Download n - Smarandache Notions Journal

Sebastián Martín Ruiz
Applications of Smarandache Function, and
Prime and Coprime Functions
0 if
C k (n1 , n 2 , L , n k ) = 
1
n1 , n2 ,L , n k
are coprime numbers
otherwise
American Research Press
Rehoboth
2002
Sebastián Martín Ruiz
Avda. De Regla, 43, Chipiona 11550 (Cadiz), Spain
[email protected]
www.terra.es/personal/smaranda
Applications of Smarandache Function, and Prime and
Coprime Functions
American Research Press
Rehoboth
2002
2
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Copyright 2002 by American Research Press
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http://www.gallup.unm.edu/~smarandache/math.htm
ISBN: 1-931233-30-6
Standard Address Number 297-5092
Printed in the United States of America
3
Contents:
Chapter 1: Smarandache Function applied to perfect numbers
Chapter 2: A result obtained using the Smarandache Function
Chapter 3: A Congruence with the Smarandache Function
Chapter 4: A functional recurrence to obtain the prime numbers
using the Smarandache prime function
Chapter 5: The general term of the prime number sequence and
the Smarandache prime function
Chapter 6:Expressions of the Smarandache Coprime Function
Chapter 7: New Prime Numbers
4
Chapter 1: Smarandache function applied to perfect
numbers
The Smarandache function is defined as follows:
S(n)= the smallest positive integer such that S(n)! is divisible by n. [1]
In this article we are going to see that the value this function takes when n
is a perfect number of the form n = 2 k −1 ⋅ (2 k − 1) , p = 2 k − 1 being a prime
number.
Lemma 1: Let n = 2 i ⋅ p when p is an odd prime number and i an integer
such that:
p 

 p
 p
 p
0 ≤ i ≤ E   + E  2  + E  3  + L + E  E (log 2 p )  = e2 ( p!)
2

2 
2 
2
where e2 ( p!) is the exponent of 2 in the prime number decomposition of
p!.
E(x) is the greatest integer less than or equal to x.
One has that S (n) = p .
Demonstration:
Given that GCD (2 i , p) = 1 (GCD= greatest common divisor) one has that
S (n) = max{S (2 i ), S ( p )} ≥ S ( p ) = p. Therefore S (n) ≥ p.
If we prove that p! is divisible by n then one would have the equality.
p!= p1
e p1 ( p!)
⋅ p2
e p2 ( p!)
L ps
e ps ( p!)
where pi is the i − th prime of the prime number decomposition of p!. It is
clear that
p1 = 2, p s = p , e ps ( p!) = 1 for which:
p!= 2
e ( p!)
2
⋅ p2
e p ( p!)
2
L p s −1
5
ep
s −1
( p!)
⋅p
From where one can deduce that:
p!
e ( p!) − i
( p!)
e ( p!)
e
=2 2
⋅ p 2 p2 L p s −1 ps −1
n
is a positive integer since e2 ( p!) − i ≥ 0 .
Therefore one has that S (n) = p
Proposition1: If n is a perfect number of the form n = 2 k −1 ⋅ (2 k − 1) with k is a
positive integer, 2 k −1 = p prime, one has that S (n) = p .
Demonstration:
For the Lemma it is sufficient to prove that k − 1 ≤ e2 ( p!) .
If we can prove that:
k − 1 ≤ 2 k −1 −
1
2
(1)
we will have proof of the proposition since:
k − 1 ≤ 2 k −1 −
1 2k −1 p
=
=
2
2
2
p
As k − 1 is an integer one has that k − 1 ≤ E   ≤ e2 ( p!)
2
1
Proving (1) is the same as proving k ≤ 2 k −1 + at the same time, since k is
2
integer, is equivalent to proving k ≤ 2 k −1 (2).
In order to prove (2) we may consider the function: f ( x) = 2 x −1 − x x real
number.
This function may be derived and its derivate is f ′( x) = 2 x −1 ln 2 − 1 .
f will be increasing when 2 x −1 ln 2 − 1 > 0 resolving x:
x > 1−
ln(ln 2)
≅ 1'5287
ln 2
In particular f will be increasing ∀ x ≥ 2 .
Therefore ∀ x ≥ 2
f ( x) ≥ f (2) = 0 that is to say 2 x −1 − x ≥ 0 ∀x ≥ 2 .
6
Therefore: 2 k −1 ≥ k ∀k ≥ 2 integer.
And thus is proved the proposition.
EXAMPLES:
6 = 2⋅3
28 = 2 2 ⋅ 7
496 = 2 4 ⋅ 31
8128= 2 6 ⋅ 127
S(6)=3
S(28)=7
S(496)=31
S(8128)=127
References:
[1] C. Dumitrescu and R. Müller: To Enjoy is a Permanent Component
of Mathematics. SMARANDACHE NOTIONS JOURNAL Vol. 9 No
1-2, (1998) pp 21-26
7
Chapter 2: A result obtained using the Smarandache
Function
Smarandache Function is defined as followed:
S(m)=The smallest positive integer so that S(m)! is divisible by m. [1]
n
Let’s see the value which such function takes for m = p p with n integer,
n≥2
and p prime number. To do so a Lemma required.
Lemma 1 ∀ m, n ∈ Ν m, n ≥ 2


