4), and the preceding case does not hold.
For primes p <= q, a(p*q) = 6 in all remaining cases.
a(p) = a(2*p) = a(3*p) = 6 for every prime p.
EXAMPLE
For n = 7*13 = 91, one has (s_7, u_7) = (2, 8) and (s_13, u_13) = (1, 15). Since 3*(8*1 + 15*2) = 114 < 358 = 3*8*15 - 2*1, the semiprime classification gives a(91) = 18. An eighteen-gon is obtained from six sides of length 7, six sides of length 26, and six sides of length 61.
For n = 5*13 = 65, the value 18 is impossible because 5 is not congruent to 1 modulo 3. Since 5 and 13 are distinct primes congruent to 1 modulo 4, the Gaussian case gives a(65) = 8. One corresponding side-length multiset is {32, 32, 50, 50, 50, 50, 66, 66}.
For n = 7*11 = 77, the value 18 is impossible because 11 is not congruent to 1 modulo 3, and the value 8 is impossible because 7 and 11 are not both congruent to 1 modulo 4. Hence a(77) = 6.
MAPLE
# See Huber link.
CROSSREFS
KEYWORD
nonn
AUTHOR
Felix Huber, Aug 15 2026
STATUS
approved