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Given a positive integer $m$ , how can I construct a positive integer $k$ such that $varphi(n)=k$ has exactly $m$ solutions?
Clash Royale CLAN TAG#URR8PPP
up vote
7
down vote
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In this article : https://en.wikipedia.org/wiki/Euler%27s_totient_function
it is stated that for every positive integer $mge 2$, there exists a postive integer $k$, such that $varphi(n)=k$ has exactly $m$ solutions in positive integers $n$.
How can I construct such a number $k$ for a given $m$ ?
I determined such a $k$ for $2le mle 100$ , but I did not find a useful pattern to see how to get a number for arbitary $m$
Here the values I found :
m k
2 10
3 44
4 4
5 8
6 12
7 32
8 36
9 40
10 24
11 48
12 160
13 396
14 2268
15 704
16 312
17 72
18 336
19 216
20 936
21 144
22 624
23 1056
24 1760
25 360
26 2560
27 384
28 288
29 1320
30 3696
31 240
32 768
33 9000
34 432
35 7128
36 4200
37 480
38 576
39 1296
40 1200
41 15936
42 3312
43 3072
44 3240
45 864
46 3120
47 7344
48 3888
49 720
50 1680
51 4992
52 17640
53 2016
54 1152
55 6000
56 12288
57 4752
58 2688
59 3024
60 13680
61 9984
62 1728
63 1920
64 2400
65 7560
66 2304
67 22848
68 8400
69 29160
70 5376
71 3360
72 1440
73 13248
74 11040
75 27720
76 21840
77 9072
78 38640
79 9360
80 81216
81 4032
82 5280
83 4800
84 4608
85 16896
86 3456
87 3840
88 10800
89 9504
90 18000
91 23520
92 39936
93 5040
94 26208
95 27360
96 6480
97 9216
98 2880
99 26496
100 34272
number-theory elementary-number-theory totient-function
asked Aug 2 at 21:30
Peter
44.8k938119
add a comment |Â
up vote
7
down vote
favorite
In this article : https://en.wikipedia.org/wiki/Euler%27s_totient_function
it is stated that for every positive integer $mge 2$, there exists a postive integer $k$, such that $varphi(n)=k$ has exactly $m$ solutions in positive integers $n$.
How can I construct such a number $k$ for a given $m$ ?
I determined such a $k$ for $2le mle 100$ , but I did not find a useful pattern to see how to get a number for arbitary $m$
Here the values I found :
m k
2 10
3 44
4 4
5 8
6 12
7 32
8 36
9 40
10 24
11 48
12 160
13 396
14 2268
15 704
16 312
17 72
18 336
19 216
20 936
21 144
22 624
23 1056
24 1760
25 360
26 2560
27 384
28 288
29 1320
30 3696
31 240
32 768
33 9000
34 432
35 7128
36 4200
37 480
38 576
39 1296
40 1200
41 15936
42 3312
43 3072
44 3240
45 864
46 3120
47 7344
48 3888
49 720
50 1680
51 4992
52 17640
53 2016
54 1152
55 6000
56 12288
57 4752
58 2688
59 3024
60 13680
61 9984
62 1728
63 1920
64 2400
65 7560
66 2304
67 22848
68 8400
69 29160
70 5376
71 3360
72 1440
73 13248
74 11040
75 27720
76 21840
77 9072
78 38640
79 9360
80 81216
81 4032
82 5280
83 4800
84 4608
85 16896
86 3456
87 3840
88 10800
89 9504
90 18000
91 23520
92 39936
93 5040
94 26208
95 27360
96 6480
97 9216
98 2880
99 26496
100 34272
number-theory elementary-number-theory totient-function
asked Aug 2 at 21:30
Peter
44.8k938119
Presumably you could find your answer by looking up Ford's 1999 paper in the Annals of Mathematics...
– Chris Custer
Aug 2 at 22:16
You could see that the results for successive numbers are quite irregular so it is to be assumed that there is no formula to determine these numbers. If it exists, it would have been established long ago I guess. However........
– Piquito
Aug 2 at 22:42
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
In this article : https://en.wikipedia.org/wiki/Euler%27s_totient_function
it is stated that for every positive integer $mge 2$, there exists a postive integer $k$, such that $varphi(n)=k$ has exactly $m$ solutions in positive integers $n$.
How can I construct such a number $k$ for a given $m$ ?
