A181189 Maximal number of elements needed to identify an abelian group of order n by testing the order of random elements.

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

Author

Isaac Lambert, Oct 10 2010

Keywords

• nonn

Examples

			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
		

Links

• Max Alekseyev, Table of n, a(n) for n = 1..10000
• R. J. Mathar, List of element order statistics for n <= 64.

Crossrefs

Cf. A000688, A005117.

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.

Extensions

Corrected and extended by Don Reble - N. J. A. Sloane, Oct 04 2023

a(1)=0 prepended by Max Alekseyev, Oct 07 2023