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

1,4

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.

Links

Table of n, a(n) for n=1..31.

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.

Formula

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

Example

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

Crossrefs

Cf. A000001, A027623.

Sequence in context: A339757 A354347 A010166 * A242627 A289502 A010165

Adjacent sequences: A186113 A186114 A186115 * A186117 A186118 A186119

Keyword

sign,hard

Author

Jonathan Vos Post, Feb 13 2011

Status

approved