I determined such a $k$ for $2le mle 100$ , but I did not find a useful pattern to see how to get a number for arbitary $m$
Here the values I found :
m k
2 10
3 44
4 4
5 8
6 12
7 32
8 36
9 40
10 24
11 48
12 160
13 396
14 2268
15 704
16 312
17 72
18 336
19 216
20 936
21 144
22 624
23 1056
24 1760
25 360
26 2560
27 384
28 288
29 1320
30 3696
31 240
32 768
33 9000
34 432
35 7128
36 4200
37 480
38 576
39 1296
40 1200
41 15936
42 3312
43 3072
44 3240
45 864
46 3120
47 7344
48 3888
49 720
50 1680
51 4992
52 17640
53 2016
54 1152
55 6000
56 12288
57 4752
58 2688
59 3024
60 13680
61 9984
62 1728
63 1920
64 2400
65 7560
66 2304
67 22848
68 8400
69 29160
70 5376
71 3360
72 1440
73 13248
74 11040
75 27720
76 21840
77 9072
78 38640
79 9360
80 81216
81 4032
82 5280
83 4800
84 4608
85 16896
86 3456
87 3840
88 10800
89 9504
90 18000
91 23520
92 39936
93 5040
94 26208
95 27360
96 6480
97 9216
98 2880
99 26496
100 34272
number-theory elementary-number-theory totient-function
asked Aug 2 at 21:30
Peter
44.8k938119
In this article : https://en.wikipedia.org/wiki/Euler%27s_totient_function
it is stated that for every positive integer $mge 2$, there exists a postive integer $k$, such that $varphi(n)=k$ has exactly $m$ solutions in positive integers $n$.
How can I construct such a number $k$ for a given $m$ ?
I determined such a $k$ for $2le mle 100$ , but I did not find a useful pattern to see how to get a number for arbitary $m$
Here the values I found :
m k
2 10
3 44
4 4
5 8
6 12
7 32
8 36
9 40
10 24
11 48
12 160
13 396
14 2268
15 704
16 312
17 72
18 336
19 216
20 936
21 144
22 624
23 1056
24 1760
25 360
26 2560
27 384
28 288
29 1320
30 3696
31 240
32 768
33 9000
34 432
35 7128
36 4200
37 480
38 576
39 1296
40 1200
41 15936
42 3312
43 3072
44 3240
45 864
46 3120
47 7344
48 3888
49 720
50 1680
51 4992
52 17640
53 2016
54 1152
55 6000
56 12288
57 4752
58 2688
59 3024
60 13680
61 9984
62 1728
63 1920
64 2400
65 7560
66 2304
67 22848
68 8400
69 29160
70 5376
71 3360
72 1440
73 13248
74 11040
75 27720
76 21840
77 9072
78 38640
79 9360
80 81216
81 4032
82 5280
83 4800
84 4608
85 16896
86 3456
87 3840
88 10800
89 9504
90 18000
91 23520
92 39936
93 5040
94 26208
95 27360
96 6480
97 9216
98 2880
99 26496
100 34272
number-theory elementary-number-theory totient-function
asked Aug 2 at 21:30
Peter
44.8k938119
asked Aug 2 at 21:30
Peter
44.8k938119
asked Aug 2 at 21:30
Peter
44.8k938119
asked Aug 2 at 21:30
Peter
44.8k938119
44.8k938119
Presumably you could find your answer by looking up Ford's 1999 paper in the Annals of Mathematics...
– Chris Custer
Aug 2 at 22:16
You could see that the results for successive numbers are quite irregular so it is to be assumed that there is no formula to determine these numbers. If it exists, it would have been established long ago I guess. However........
– Piquito
Aug 2 at 22:42
add a comment |Â
Presumably you could find your answer by looking up Ford's 1999 paper in the Annals of Mathematics...
– Chris Custer
Aug 2 at 22:16
You could see that the results for successive numbers are quite irregular so it is to be assumed that there is no formula to determine these numbers. If it exists, it would have been established long ago I guess. However........
– Piquito
Aug 2 at 22:42
Presumably you could find your answer by looking up Ford's 1999 paper in the Annals of Mathematics...
– Chris Custer
Aug 2 at 22:16
Presumably you could find your answer by looking up Ford's 1999 paper in the Annals of Mathematics...
– Chris Custer
Aug 2 at 22:16
You could see that the results for successive numbers are quite irregular so it is to be assumed that there is no formula to determine these numbers. If it exists, it would have been established long ago I guess. However........
– Piquito
Aug 2 at 22:42
You could see that the results for successive numbers are quite irregular so it is to be assumed that there is no formula to determine these numbers. If it exists, it would have been established long ago I guess. However........
– Piquito
Aug 2 at 22:42
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
This is a theorem of Kevin Ford in 1999, published in Annals of Mathematics. It is a quite difficult result, and the proof does not go by finding any pattern on the function $A(m)$ that assigns the $k$ to every $m$. It was conjectured long time ago by Sierpiński, and proved using a very strong conjecture (the H-Hypothesis) by Schinzel in 1961. It was a bit of a surprise that it could be proved unconditionality.
Of course, this does not mean that there is no "elementary proof" on this result.
