cp's OEIS Frontend

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

%I A181189 #28 May 17 2024 09:56:24
%S A181189 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,
%T A181189 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,
%U A181189 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
%N A181189 Maximal number of elements needed to identify an abelian group of order n by testing the order of random elements.
%H A181189 Max Alekseyev, <a href="/A181189/b181189.txt">Table of n, a(n) for n = 1..10000</a>
%H A181189 R. J. Mathar, <a href="/A181189/a181189.pdf">List of element order statistics for n <= 64</a>.
%F A181189 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.
%e A181189 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.
%e A181189 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
%e A181189         [4,2,2] 1:1 2:7 4:8
%e A181189         [4,4]   1:1 2:3 4:12
%e A181189     The sample 1:1, 2:3, 4:8 (12 in total) won't distinguish those two. - _Don Reble_, Oct 04 2023
%Y A181189 Cf. A000688, A005117.
%K A181189 nonn
%O A181189 1,4
%A A181189 _Isaac Lambert_, Oct 10 2010
%E A181189 Corrected and extended by _Don Reble_ - _N. J. A. Sloane_, Oct 04 2023
%E A181189 a(1)=0 prepended by _Max Alekseyev_, Oct 07 2023