Problems & Puzzles:
Problems
Problem 49.
Sierpinskilike
numbers
Let's remember some known concepts:
A Sierpinski Number
is an odd integer k such that k.2^{n}+1 is composite for any integer
value of n. In 1962, John Selfridge discovered the Sierpinski
number k = 78557, which is now believed to be in fact the smallest such
number.
A Riesel Number is
an odd integer k such that k.2^{n}1 is composite for any integer value
of n. In 1956 Riesel showed that k = 509203 has this property.
A Brier Number is
an odd integer k such that k.2^{n}+1 and k.2^{n}1
are composite for any value of n. Eric Brier computed the first one in
September, 1998; but Yves Gallot computed, January 2002 (See
Problem 29) the smallest known Brier number:
878503122374924101526292469 (27 digits).
Well, a few days ago, Thomas Masser
wrote to me to inform that he had discovered empirically certain numbers
Sierpinskialike but related to the twin numbers k*2^n+/1. Here are the
basic Masser lines:
I recently found three values (k=79, 269, 1527) so
that 3k*2^n+/1 never forms a twin prime pair. The "covering sets" for
these values are
79: {5, 7, 13, 17, 241}
269: {5, 7, 13, 19, 37, 73}
1521: {5, 7, 13, 17, 241}
As in the Sierpinski and Riesel Problems, one could
conjecture that 79 is the smallest such k. There are eleven values of k<79
that exhibit no twin prime pairs for n<=8000. These values are {37, 41,
49, 51, 53, 57, 61, 63, 71, 73, 75}. Finding twin prime pairs for these
values could be an arduous task. Has anyone ever considered this problem?
The same day I wrote to Eric Brier and Wilfrid Keller,
asking if they was knowing something about the Masser issue, and 24 hours
later Keller responded the following lines:
The problem and results proposed by Tom Masser have
been known for many years.
In my files, instead of k and 3k*2^n+/1 I used the multipliers h=3k
for h*2^n+/1, keeping in mind that h must be divisible by 3.
Many, many years ago I had produced a list of those h never giving a twin
prime pair (with corresponding "covering sets"), which is appended below
for h<10^6. The first entries, 237=3*79, 807=3*269, 4581=3*1527 coincide
with Masser's findings.
> There are eleven values of k<79 that exhibit no
> twin prime pairs for n<=8000. These values are
> {37, 41, 49, 51, 53, 57, 61, 63, 71, 73, 75}.
> Finding twin prime pairs for these values could
> be an arduous task.
The eleven values would be h=111, 123, 147, 153, 159, 171, 183, 189, 213,
219, 225 in "my" notation.
However, my computations showed that 147*2^n+/1 are both prime for
n=44, and that 213*2^n+/1 are both prime for n=36, so h=147 and
h=213 may be removed from the list.
Drastically extending the limit of n<=8000, I can assure that for each
of h=111, 123, 153, 159, 171, 183, 189, 219, 225 no twin prime pair
exists for n<=140000. This follows from existing tables and some minor
verifications I just did to thoroughly establish that high limit.
I agree that finding a twin prime pair for any of
the remaining h would be "an arduous task", if not almost impossible.
That's the reason why an attempted coordinated effort to attack the
Sierpinski analogue for twin primes (Chris Nash, 1998) has quickly been
given up (to my recollection).
Please feel free to forward this message to Tom Masser and Eric Brier, if
you like.
The next day Keller added:
> Are these effortsresults somewhere in a web
page available to public?
Obviously not. They seem to have been withdrawn long ago. Maybe Chris
(Nash) would tell you if he has preserved them somewhere.
Actually, that coordinated search for twin pairs concerned two different
sorts of pairs: twin primes, and also Sophie Germain pairs of the form
h*2^(n+1)1, h*2^n1 (again, h must be a multiple of 3).
Here the smallest h never giving a pair of Sophie Germain primes is
h=807, with covering set {5, 7, 13, 19, 37, 73}. Again, one has
determined the possible "candidates" h<807 without known prime pair so
far (32 values of h), as follows.
