A186116 Number of nonisomorphic rings with n elements minus number of groups of order n.

0, 1, 1, 9, 1, 2, 1, 47, 9, 2, 1, 17, 1, 2, 3, 376, 1, 17, 1, 17, 2, 2, 1, 89, 9, 2, 54, 18, 1, 4, 1

Offset: 1

Author

Jonathan Vos Post, Feb 13 2011

Keywords

• sign
• hard

Comments

a(p) = 1 for p prime, as there is a unique group of order p (the cyclic group), and 2 nonisomorphic rings with p elements, so 2 - 1 = 1.

a(p^2) = 9 for p prime, as there are 11 mutually nonisomorphic rings of order p^2 [Raghavendran, p. 228] and 2 groups of order p^2, so 11 - 2 = 9.

a(p^3) = 3*p+45 for p an odd prime, as there are 3*p+50 nonisomorphic rings with p^3 elements [R. Ballieu, Math. Rev. 9, 267; see also Math. Rev. 51#5655]; see also Antipkin, and 5 nonisomorphic groups of order p^3.

The first unknown value as of Feb 13, 2011 is a(32). Then a(64) is unknown.

In a sense, this measures the excess in combinatorial structures available by moving from one binary operation to two binary operations, and moving from the group axioms to the ring axioms.

Examples

			a(1) = 0 because there is a unique ring with 1 element, and a unique group of order 1, so 1 - 1 = 0.
		

Links

• V. G. Antipkin and V. P. Elizarov, Rings of order p^3, Sib. Math. J. vol. 23, no 4. (1982) pp 457-464, MR0668331 (84d:16025)
• R. Raghavendran, Finite associative rings, Compositio Mathematica, vol. 21, no. 2 (1969) pp 195-229.

Crossrefs

Cf. A000001, A027623.

Formula

a(n) = A027623(n) - A000001(n).