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A181189
Maximal number of elements needed to identify an abelian group of order n by testing the order of random elements.
2
0, 0, 0, 3, 0, 0, 0, 5, 4, 0, 0, 7, 0, 0, 0, 13, 0, 7, 0, 11, 0, 0, 0, 13, 6, 0, 10, 15, 0, 0, 0, 29, 0, 0, 0, 19, 0, 0, 0, 21, 0, 0, 0, 23, 16, 0, 0, 37, 8, 11, 0, 27, 0, 19, 0, 29, 0, 0, 0, 31, 0, 0, 22, 61, 0, 0, 0, 35, 0, 0, 0, 37, 0, 0, 16, 39, 0, 0, 0, 61, 64, 0, 0, 43, 0, 0, 0, 45, 0, 31
OFFSET
1,4
FORMULA
For all squarefree n, a(n)=0, since there is only one abelian group of order n. Hence the group is trivially known without any checking.
EXAMPLE
For n=20, by the fundamental theorem of finite abelian groups, the group is either Z20 or Z10 x Z2. At worst, you could choose the identity, 1 element of order 2, 4 elements of order 5, and 4 elements of order 10. Then you still wouldn't know which group you have. But the order of the next element you choose will determine the group you have. So a(20)=11.
The previous value was a(16) = 9; It should be 13. Two of the size-16 groups have shapes [4,2,2] and [4,4], with element-orders:quantities
[4,2,2] 1:1 2:7 4:8
[4,4] 1:1 2:3 4:12
The sample 1:1, 2:3, 4:8 (12 in total) won't distinguish those two. - Don Reble, Oct 04 2023
CROSSREFS
Sequence in context: A330734 A081805 A333784 * A327889 A221702 A084681
KEYWORD
nonn
AUTHOR
Isaac Lambert, Oct 10 2010
EXTENSIONS
Corrected and extended by Don Reble - N. J. A. Sloane, Oct 04 2023
a(1)=0 prepended by Max Alekseyev, Oct 07 2023
STATUS
approved