Values of h without known prime pair:
h = 39, 183, 213, 219, 273, 279, 333, 351, 387, 393, 399, 417, 429, 471,
531, 543, 561, 567, 573, 591, 597, 603, 639, 681, 687, 693, 699, 723, 753,
759, 771, 795
To eliminate one of these candidates h, one has "only" to find primes
h*2^n1 for two consecutive exponents n. I recall that Chris' search had
reached substantial limits in this case. After 6 years, with improved
hardware and software, they might be raised considerably these days.
Note that h=807 is quite exceptional: it never produces a Sophie Germain
pair, nor does it produce a twin prime pair.
Summarizing, we have six kinds of constants k, related
to the numbers k*2^n+1 and/or k*2^n1, such that, these numbers, or pairs of
these numbers, are composites for all n, in certain specific manner, as seen
in the following scheme:
k numbers
(smallest proven known, smallest uncertain candidate) 
N numbers 
Condition 
Sierpinski (78557,
4847) 
N=k*2^n+1 
N is never prime 
Riesel (509203, 2293) 
N=k*2^n1 
N is never prime 
Brier [Combined
Sierpinski &
Riesel] (878503122374924101526292469,
2152690373) See
Problem 30 
N1=k*2^n+1 &
N2=k*2^n1 
N1 & N2 are never
primes 
No Twins (237, 111) 
N1=k*2^n+1 & N2=k*2^n1 
N1 & N2 are never
primes for the same n 
No Sophie Germain
(807, 39) 
N1=k*2^n1 &
N2=k*2^(n+1)1 
N1 & N2 are never primes for the same n 
Combined No Twins & No
SG (807, 183) 
N1=k*2^n+1 & N2=k*2^n1; N2 &
N2=k*2^(n+1)1 
N1, N2 & N3 are never primes for the
same n 
No Cunningham2 (32469, 51)
Added line after the kind submissions from V. Liskovets &
W. Keller (see below) 
N1=k*2^n+1 & N2=k*2^(n+1)+1 
N1 & N2 are never primes for the same n 
Questions:
1. Find a twin prime of the
type k*2^n+/1 for any of the following k values: 111, 123, 153, 159, 171,
183, 189, 219, 225 ^{
(*)}
2. Find a SG prime pair of
the type k*2^n1 & k*2^(n+1)1 for any of the following k values: 39, 183,
213, 219, 273, 279, 333, 351, 387, 393, 399, 417, 429, 471, 531, 543, 561,
567, 573, 591, 597, 603, 639, 681, 687, 693, 699, 723, 753, 759, 771, 795
^{(*)}
3. Are there any Sierpinski,
Riesel or Brier numbers that are divisible by three? Any of these would
automatically be a NeverTwinmultiplier (question posed by
Masser).
_________
^{(*) }
If you decide to search a bit for the primes asked in the
questions 1 & 2, a good starting point for n could be stated
consulting for the largest prime available in the Chris Caldwell
primes database, for the corresponding form and k multiplier (see this
page). For example
inputting '111*2^' in the 'Mathematical description' box and checking the
'all verified primes' in the 'Type' option, we can see that
111 · 2^{147346}1
is the largest prime found
with this multiplier. On top of that
"Wilfrid Keller can assert that any such twin
prime pair must have n > 160000"
Comment: Please take note that
the search for Q1 & Q2 can be really hard and disappointing or boring for a long time;
but of course that in getting just one of asked targets this will put your
name in the Top twenties list of primes of Twin primes or SG primes. One
more thing, if you find a solution to Q1, for a small value of k, you also
increases your probability of finding a Fermat Divisor (thus, getting a
double prize, if you get it...)
Solution:
Wilfrid Keller sent (3/5/04) a solution to Q3.
Concerning Question 3, you asked "how hard" I would
estimate it might be to solve it.
Well, trying to give an "estimate" as realistic as
possible, I have struggled with the problem for many hours in order to get
more deeply acquainted with it.
As a result, I can now give you a quite precise
answer, since in the process I have finally found examples for Sierpinski
and Riesel (I didn't look for Brier so far), which show that there are
infinitely many Sierpinski numbers and (by "duality") infinitely many
Riesel numbers divisible by three!
