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A370360
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Number of labeled semisimple rings with n elements.
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0
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1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 6227020800, 87178291200, 1307674368000, 24409921536000, 355687428096000, 6402373705728000, 121645100408832000, 2432902008176640000, 51090942171709440000, 1124000727777607680000
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OFFSET
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1,2
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COMMENTS
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Using the Artin-Wedderburn theorem, a finite semisimple ring is a product of matrix algebras over finite field. In particular, if n is squarefree then any semisimple ring of cardinal n is commutative. One can be more precise, indeed all semisimple rings with n elements are commutative if and only if the only 4th power that divides n is 1.
The analogous sequences for abelian groups and cyclic groups are A034382 and A034381, respectively.
In the case of commutative semisimple rings, we get the factorial numbers.
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LINKS
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FORMULA
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If n is squarefree then we have a(n) = n!. More precisely, a(n) = n! if and only if the only 4th power that divides n is 1. In particular, n=16 is the smallest n such that a(n) is different from n!.
If n and m are relatively prime, then a(n*m) = (n*m)!*a(n)*a(m)/(n!*m!).
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EXAMPLE
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For n=4, we have two possible rings: F_4 and F_2 X F_2. We use the notation F_q to denote the finite field with q elements. To compute a(4) we need to know how many ring automorphisms F_4 and F_2 X F_2 admit. For F_4, we have that Aut(F_4) is generated by the Frobenius morphism, hence we have 2 automorphisms. For F_2 X F_2, the only nontrivial automorphism is exchanging the two coordinates, hence we also have 2 automorphisms. Hence:
a(4) = 24/2 + 24/2 = 24.
We can compute a(2^k) for some small values of k:
a(4) = 4! = 24,
a(8) = 8!,
a(16) = 16! + 16!/6,
a(32) = 32! + 32!/6,
a(64) = 64! + 64!/12 + 64!/12,
a(128) = 128! + 128!/36 + 128!/18 + 128!/12,
...
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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