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%I #52 Feb 16 2024 01:20:46
%S 1,3,5,7,8,9,10,11,13,15,17,18,19,21,23,24,26,28,29,30,31,33,34,35,37,
%T 39,40,41,43,45,47,49,51,52,53,55,56,57,58,59,61,63,65,67,69,70,71,72,
%U 73,74,76,77,78,79,82,83,84,85,87,88,89,90,91,93,95,97,98,99,101,102,103,104
%N Numbers k such that sigma_0(k) divides phi(k).
%C In other words, numbers k such that d(k) divides phi(k).
%C From _Enrique Pérez Herrero_, Aug 11 2010: (Start)
%C sigma_0(k) divides phi(k) when:
%C k is an odd prime: A065091;
%C k is an odd squarefree number: A056911;
%C k = 2^m, where m <> 1 is a Mersenne number (A000225).
%C If d divides (p-1), with p prime, then p^(d-1) is in this sequence, as are p^(p-1), p^(p-2) and p^(-1+p^n).
%C (End)
%C phi(n) and d(n) are multiplicative functions, so if m and n are coprime and both of them are in this sequence then m*n is also in this sequence. - _Enrique Pérez Herrero_, Sep 05 2010
%C From _Bernard Schott_, Aug 14 2020: (Start)
%C The corresponding quotients are in A289585.
%C About the 3rd case of _Enrique Pérez Herrero_'s comment: if k = 2^M_m, where M_m = 2^m - 1 is a Mersenne number >= 3 (A000225), then the corresponding quotient phi(k)/d(k) is the integer 2^(2^m-m-2) = A076688(m); hence, these numbers k, A058891 \ {2}, form a subsequence. (End)
%H Enrique Pérez Herrero, <a href="/A020491/b020491.txt">Table of n, a(n) for n = 1..5000</a>
%H Psychedelic Geometry Blogspot, <a href="http://psychedelic-geometry.blogspot.com/2009/02/fermat-and-mersenne-numbers-conjecture_14.html">Fermat and Mersenne Numbers Conjecture-(2)</a>
%t Select[ Range[ 105 ], IntegerQ[ EulerPhi[ # ]/DivisorSigma[ 0, # ] ]& ]
%o (PARI) isok(k) = !(eulerphi(k) % numdiv(k)); \\ _Michel Marcus_, Aug 10 2020
%Y Cf. A000005, A000010.
%Y Complement of A015733. [_Enrique Pérez Herrero_, Aug 11 2010]
%Y Cf. A058891, A076688, A289585.
%K nonn
%O 1,2
%A _David W. Wilson_
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