|
|
A049237
|
|
Quotient n/phi(n) for n in A007694.
|
|
7
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
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
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)
|
|
REFERENCES
|
Sárközy A. and Suranyi J., Number Theory Problem Book (in Hungarian), Tankonyvkiado, Budapest, 1972.
|
|
LINKS
|
|
|
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, ...
|
|
EXAMPLE
|
For powers of 2 the quotient is 2.
a(95) = 124416/phi(124416) = 124416/41472 = 3.
|
|
MATHEMATICA
|
|
|
PROG
|
(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
(PARI) lista(NN) = for(n=1, NN, if(n%eulerphi(n)==0, print1(n/eulerphi(n), ", "))); \\ Jinyuan Wang, Jul 31 2019
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|