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%I #18 Mar 31 2012 13:22:29
%S 4,6,10,14,20,22,26,28,34,38,42,44,46,50,58,62,74,75,76,78,82,86,90,
%T 92,94,106,114,118,122,124,125,134,135,142,146,158,166,172,178,186,
%U 188,194,202,204,206,214,218,222,226,236,254,258
%N Numbers k such that A000001(k) > 1 and A000001(k) | k.
%C Numbers k such that the number of groups of order k (when greater than 1) divides the group order k. I require a proper divisor > 1 because trivially for any p there is 1 group (the cyclic group) of order p, and 1 | p. Even semiprimes A100484 are a proper subset, because when k = p*q for primes p and q, then A000001(k) = 1 if gcd(p,q-1) = 1, 2 if gcd(p,q-1) = p, and (p < q).
%H Nathaniel Johnston, <a href="/A184335/b184335.txt">Table of n, a(n) for n = 1..324</a>
%e a(11) = 42 is in the sequence because there are 6 nonisomorphic groups of order 42, and 42/6 = 7.
%e a(18) = 75 is the first odd value, because there are 5 nonisomorphic groups of order 75, and 75/5 = 15. The next odd value is 125.
%Y Cf. A000001, A100484.
%K nonn,easy
%O 1,1
%A _Jonathan Vos Post_, Feb 13 2011
%E a(44) - a(52) from _Nathaniel Johnston_, Apr 26 2011
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