A289585 Quotients as they appear as k increases when tau(k) divides phi(k).

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

Author

Bernard Schott, Jul 08 2017

Keywords

• nonn

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

Examples

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

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..1000

Crossrefs

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

Cf. A000225, A058891, A076688.

Cf. A005097 (a subsequence).

Cf. A290634 (complement).

Programs

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 *)

PARI

lista(nn) = {for (n=1, nn, q = eulerphi(n)/numdiv(n); if (denominator(q)==1, print1(q, ", ")););} \\ Michel Marcus, Jul 10 2017

Formula

a(n) = A000010(A020491(n)) / A000005(A020491(n)). - David A. Corneth, Jul 09 2017