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%I A289585 #60 Aug 25 2020 06:37:25
%S A289585 1,1,2,3,1,2,1,5,6,2,8,1,9,3,11,1,3,2,14,1,15,5,4,6,18,6,2,20,21,4,23,
%T A289585 14,8,4,26,10,3,9,7,29,30,6,12,33,11,3,35,2,36,9,6,15,3,39,10,41,2,16,
%U A289585 14,5,44,2,18,15,18,48,7,10,50,4,51,6,6,13,53,3,54,5,18,56,22,12,24,2
%N A289585 Quotients as they appear as k increases when tau(k) divides phi(k).
%C A289585 Numbers k such that tau(k) divides phi(k) are in A020491.
%C A289585 Only for seven integers which are in A020488, we have a(n) = 1.
%C A289585 The integers such that a(n) = 2, 3, 4 are respectively in A062516, A063469, A063470.
%C A289585 When p is an odd prime then phi(p) = p-1, tau(p) = 2, so phi(p)/tau(p) = (p-1)/2 and A005097 is an infinite subsequence.
%C A289585 For k = A058891(m+1), that is 2^A000225(m), with m>=2, the corresponding quotient phi(k)/tau(k) is the integer A076688(m). - _Bernard Schott_, Aug 15 2020
%H A289585 Robert G. Wilson v, <a href="/A289585/b289585.txt">Table of n, a(n) for n = 1..1000</a>
%F A289585 a(n) = A000010(A020491(n)) / A000005(A020491(n)). - _David A. Corneth_, Jul 09 2017
%e A289585 a(10) = 2 because A020491(10) = 15 and phi(15)/tau(15) = 8/4 = 2.
%p A289585 for n from 1 to 50 do q:=phi(n)/tau(n);
%p A289585 if q=floor(q) then print(n,q,phi(n),tau(n)) else fi; od:
%t A289585 f[n_] := Block[{d = EulerPhi[n]/DivisorSigma[0, n]}, If[ IntegerQ@d, d, Nothing]]; Array[f, 120] (* _Robert G. Wilson v_, Jul 09 2017 )
%o A289585 (PARI) lista(nn) = {for (n=1, nn, q = eulerphi(n)/numdiv(n); if (denominator(q)==1, print1(q, ", ")););} \\ _Michel Marcus_, Jul 10 2017
%Y A289585 Cf. A000010, A000005, A020491, A015733, A020488, A058891, A062516, A063469, A063470, A175667, A112955, A112954.
%Y A289585 Cf. A000225, A058891, A076688.
%Y A289585 Cf. A005097 (a subsequence).
%Y A289585 Cf. A290634 (complement).
%K A289585 nonn
%O A289585 1,3
%A A289585 _Bernard Schott_, Jul 08 2017

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