The idea was the one I had told you the other day:
to find a covering set not containing the prime 3, and an actual covering
based on it, and then determine the sought numbers by applying the Chinese
Remainder Theorem.
I thought that a covering set published by R. G.
Stanton in 1980 (with 16 primes) could be used, but I failed to establish
a real covering with the listed primes. In fact I am now doubtful if this
was a correct example of a covering set. However, adding three primes to
the given example, I was able to construct a covering using 19 primes,
which are all divisors of 2^720  1.
The resulting Sierpinski and Riesel numbers have 34
digits each, and they are:
Sierpinski k1 = 4423988484893019254861842753686117
Riesel k2 = 3674515543532961928874271620966553
In this connection, the number P =
2.3.5.7.11.13.17.19.31.37.41.61.73.97.109.151.181.241.257.331.673 =
8098504028425981183736114374652670 is essential, because every number k =
k1 + h.P (h = 0, 1, 2, ...) is also a Sierpinski number, and every k = k2
+ h.P is also a Riesel number (infinitely many). Note that once k1 was
found, k2 = P  k1 was the appropriate Riesel number.
The factors 5.7.11.13... of P are just the prime
components of the covering set, while the additional factor 2 ensures that
every other k is also odd, and the factor 3 ensures that every k is also a
multiple of 3.
Followup question: Would someone be able to
construct a covering with a smaller P (which would lead to smaller
Sierpinski and Riesel numbers)? Especially, would someone be able to
exhibit a covering set (not including the prime 3) with a smaller number
of primes?
***
Thomas Masser wrote (11/5/04)
I have been working on Prof. Keller's new question on
Problem 49 and recently found a 28 digit Sierpinski number,
a multiple of three, with a smaller covering set than the example
cited by Prof. Keller. Here are the details:
k=1139634035240881514319907179
CS={5,7,11,13,17,19,31,41,61,73,97,109,151,181,241,331,673}
I spent several hours looking for a smaller example than
the 28 digit Sierpinski number. It might be possible to
find a smaller number, but that will probably require a
smaller covering set.
I found the reduced covering set
{5,7,11,13,17,19,31,41,61,73,97,109,151,181,241,331,673}
by studying and confirming the example Keller provided. In
the process of this verification, I realized that the
numbers 37 and 257 listed in his covering set were
unnecessary. The reduced covering set was then used to
find the smaller Sierpinski number.
***
On May 29, Thomas Masser wrote again:
I recently found an even smaller covering set (15
primes), not containing 3, for Sierpinski/Riesel numbers. The
covering set is:
CS: {5 7 11 13 17 19 31 37 41 61 73 109 151 241 331}
After communicating this result to Prof. Keller,
he determined the smallest Sierpinski(k1) and smallest Riesel(k2)
numbers, divisible by 3, covered by the set above:
k1 = 4563308668738348027581
k2 = 5785418219844359742627
So we now have 22digit numbers as the smallest
known Sierpinski/Riesel numbers divisible by three. It may be
possible to further refine this result, by finding an even smaller
covering set; at any rate, I think we are getting close to the smallest
provable (with covering set) Sierpinski/Riesel numbers divisible by
three.
...
I found the new covering set after Prof.
Keller asked me if I could confirm
Stanton's claim that
C = {5,7,11,13,17,19,31,37,41,61,73,97,109,151,241,257}
is a covering set. I could not prove that the above set is
a covering set, but by including 331 in the set, a covering
set was obtained. After further study of this new covering set, I
realized the primes 97 and 257 were unnecessary.
Perhaps if someone is very clever they can find a covering
set not containing 3 with less than 15 primes. So far, I
have not been able to find any smaller covering sets.
***
Valery Liskovets wrote (November 3, 2004):
Prob_49... contains plenty of highly interesting new facts and data.
But in essence it is very CLOSE to your prob_36, where several similar
questions including
 Sierpinskilike numbers and
 Sierpinski numbers divisible by 3
are considered as well.
