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A186116
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Number of nonisomorphic rings with n elements minus number of groups of order n.
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2
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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
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OFFSET
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1,4
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COMMENTS
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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.
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LINKS
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FORMULA
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EXAMPLE
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a(1) = 0 because there is a unique ring with 1 element, and a unique group of order 1, so 1 - 1 = 0.
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CROSSREFS
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KEYWORD
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sign,hard
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AUTHOR
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STATUS
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approved
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