A357602 a(n) is the number of n-gons in A000940 that are asymmetric.

0, 0, 0, 1, 15, 121, 1026, 8696, 81515, 827282, 9200052, 111138573, 1452810216, 20431516128, 307685506112, 4940087544198, 84241712837259, 1520563450896396, 28963118215975362, 580578881769905815, 12217399194536594927, 269291840875920127842, 6204484016841896754800

Offset

3,5

Links

Ludovic Schwob, Table of n, a(n) for n = 3..100

Formula

a(n) = A000940(n) - A342533(n).

a(p) = ((((p-1)! + 1)/p) + p - 2 - (2^((p-1)/2)*((p-1)/2)!))/4 for prime p. See A007619.

Example

Of the A000940(6) = 12 hexagons, 11 have symmetry and 1 is asymmetric, so a(6)=1.

Prog

(SageMath)
def a357602(n):
    s = 0
    for d in divisors(n):
        if d!=2 or n%2==1:
            s += moebius(d) * euler_phi(d) * factorial(n/d) * d^(n/d)
        p = -moebius(d) * d^2 * euler_phi(d) * factorial((n//d)//2) * (2*d)^((n//d)//2)
        if d==1 and n%2==0:
            s += ((n/2)^2+3*n/2+1) * -factorial(n/2) * 2^(n/2)
        elif d==2:
            if n%4==0:
                s += (n/4)^2 * (3+4/n) * p
            else:
                s += n^2/2 * p
        elif (n//d)%2==0:
            if d%2==0:
                s += (n/d/2)^2 * (1+4/n) * p
            else:
                s += (n/d/2)^2 * (1+2/n) * p
        else:
            s += (n/d)^2 * p
    return s/(4*n^2)
# Ludovic Schwob, Jan 07 2026

Crossrefs

Cf. A000940, A342533.

Cf. A064852 (cycles in A002619 with no rotational symmetry), A324513 (polygons in A000939 with no rotational symmetry).

Sequence in context: A081079 A138424 A279267 * A120794 A264377 A038743

Adjacent sequences: A357599 A357600 A357601 * A357603 A357604 A357605

Keyword

nonn

Author

Ian Mooney, Oct 05 2022

Extensions

Terms a(14) and beyond from Ludovic Schwob, Jan 07 2026

Status

approved