Problems & Puzzles:
Problems
Problem 50.
Is there
a Cullen prime Cn with n prime ?
Patrick Capelle sent the
following problem for our pages:
The context :
Cn = n*2n+1 are called Cullen
numbers.
Wn = n*2n-1 are called Woodall numbers.
There are Woodall primes Wn with n prime.
Examples : n = 3, 751, 12379.
See
http://www.prothsearch.net/woodall.html
But what about the Cullen primes ?
Questions :
1. Is there a Cullen prime Cn
with n prime ?
2. Are there arguments against the
existence of such a Cullen prime ?
____
References :
http://mathworld.wolfram.com/CullenNumber.html
http://primes.utm.edu/glossary/page.php?sort=Cullens
http://primes.utm.edu/top20/page.php?id=6
http://www.prothsearch.net/cullen.html

Contributions came from Daniele Degiorgi, Enoch Haga &
Farideh Firoozbakht.
***
Daniele wrote:
May be it is
unrelated, but I noted that for some primes p C_p has a common
factor with p+2.
For example
C_3=25 is multiple of 5, C_13 is divisible by 3 like 15, and C_47 is
divisible by 7 like 49.
Within the
first 50 primes, only for p = 53, 113 and 233 p+2 is relatively
prime with C_p. In this case, the smaller divisor of C_p are
respectively 5591, 241 and 17389.
Anyway, for
all 10000 odd primes p < 104743 there are only 1879 for which C_p
is prime with p+2 and for each p such that p+2 is multiple of 3,
C_p is also multiple of 3.
The latter
property, i.e. If p is of the form 3k-2, then C_p is divisible by 3
Should be
easy to prove just using the little Fermat theorem
Enoch wrote:
As you note, this subject is covered in Ribenboim, The
New Book of Prime
Number Records, pp 360-361. He has an apparent typo, 5611 rather than
the
correct 6611... Hooley in 1976 gave a formula for the number of Cullen
numbers <= x which are prime (page 361)... Hooley's formula implies an
infinite number of Cullen primes. The question is open.
***
Farideh wrote:
For many kind of primes p we can prove that C_p is
composite.
Some of them are as follows.
1. If both numbers p & p+2 are prime then p+2
divides C_p, so C_p is
composite.
Proof :
C_p = p*2^p + 1 = ((p+2)-2)*2^p + 1 = (p+2)*2^p - (2^(p+1) -1)
and since (p+2) is an odd prime by FLT (Ferrmat's
Little Theorem) p+2
divides 2^(p+1) -1. So p+2 divides (p+2)*2^p -
(2^(p+1) -1) = C_p.
2. we know if p > 3 is prime then p is of the
form 6k-1 or 6k+1,
for the second case C_p is composite because 3
divides C_p.
Proof : p
= 6k+1, C_p = (6k+1)*2^(6k+1)+1 so, C_p = 1*(-1)^(6k+1)+1 = 0
(mod 3)hence 3 divides C_p.
3. If p is of the form 6k-1 and k = 3 (mod 10) or
k = 4 (mod 10) then 5 divides C_p,
so C_p is composite.
Proof :
p = 6k-1, C_p = (6k-1)*2^(6k-1)+1
Case 1. k = 3 (mod 10); k = 10m+3; C_p =
(6(10m+3)-1)*2^(6(10m+3)-1)+1,
hence,
C_p = (60m+17)*2^(60m+17)+1
= (60m+15+2)*2^(60m+16+1)+1
= (60m+15+2)*4^(30m+8)*2+1
so, C_p = 2*(-1)^(30m+8)*2+1 = 2*1*2+1= 0 (mod
5) or 5 divides C_p.
Case 2. k = 4 (mod 10); k = 10m+4; C_p =
(6(10m+4)-1)*2^(6(10m+4)-1)+1,
hence,
C_p = (60m+23)*2^(60m+23)+1
= (60m+20+3)*2^(60m+22+1)+1
= (60m+20+3)*4^(30m+11)*2+1
so, C_p = 3*(-1)^(30m+11)*2+1 =3*(-1)*2+1 =
0 (mod 5) or 5 divides C_p.
The proofs of the following cases are easy
and similar to Number 3.
4. If p is of the form 6k-1 and k = 1 (mod 7)
then 7 divides C_p.
5. If p is of the form 6k-1 and k = i (mod
5*11) where i is in the set {8,12, 20, 26, 54}
"6. If p is of the form 6k-1
and k = i (mod 4*13) where i is in the set
{2, 7, 28, 33} then 13 divides C_p."
7. If p is of the form 6k-1 and k = i (mod
4*17) where i is in the set {13, 26, 27, 48}
then 17 divides C_p.
8. If p is of the form 6k-1 and k =
i (mod 3*19) where i is in the set {3, 20, 25} then
19 divides C_p.
9. If p is of the form 6k-1 and k = i (mod
11*23) where i is in the set {1, 14, 18, 21, 46,
48, 72, 88, 107, 170, 218} then
23 divides C_p.
10, If p is of the form 6k-1 and k =
i (mod 14*29) where i is in the set {16, 23, 31, 66,
71, 75, 138, 140, 165, 189, 293, 316,
384, 396} then 29 divides C_p.
...
And finally it seems that we can say,
For each prime q, q > 3, there
exist a number j(q) and a non-empty set
A(q) where j(q) divides q-1 and
A(q) is a subset of {1, 2, 3, ..., j(q)*q-1}
such that If p is of the form
6k-1 and k = i (mod j(q)*q) where i is in the
set A(q) then q divides C_p.
Later she added:
I have found the following other cases that C_n is
composite, but for
some of these cases n is composite. The proofs of
all of them are easy.
1. If n+1 is an odd prime then n+1 divides C_n,
so C_n is composite.
Numbers n such that both numbers n+1 & C_n/(n+1)
are prime:
2, 4, 16, 5692, ...
2. If n+2 is an odd prime then n+2 divides C_n,
so C_n is composite.
(In my previous email I wrote the special
case that n is also a prime.)
Numbers n such that both numbers n+2 & C_n/(n+2) are
prime:
3, 5, 9, 11, 21, 35, 101, 165, 191, 459, 671,
...
3. If 2n-1 is prime and 4 | (n-2)(n-3) , namely n =
2 (mod 4) or n = 3 (mod 4)
then 2n-1 divides C_n, so C_n is composite.
Numbers n such that both numbers n+2 & C_n/(n+2) are
prime: 2, 3, ....
4. If (2n+1)/3 is prime and 4 | (n-1)(n-2) , namely
n = 1 (mod 4) or n = 2 (mod 4)
then (2n+1)/3 divides C_n, so C_n is composite.
5. If n/2+2 is prime then n/2+2 divides C_n, so C_n
is composite.
6. If 6 | (n-1)(n-2) , namely n = 1 (mod 6) or n
= 2 (mod 6) then 3 divides
C_n, so C_n is composite.
(In my previous email I wrote the special
case that n is also a prime.)
Numbers n such that 1/3*C_n is prime: 2, 8,
158, 2150, 3698, 8330, ...
***
n = 6 is an interesting example for the
three cases 1, 3 & 5:
6+1 = 7 is an odd prime so 7
divides C_6,
6/2+2 = 5 is prime so 5
divides C_6,
2*6-1 = 11 is prime and 6 = 2 (mod 4) so 11
divides C_6,
***
Anton Vrba added:
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