n +1
n
 m n +1 − m n + m 
 m n +1 − m n + m 
 m −m +m 
L
E
+
+
E
+
mn = E



 E  logm  m n +1 −m n + m   
m
m2




m

where E(x) gives the greatest integer less than or equal to x.
Proof:
Let’s see in the first place the value taken by


E  log m  m n +1 − m n + m  



.
If n ≥ 2 : m n +1 − m n + m < m n +1 and therefore
log m (m n +1 − m n + m ) < log m m n +1 = n + 1 .
And if m ≥ 2 :
mm n ≥ 2m n ⇒ m n +1 ≥ 2m n ⇒ m n +1 + m ≥ 2m n ⇒ m n+1 − m n + m ≥ m n
⇒ log m m n +1 − m n + m ≥ log m m n = n ⇒ E log m m n +1 − m n + m ≥ n
(
[
)
)]
(
As a result: n ≤ E [log m (m n +1 − m n + m )] < n + 1 therefore:
[
)]
(
E log m m n +1 − m n + m = n if n, m ≥ 2
 m n +1 − m n + m 

mk


Now let’s see the value which it takes for 1 ≤ k ≤ n : E 
 m n +1 − m n + m 
1 
 n +1−k
− m n− k + k −1 
E
 = E m
k
m
m 



8
 m n +1 − m n + m 
n
n −1
 = m − m +1
k
m


If k=1: E 
 m n +1 − m n + m 
n +1− k
− m n− k
 =m
k
m


If 1 < k ≤ n : E 
Let’s see what is the value of the sum:
k=1
mn -mn-1
…
k=2
mn-1 -mn-2
k=3
mn-2
…
…
…
m2
-m
+1
-mn-3
.
.
.
k=n-1
.
.
.
k=n
m
-1
Therefore:
n
 m n +1 − m n + m 
n
=m
k
m


∑ E
k =1
Proposition:
∀
m, n ≥ 2
p prime number ∀n ≥ 2 :
n
S ( p p ) = p n +1 − p n + p
Proof:
Having e p (k ) = exponent of the prime number p in the prime decomposition
of k.
We get:


