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

1,3

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

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

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

Cf. A000225, A058891, A076688.

Cf. A005097 (a subsequence).

Cf. A290634 (complement).

Sequence in context: A260451 A328749 A036848 * A128864 A106798 A214640

Adjacent sequences: A289582 A289583 A289584 * A289586 A289587 A289588

Keyword

nonn

Author

Bernard Schott, Jul 08 2017

Status

approved