A049237 Quotient n/phi(n) for n in A007694.

1, 2, 2, 3, 2, 3, 2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2

Offset: 1

Author

Labos Elemer

Keywords

• nonn

Comments

Here phi(n) denotes Euler's totient function A000010.

As n increases, the proportion of 3's seems to approach 100 percent (it is 40 percent for the first 10 results; 82 percent for 100 results; 87.5 percent for 200 results while up to 200 million, for the first 235 results, is 88.51 percent). - Zoltan Galantai, Jul 28 2019

From Bernard Schott, Jul 30 2019: (Start)

According to [Ecker and Beslin], the quotients n/phi(n) when phi(n) divides n can take only 3 distinct values:

n/phi(n) = 1 iff n = 1,

n/phi(n) = 2 iff n = 2^w, w >= 1,

n/phi(n) = 3 iff n = 2^w * 3^u, w >= 1, u >= 1.

The previous comment follows because between 2^k and 2^(k+1) there are two consecutive integers for which n/phi(n) = 2, and there are floor(k*(log(2)/log(3))) integers of the form 2^b*3^c (b and c>=1) for which n/phi(n) = 3. (End)

Examples

			For powers of 2 the quotient is 2.
a(95) = 124416/phi(124416) = 124416/41472 = 3.
		

References

• Sárközy A. and Suranyi J., Number Theory Problem Book (in Hungarian), Tankonyvkiado, Budapest, 1972.

Links

• Jinyuan Wang, Table of n, a(n) for n = 1..10000
• Michael W. Eckert and Scott J. Beslin, Find all positive integers n such that phi(n) divides n, AMM, Vol. 93, No 8, Oct. 86, 656-657, E 3037.

Crossrefs

Cf. A000010, A007694, A062356.

Programs

Magma

v:=[m:m in [1..150000]|m mod EulerPhi(m) eq 0];[v[k]/EulerPhi(v[k]):k in [1..#v]]; // Marius A. Burtea, Jul 28 2019

Mathematica

Select[#/EulerPhi@ # & /@ Range[10^6], IntegerQ] (* Michael De Vlieger, Jul 02 2016 *)

PARI

lista(NN) = for(n=1,NN,if(n%eulerphi(n)==0,print1(n/eulerphi(n),", "))); \\ Jinyuan Wang, Jul 31 2019

Formula

n/phi(n) is an integer iff n = 1 or n = 2^w*3^u for w = 1, 2, ... and u = 0, 1, 2, ...

Extensions

Edited by M. F. Hasler, Jul 02 2016