A337518 Number of non-isomorphic graphs on n unlabeled nodes modulo 3.

1, 1, 2, 1, 2, 1, 0, 0, 1, 0, 2, 2, 0, 0, 1, 2, 0, 1, 0, 2, 1, 2, 0, 0, 2, 0, 0, 1, 1, 1, 0, 0, 2, 0, 1, 0, 2, 1, 0, 0, 2, 2, 0, 2, 0, 0, 2, 1, 2, 1, 0, 1, 0, 1, 1, 1, 0, 1, 2, 2, 1, 0, 1, 2, 0, 1, 2, 1, 1, 1, 1, 0, 0, 2, 1, 1, 2, 0, 2, 0, 0, 0, 2, 1, 2, 2, 1

Offset: 0

Author

Drake Thomas, Nov 21 2020

Keywords

• nonn

Comments

For the mod-2 case, the sequence is eventually constant, because there are an even number of graphs on n vertices for n>4. (In fact, the number of factors of 2 in A000088(n) is asymptotically n/2; see Cater and Robinson in the Links section.)

Examples

			For n = 4, there are 11 graphs on 4 nodes up to isomorphism, so a(4) = 2 = 11 mod 3.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..98
• Steven C. Cater and Robert W. Robinson, Exponents of 2 in the numbers of unlabeled graphs and tournaments, Congressus Numerantium, 82 (1991), pp. 139-155.

Crossrefs

Cf. A000088, A007149.

Programs

Python

from itertools import combinations
from math import prod, factorial, gcd
from fractions import Fraction
from sympy.utilities.iterables import partitions
def A337518(n): return int(sum(Fraction(1<>1)*r+(q*r*(r-1)>>1) for q, r in p.items()),prod(q**r*factorial(r) for q, r in p.items()))%3 for p in partitions(n))) % 3 # Chai Wah Wu, Jul 02 2024

Formula

a(n) = A000088(n) mod 3.