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

1,4

Links

Max Alekseyev, Table of n, a(n) for n = 1..10000

R. J. Mathar, List of element order statistics for n <= 64.

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

Cf. A000688, A005117.

Sequence in context: A330734 A081805 A333784 * A327889 A221702 A084681

Adjacent sequences: A181186 A181187 A181188 * A181190 A181191 A181192

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