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%I #20 May 14 2023 23:47:36
%S 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,
%T 1
%N Number of nonisomorphic rings with n elements minus number of groups of order n.
%C 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.
%C 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.
%C 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.
%C The first unknown value as of Feb 13, 2011 is a(32). Then a(64) is unknown.
%C 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.
%H V. G. Antipkin and V. P. Elizarov, <a href="http://dx.doi.org/10.1007/BF00968650">Rings of order p^3</a>, Sib. Math. J. vol. 23, no 4. (1982) pp 457-464, <a href="http://www.ams.org/mathscinet-getitem?mr=668331">MR0668331 (84d:16025)</a>
%H R. Raghavendran, <a href="http://www.numdam.org/item?id=CM_1969__21_2_195_0">Finite associative rings</a>, Compositio Mathematica, vol. 21, no. 2 (1969) pp 195-229.
%F a(n) = A027623(n) - A000001(n).
%e a(1) = 0 because there is a unique ring with 1 element, and a unique group of order 1, so 1 - 1 = 0.
%Y Cf. A000001, A027623.
%K sign,hard
%O 1,4
%A _Jonathan Vos Post_, Feb 13 2011
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