 k 
 k 
k
k
e p (k ) = E   + E  2  + E  3  + L + E  E (log p k ) 
p

 p
p 
p 


9
And using the lemma we have
[(
ep p
n +1


 p n +1 − p n + p 
 p n +1 − p n + p 
p n +1 − p n + p 

n
− p + p ! = E
 + E
 + L + E  E  log  p n +1 − p n + p    = p
2
p
p
p




 
 p  

)]
n
Therefore:
(p
n +1
p
And :
) ∈Ν
− pn + p !
pn
and
(p
n +1
p
( )= p
S pp
n
n +1
) ∉Ν
− pn + p −1 !
pn
− pn + p
References:
[1] C. Dumitrescu and R. Müller: To Enjoy is a Permanent Component of
Mathematics. SMARANDACHE NOTIONS JOURNAL VOL 9:, No. 1-2
(1998) pp 21-26.
10
Chapter 3: A Congruence with the Smarandache function
Smarandache’s function is defined thus:
S(n)= is the smallest integer such that S(n)! is divisible by n. [1]
In this article we are going to look at the value that has S(2k –1) (mod k)
For all integer, 2 ≤ k ≤ 97 .
k
S(2k-1)
S(2k-1) (mod k)
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
3
7
5
31
7
127
17
73
31
89
13
8191
127
151
257
131071
73
524287
41
337
683
178481
241
1801
8191
262657
127
2089
331
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
15
1
1
11
k
S(2k-1)
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
2147483647
1
65537
1
599479
1
131071
1
122921
1
109
1
616318177
1
524287
1
121369
1
61681
1
164511353
1
5419
1
2099863
1
2113
1
23311
1
2796203
1
13264529
1
673
1
4432676798593
1
4051
1
131071
1
8191
27
20394401
1
262657
1
201961
1
15790321
1
1212847
1
3033169
1
3203431780337
1
1321
1
2305843009213693951 1
2147483647
1
649657
1
6700417
1
145295143558111
1
599479
1
761838257287
1
131071
35
12
S(2k-1) (mod k)
k
S(2k-1)
S(2k-1) (mod k)
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
10052678938039
122921
212885833
38737
9361973132609
616318177
10567201
525313
581283643249112959
22366891
1113491139767
4278255361
97685839
8831418697
1
1
1
1
1
1
1
1
1
1
1
1
1
1
57912614113275649087721
1
14449
1
9520972806333758431 1
2932031007403
1
9857737155463
1
2931542417
1
618970019642690137449562111 1
18837001
1
23140471537
1
2796203
47
658812288653553079 1
165768537521
1
30327152671
1
22253377
1
13842607235828485645766393 1
One can see from the table that there are only 4 exceptions for 2 ≤ k ≤ 97
13
We can see in detail the 4 exceptions in a table:
k=28=22—7
k=52=22—13
k=68=22—17
k=92=22—23
S(228-1)h15 (mod 28 )
S(252-1)h27 (mod 52 )
S(268-1)h35 (mod 68 )
S(292-1)h47 (mod 92)
One can observe in these 4 cases that k=22p with p is a prime and more
k
2
over S (2 k − 1) ≡ + 1 (mod k )
UNSOLVED QUESTION:
One can obtain a general formula that gives us, in function of k the value
S (2 k − 1) (mod k ) for all positive integer values of k?.
Reference:
[1] Smarandache Notions Journal, Vol. 9, No. 1-2, (1998), pp. 21-26.
14
Chapter 4: A functional recurrence to obtain the prime numbers
using the Smarandache prime function.
Theorem: We are considering the function:
For n integer:
  i   i   i − 1 


2
−
  ∑    − 


2n
m
  j =1   j   j  

F ( n) = n + 1 + ∑ ∏  −  −

i
m = n +1 i = n +1
 

 
 

one has: p k +1 = F ( p k ) for all k ≥ 1 where {p k }k ≥1 are the prime numbers and
x  is the greatest integer less than or equal to x.
Observe that the knowledge of p k +1 only depends on knowledge of p k and
the knowledge of the fore primes is unnecessary.
Proof:
Suppose that we have found a function P(i ) with the following property:
1 if i is composite
P(i ) = 
0 if i is prime
This function is called Smarandache prime function.(Ref.)
Consider the following product:
m
∏ P(i)
i = pk +1
If p k < m < p k +1
m
∏ P(i) = 1 since i : p k +1 ≤ i ≤ m are all composites.
i = pk +1
15
m
If m ≥ p k +1
∏ P (i) = 0
i = p k +1
since P( p k +1 ) = 0
Here is the sum:
2 pk
∑
m
pk +1 −1
∏ P(i) = ∑
m = pk +1 i = pk +1
2 pk
m
∏ P(i) + ∑
m = pk +1 i = pk +1
m
pk +1 −1
∏ P(i ) = ∑ 1 =
m = pk +1 i = pk +1
m = pk +1
= p k +1 − 1 − ( p k + 1) + 1 = p k +1 − p k − 1
The second sum is zero since all products have the factor P ( p k +1 ) = 0.
Therefore we have the following recurrence relation:
2 pk
p k +1 = p k + 1 + ∑
m
∏ P(i )
m = pk +1 i = pk +1
Let’s now see we can find P(i ) with the asked property.
Consider:
 i   i − 1 1 si j | i
 j  −  j  = 0 si j not | i
  