In fact lemma 1.1 in that paper by Kevin Ford shows that, if $A(m)=k$ and $p$ is a prime such that $p>2m+1$, $2p+1$ and $2mp+1$ are prime and $dp+1$ are composite for all $dmid 2m$ except $d=2$ and $d=2m$, then $A(2mp)=k+2$.
This can be used to find a number $m$ for a given explcit $k$ such that $A(m)=k$, although you will not give the smallest one, and even it can be quite big.
answered Aug 2 at 22:40
xarles
83548
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
This is a theorem of Kevin Ford in 1999, published in Annals of Mathematics. It is a quite difficult result, and the proof does not go by finding any pattern on the function $A(m)$ that assigns the $k$ to every $m$. It was conjectured long time ago by Sierpiński, and proved using a very strong conjecture (the H-Hypothesis) by Schinzel in 1961. It was a bit of a surprise that it could be proved unconditionality.
Of course, this does not mean that there is no "elementary proof" on this result.
In fact lemma 1.1 in that paper by Kevin Ford shows that, if $A(m)=k$ and $p$ is a prime such that $p>2m+1$, $2p+1$ and $2mp+1$ are prime and $dp+1$ are composite for all $dmid 2m$ except $d=2$ and $d=2m$, then $A(2mp)=k+2$.
This can be used to find a number $m$ for a given explcit $k$ such that $A(m)=k$, although you will not give the smallest one, and even it can be quite big.
answered Aug 2 at 22:40
xarles
83548
add a comment |Â
up vote
4
down vote
accepted
This is a theorem of Kevin Ford in 1999, published in Annals of Mathematics. It is a quite difficult result, and the proof does not go by finding any pattern on the function $A(m)$ that assigns the $k$ to every $m$. It was conjectured long time ago by Sierpiński, and proved using a very strong conjecture (the H-Hypothesis) by Schinzel in 1961. It was a bit of a surprise that it could be proved unconditionality.
Of course, this does not mean that there is no "elementary proof" on this result.
In fact lemma 1.1 in that paper by Kevin Ford shows that, if $A(m)=k$ and $p$ is a prime such that $p>2m+1$, $2p+1$ and $2mp+1$ are prime and $dp+1$ are composite for all $dmid 2m$ except $d=2$ and $d=2m$, then $A(2mp)=k+2$.
This can be used to find a number $m$ for a given explcit $k$ such that $A(m)=k$, although you will not give the smallest one, and even it can be quite big.
answered Aug 2 at 22:40
xarles
83548
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
This is a theorem of Kevin Ford in 1999, published in Annals of Mathematics. It is a quite difficult result, and the proof does not go by finding any pattern on the function $A(m)$ that assigns the $k$ to every $m$. It was conjectured long time ago by Sierpiński, and proved using a very strong conjecture (the H-Hypothesis) by Schinzel in 1961. It was a bit of a surprise that it could be proved unconditionality.
Of course, this does not mean that there is no "elementary proof" on this result.
In fact lemma 1.1 in that paper by Kevin Ford shows that, if $A(m)=k$ and $p$ is a prime such that $p>2m+1$, $2p+1$ and $2mp+1$ are prime and $dp+1$ are composite for all $dmid 2m$ except $d=2$ and $d=2m$, then $A(2mp)=k+2$.
This can be used to find a number $m$ for a given explcit $k$ such that $A(m)=k$, although you will not give the smallest one, and even it can be quite big.
answered Aug 2 at 22:40
xarles
83548
This is a theorem of Kevin Ford in 1999, published in Annals of Mathematics. It is a quite difficult result, and the proof does not go by finding any pattern on the function $A(m)$ that assigns the $k$ to every $m$. It was conjectured long time ago by Sierpiński, and proved using a very strong conjecture (the H-Hypothesis) by Schinzel in 1961. It was a bit of a surprise that it could be proved unconditionality.
Of course, this does not mean that there is no "elementary proof" on this result.
In fact lemma 1.1 in that paper by Kevin Ford shows that, if $A(m)=k$ and $p$ is a prime such that $p>2m+1$, $2p+1$ and $2mp+1$ are prime and $dp+1$ are composite for all $dmid 2m$ except $d=2$ and $d=2m$, then $A(2mp)=k+2$.
This can be used to find a number $m$ for a given explcit $k$ such that $A(m)=k$, although you will not give the smallest one, and even it can be quite big.
answered Aug 2 at 22:40
xarles
83548
answered Aug 2 at 22:40
xarles
83548
answered Aug 2 at 22:40
xarles
83548
answered Aug 2 at 22:40
xarles
83548
83548
add a comment |Â
add a comment |Â
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Presumably you could find your answer by looking up Ford's 1999 paper in the Annals of Mathematics...
– Chris Custer
Aug 2 at 22:16
You could see that the results for successive numbers are quite irregular so it is to be assumed that there is no formula to determine these numbers. If it exists, it would have been established long ago I guess. However........
– Piquito
Aug 2 at 22:42