OFFSET
1,3
COMMENTS
Numbers k such that tau(k) divides phi(k) are in A020491.
Only for seven integers which are in A020488, we have a(n) = 1.
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.
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
LINKS
Robert G. Wilson v, Table of n, a(n) for n = 1..1000
FORMULA
EXAMPLE
a(10) = 2 because A020491(10) = 15 and phi(15)/tau(15) = 8/4 = 2.
MAPLE
for n from 1 to 50 do q:=phi(n)/tau(n);
if q=floor(q) then print(n, q, phi(n), tau(n)) else fi; od:
MATHEMATICA
f[n_] := Block[{d = EulerPhi[n]/DivisorSigma[0, n]}, If[ IntegerQ@d, d, Nothing]]; Array[f, 120] (* Robert G. Wilson v, Jul 09 2017 )
PROG
(PARI) lista(nn) = {for (n=1, nn, q = eulerphi(n)/numdiv(n); if (denominator(q)==1, print1(q, ", ")); ); } \\ Michel Marcus, Jul 10 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Bernard Schott, Jul 08 2017
STATUS
approved