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A289585
Quotients as they appear as k increases when tau(k) divides phi(k).
5
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, 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, 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
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.
The integers such that a(n) = 2, 3, 4 are respectively in A062516, A063469, A063470.
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
FORMULA
a(n) = A000010(A020491(n)) / A000005(A020491(n)). - David A. Corneth, Jul 09 2017
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
Cf. A005097 (a subsequence).
Cf. A290634 (complement).
Sequence in context: A260451 A328749 A036848 * A128864 A106798 A214640
KEYWORD
nonn
AUTHOR
Bernard Schott, Jul 08 2017
STATUS
approved