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%I #22 Dec 15 2017 17:35:18
%S 1,2,39,40,58,65,93,101,145,149,150,159,160,163,164,166,214,231,232,
%T 235,236,238,254,329,331,332,333,353,355,356,358,362,363,364,366,393,
%U 401,403,404,405,407,408,413,414,419,420,422,423,424,425,427,428,537
%N Numbers n such that Mertens's function of n (A002321) is divisible by phi(n).
%C Except for the initial term, this sequence is the same as A028442, the n for which Mertens's function M(n) is zero. Because phi(n) >= sqrt(n) and M(n) < sqrt(n) for all known n, phi(n) does not divide M(n), except possibility for some extremely large n. Research project: find the least n > 1 with M(n) not zero and phi(n) divides M(n). - _T. D. Noe_, Jul 28 2005
%H Harry J. Smith, <a href="/A062982/b062982.txt">Table of n, a(n) for n = 1..1000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/MertensConjecture.html">Mertens Conjecture</a>
%t Select[Range[500], Mod[Plus @@ MoebiusMu[Range[#]], EulerPhi[#]] == 0 &] (* _Carl Najafi_, Aug 17 2011 *)
%o (PARI) M(n)=sum(k=1,n,moebius(k)); j=[]; for(n=1,1500, if(Mod(M(n),eulerphi(n))==0,j=concat(j,n))); j
%o (PARI) { n=m=0; for (k=1, 10^9, m+=moebius(k); if (m%eulerphi(k)==0, write("b062982.txt", n++, " ", k); if (n==1000, break)) ) } \\ _Harry J. Smith_, Aug 15 2009
%Y Cf. A002321.
%K nonn
%O 1,2
%A _Jason Earls_, Jul 25 2001
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