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%I #68 Jun 26 2022 03:08:53
%S 6,10,14,18,21,22,26,30,34,36,38,39,42,46,50,54,55,57,58,62,66,70,74,
%T 78,82,86,90,93,94,98,100,102,105,106,108,110,111,114,118,122,126,129,
%U 130,134,138,142,146,147,150,154,155,158,162,165,166,170,174,178,180
%N Numbers k for which there exists a group of order k that cannot be generated by A051903(k) elements.
%C For such k, it follows from the MathOverflow thread in Links that 1 + A051903(k) generators suffice.
%C Lists all k such that A059829(k) != A051903(k). Also k such that A059829(k) = 1 + A051903(k).
%D R. Guralnick, A bound for the number of generators of a finite group, Arch. Math. 53 (1989), 521-523.
%D A. Lucchini, A bound on the number of generators of a finite group, Arch. Math. 53, (1989), 313-317.
%H MathOverflow, <a href="https://mathoverflow.net/questions/225172/">How large can the smallest generating set of a group G of order n be?</a>
%e k=20 is 2^2*5, so maximal exponent is 2. All five groups of order 20 can be generated by 2 elements. So 20 does NOT belong here.
%e On the other hand, k=21 is 3*7, so maximal exponent is 1. But there exists a group of order 21 that cannot be generated by 1 element. Therefore 21 belongs in this sequence.
%Y Cf. A051903, A059829.
%K nonn
%O 1,1
%A _Jeppe Stig Nielsen_, Apr 25 2020
%E More terms from _Jinyuan Wang_, Jun 26 2022
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