%I #60 Aug 25 2020 06:37:25

%S 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 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 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 Quotients as they appear as k increases when tau(k) divides phi(k).

%C Numbers k such that tau(k) divides phi(k) are in A020491.

%C Only for seven integers which are in A020488, we have a(n) = 1.

%C The integers such that a(n) = 2, 3, 4 are respectively in A062516, A063469, A063470.

%C 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 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 Robert G. Wilson v, <a href="/A289585/b289585.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) = A000010(A020491(n)) / A000005(A020491(n)). - _David A. Corneth_, Jul 09 2017

%e a(10) = 2 because A020491(10) = 15 and phi(15)/tau(15) = 8/4 = 2.

%p for n from 1 to 50 do q:=phi(n)/tau(n);

%p if q=floor(q) then print(n,q,phi(n),tau(n)) else fi; od:

%t 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 (PARI) lista(nn) = {for (n=1, nn, q = eulerphi(n)/numdiv(n); if (denominator(q)==1, print1(q, ", ")););} \\ _Michel Marcus_, Jul 10 2017

%Y Cf. A000010, A000005, A020491, A015733, A020488, A058891, A062516, A063469, A063470, A175667, A112955, A112954.

%Y Cf. A000225, A058891, A076688.

%Y Cf. A005097 (a subsequence).

%Y Cf. A290634 (complement).

%K nonn

%O 1,3

%A _Bernard Schott_, Jul 08 2017