 
We deduce of this relation:
i i 
 i − 1
d (i ) = ∑   − 

j =1  j 
 j 
where d (i) is the number of divisors of i .
16
j = 1,2,L , i i ≥ 1
If i is prime d (i ) = 2 therefore:
 d (i ) − 2 
− −
 = 0
i

If i is composite d (i ) > 2 therefore:
0<
d (i ) − 2
 d (i ) − 2 
< 1 ⇒ − −
 =1
i
i


Therefore we have obtained the Smarandache Prime Function P(i ) which is:

 i   i   i − 1 
 − 2
 ∑    − 


j =1  j 
 j 

P(i ) = − − 

i




i ≥ 2 integer
With this, the theorem is already proved .
References:
[1] E. Burton, “Smarandache Prime and Coprime functions”.
www.gallup.unm.edu/~Smarandache/primfnct.txt
[2]F. Smarandache, “Collected Papers”, Vol II 200, p.p. 137,
Kishinev University Press, Kishinev, 1997.
17
Chapter 5: The general term of the prime number
sequence and the Smarandache prime function.
Let is consider the function d(i) = number of divisors of the positive integer
number i . We have found the following expression for this function:
i
i
 i − 1
d (i ) = ∑ E   − E 

k =1  k 
 k 
“E(x) =Floor[x]”
We proved this expression in the article “A functional recurrence to obtain
the prime numbers using the Smarandache Prime Function”.
We deduce that the folowing function:
 d (i ) − 2 
G (i ) = − E −

i

This function is called the Smarandache Prime Function (Reference)
It takes the next values:
0 if
G (i ) = 
1 if
i is
prime
i is composite
Let is consider now π (n) = number of prime numbers smaller or equal than
n.
It is simple to prove that:
n
π (n) = ∑ (1 − G (i))
i=2
18
Let is have too:
If
If
 π (k ) 
E
=0
 n 
 π (k ) 
E
 =1
 n 
1 ≤ k ≤ pn − 1 ⇒
Cn ≥ k ≥ pn ⇒
We will see what conditions have to carry C n .
Therefore we have the following expression for p n n-th prime number:
Cn
 π (k ) 
p n = 1 + ∑ (1 − E 
)
k =1
 n 
If we obtain C n that only depends on n , this expression will be the general
term of the prime numbers sequence, since π is in function with G and G
does with d(i) that is expressed in function with i too. Therefore the
expression only depends on n.
Let is consider C n = 2(E (n log n ) + 1)
Since p n ≈ n log n from of a certain n0 it will be true that
(1) p n ≤ 2( E (n log n) + 1)
If n0 it is not too big, we can prove that the inequality is true for smaller or
equal values than n0 .
It is necessary to that:
 π (2( E (n log n) + 1)) 
E
 =1
n


If we check the inequality:
(2)
π (2( E (n log n) + 1)) < 2n
19
We will obtain that:
π (C n )
 π (C n ) 
 π (C n ) 
≤ 1 ; C n ≥ pn ⇒ E 
< 2 ⇒ E
 =1

n
 n 
 n 
We can experimentaly check this last inequality saying that it checks for a
lot of values and the difference tends to increase, wich makes to think that
it is true for all n .
Therefore if we prove that the (1) and (2) inequalities are true for all n
which seems to be very probable; we will have that the general term of the
prime numbers sequence is:

 
 j
 

 k 
 s∑=1( E ( j / s ) − E (( j − 1) / s )) − 2    

 ∑ 1 + E −
 
j

 j =2 

 
2 ( E ( n log n ) +1) 
 

   
pn = 1 +
1 − E 
∑

n
k =1













 