Therefore in prob_49 it would be natural, I think, to refer to prob_36 and
also:
 to extend the table by the following ROW 7, with data taken from
prob_036 (similar to row 5):
No Cunningham2 pair (66741?, 51) N1=k*2^n+1 & N2=k*2^(n+1)+1
N1 & N2 are never primes for the same n
 to add the corresponding (intriguing but hopeless) PROBLEM (similar to
Q2):
Q2*: Find a Cunningham2 prime pair of the type
k*2^n+1 & k*2^(n+1)+1 for any of the following k values: 51, 87, 93, 111,
123, 129, 213, 279,...(*).
Notice also that 213 and 279 are presented in BOTH lists in Q2 and Q2*.
Hopefully Gallot's 66741 above is much greater than minimal provable k.
But WHICH least provable value is known presently? Apparently, something
significant is known to Professor W. Keller and/or other experts...
This question is particularly interesting for me because:
 I refer to your prob_036 (mentioning in particular 51, 87, 93 and
66741) and to Keller's site http://www.prothsearch.net/riesel.html in my
recent ARTICLE "Some identities for enumerators of circulant graphs", J.
Algebr. Combin., v.18, No 3 (2003), 189209, MR2036114 (2004m:05135)
(available upon request), where a fairly unexpected (though modest)
combinatorial role of Cunningham2 pairs is established and discussed.
One day after, Wilfrid Keller wrote:
I am writing to you with regard to the recent note by Valery Liskovets.
I just wanted to tell you that I would strongly support his suggestion to
adding a line to the table for Problem 49, like this:
> No Cunningham2 pair (66741??, 51)
> N1=k*2^n+1 & N2=k*2^(n+1)+1
> N1 & N2 are never primes for the same n
I am happy (but also unhappy, on the other hand) to be able to remove
Valery's question marks, since I found the smaller k=32469. Unhappy
I am because this is probably the smallest "provable" example.
To give you some more details, here is what I wrote to Valery after
completing my search:

Thank you very much for letting me have a copy of your message concerning
Problems 36 and 49 of Carlos Rivera's Prime puzzles site.
I completely agree with you that in the table of "Sierpinskilike" numbers
the case of (nonexistent) pairs
N1=k*2^n+1 & N2=k*2^(n+1)+1 should reasonably be in cluded, as a natural
analogue to N1=k*2^n1 &
N2=k*2^(n+1)1 (Sophie Germain).
I have to admit that I hadn't pay enough attention to Problem 36 before,
so I was glad to learn on this occa sion of Yves Gallot's examples of
multipliers k such that k*2^n+1 (or k*2^n1) is composite for all even n
(or for all odd n).
Although I hadn't considered the "Cunningham2" case so far, it appeared
to me (as you were also expecting) that k=66741 was unlikely to be the
smallest provable Sierpinskilike number in this case.
In order to force k*2^n+1 to never produce primes for consecutive values
of n, the most simple way is to re quire that n has always the same
parity. But there might exist more involved "coverings" leading to the
same result. In the case of Sophie Germain the "simple way" led to
k=39939, while k=807 is the smallest proven Sierpinskilike number we
know.
By analogy, I hoped to find a much smaller k for "Cunningham2" also, but
after some careful calculation I dare to assert that probably only k=32469
is the smallest provable example. Altogether, I found 6 such numbers less
than k=66741, which I give you along with the corresponding covering set
of small primes:
32469 {5,7,13,19,37,73}
42717 {5,7,13,17,241}
45009 {5,7,13,19,37,73}
57669 {5,7,13,37,73,109}
64377 {5,7,13,17,241}
64893 {5,7,13,17,241}
Looking at the "distribution" in the interval [3..66741] it's just
unfortunate that a smaller example didn't emerge.
As a result, your list of "undecided" values of k = 51, 87, 93, 111, 123,
129, 213, 279,... becomes unpleasantly long (about 1500 values, I guess).
*** Arkadiusz Wesolowski wrote (July, 2010):
I found another pair (k1, k2).
Sierpinski k1 = 1934207927446756722099
Riesel k2 = 1636996998454706441457
The integer k1 has the same covering set as the integer k2:
{5, 7, 11, 13, 17, 19, 31, 37, 41, 61, 73, 109, 151, 241, 331}.
***Arkadiusz Wesolowski wrote (July, 2017):
Here is my best solution:
Sierpinski k1 = 21484572547591559649
Riesel k2 = 7148695169714208807
***