Reference:
[1] E. Burton, “Smarandache Prime and Coprime Functions”
Http://www.gallup.unm.edu/~Smarandache/primfnct.txt
[2] F. Smarandache, “Collected Papers”, Vol. II, 200 p.,p.137, Kishinev
University Press.
20
Chapter 6: Expressions of the Smarandache Coprime
Function
Smarandache Coprime function is defined this way:
0 if
C k (n1 , n 2 , L , n k ) = 
1
n1 , n 2 , L , n k
are coprime numbers
otherwise
We see two expressions of the Smarandache Coprime Function for k=2.
EXPRESSION 1:
 n n − lcm(n1 , n 2 ) 
C 2 ( n1 , n2 ) = −  − 1 2

n1n 2


x 
=the biggest integer number smaller or equal than x.
If n1 , n2 are coprime numbers:
lcm(n1 , n2 ) = n1 n2 therefore:
 0 
C 2 (n1 , n2 ) = − 
=0
 n1 n2 
If n1 , n2 aren’t coprime numbers:
lcm( n1 , n2 ) < n1 n2 ⇒ 0 <
n1n2 − lcm( n1 , n2 )
< 1 ⇒ C 2 (n1 , n 2 ) = 1
n1 n2
EXPRESSION 2:


∏ d − d' 
 d∏
|n1 d '|n2

 d >1 d '>1
C 2 (n1 , n 2 ) = 1 + −

 ∏ ∏ (d + d ' ) 

 d |n1 d |n2


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If n1 , n2 are coprime numbers then d ≠ d ' ∀d , d ' ≠ 1
∏ ∏ d − d'
⇒0<
d |n1 d '|n2
d >1 d '>1
∏ ∏ (d + d ' )
< 1 ⇒ C 2 (n1 , n 2 ) = 0
d |n1 d |n2
If n1 , n2 aren’t coprime numbers ∃ d = d ' d > 1, d ' > 1 ⇒ C 2 (n1 , n2 ) = 1
EXPRESSION 3:
Smarandache Coprime Function for k ≥ 2 :


1
− 1
C k (n1 , n 2 , L , nk ) = − 
 GCD(n1 , n 2 ,L , n k ) 
If n1 , n2 , L, nk are coprime numbers:
GCD(n1 , n 2 , L , n k ) = 1 ⇒ C k (n1 , n 2 , L , nk ) = 0
If n1 , n2 , L, nk aren’t coprime numbers: GCD(n1 , n2 , L, nk ) > 1
0<
1
 1

< 1 ⇒ −
− 1 = 1 = C k (n1 , n2 ,L , n k )
GCD
 GCD 
References:
1. E. Burton, “Smarandache Prime and Coprime Function”
2. F. Smarandache, “Collected Papers”, Vol II 22 p.p. 137,Kishinev
University Press.
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Chapter 7: New Prime Numbers
I have found some new prime numbers using the PROTH program of Yves
Gallot.
This program in based on the following theorem:
Proth Theorem (1878):
Let N = k ⋅ 2 n + 1 where k < 2 n . If there is an integer number a so that
N −1
a 2
≡ −1 (mod N ) therefore N is prime.
The Proth progam is a test for primality of greater numbers defined as
k ⋅ b n + 1 or k ⋅ b n − 1 . The program is made to look for numbers of less
than 5.000000 digits and it is optimized for numbers of more than 1000
digits..
Using this Program, I have found the following prime numbers:
3239 ⋅ 212345 + 1
with 3720 digits
with 3721 digits
with 3721 digits
with 3713 digits
7551⋅ 212345 + 1
7595 ⋅ 2
12345
+1
9363 ⋅ 212321 + 1
a = 3, a = 7
a = 5, a = 7
a = 3, a = 11
a = 5, a = 7
Since the exponents of the first three numbers are Smarandache number
Sm(5)=12345 we can call this type of prime numbers, prime numbers
of Smarandache .
Helped by the MATHEMATICA progam, I have also found new prime
numbers which are a variant of prime numbers of Fermat. They are the
following:
n
n
n
n
2 2 ⋅ 3 2 − 2 2 − 3 2 for n=1, 4, 5, 7 .
It is important to mention that for n=7 the number which is obtained has
100 digits.
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Chris Nash has verified the values n=8 to n=20, this last one being a
number of 815.951 digits, obtaining that they are all composite. All of them
have a tiny factor except n=13.
References:
1. Micha Fleuren, “Smarandache Factors and Reverse Factors”,
Smarandache Notions Journal, Vol. 10,
www.gallup.unm.edu/~smarandache/
2. Chris Caldwell, The Prime Pages, www.utm.edu/research/primes
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A book for people who love numbers:
Smarandache Function applied to perfect numbers, congruences.
Also, the Smarandache Prime and Coprime functions in connection with the
expressions of the prime numbers.
$5.95
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