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Problems & Puzzles: Puzzles
From April 18 to 25, 2026, contributions came from Arkadiusz Wesolowski, Emilia Gurisatti *** Arkadiuz wrote:
Answer to Q3.
Let k be a Sierpiński number that has two covering sets, namely
A and B, that are disjoint,
and let U be the union of these two sets. If k > (p*q^2 - 1)/2, where p
is the second largest prime in U and q is the largest prime in U, then
k*2^n + 1 has at least three distinct
prime factors for all positive integers n.
Example:
k = 132430318140059718398963877387
223179202469722051481175208904
437986485685550827555118149321
Covering sets:
A = {5, 7, 11, 13, 17, 19, 31, 37, 41, 61, 73, 109, 151, 241, 331},
B = {3, 23, 29, 43, 67, 71, 89, 97, 113, 127, 257, 281, 337, 353, 397,
433, 449, 577, 641, 673, 683, 1153, 1429, 2113, 2689, 5153, 5419, 6337,
7393, 20857, 38737, 65537, 86171, 122921, 312709, 599479, 6700417,
7416361, 15790321}.
k*2^1 + 1 has both 3 and 31 as prime factors.
k*2^2 + 1 has both 7 and 127 as prime factors.
k*2^3 + 1 has both 3 and 61 as prime factors.
k*2^4 + 1 has both 5 and 29 as prime factors.
k*2^5 + 1 has both 3 and 7 as prime factors.
...
Analogously, let k be a Riesel number that has two covering sets, namely
A and B, that are disjoint, and let U be the union of these two sets. If
k > (p*q^2 + 1)/2, where p is the second largest prime in U and q is the
largest prime in U, then k*2^n - 1 has at least three distinct
prime factors for all positive integers n.
Example:
k = 216138667663290358832800493328
766724738215605206192616152897
806049227643609138472430270365
Covering sets are the same as above.
k*2^1 - 1 has both 3 and 7 as prime factors.
k*2^2 - 1 has both 31 and 127 as prime factors.
k*2^3 - 1 has both 3 and 5 as prime factors.
k*2^4 - 1 has both 7 and 29 as prime factors.
k*2^5 - 1 has both 3 and 17 as prime factors.
...
It is easy to show that there exist infinitely many Sierpiński and
Riesel numbers with this property.
This statement is trivial.
We assume that k > (p*q^2 - 1)/2 to exclude the cases where, for
some odd k, k*2 + 1 has two distinct prime factors and at least one
of those factors is a prime power.
It can be show by constructing a fixed arithmetic progression modulo 2*d, where d is the product of the primes in U. *** Emilia wrote: I have found a Brier number that has at least two covering sets.k = 773553855548262449239841 (24 digits) Covering set 1: C1 = [3, 5, 17, 97, 241, 257, 673] P1 = 1031043870735 e_i1 = [2, 4, 8, 48, 24, 16, 48] e1 = 48 C2 = [3, 7, 11, 13, 19, 31, 37, 41, 61, 73, 109, 151] P2 = 196658330577361653 e_i2 = [2, 3, 10, 12, 18, 5, 36, 20, 60, 9, 36, 15] e2 = 180 C = [3, 5, 7, 11, 13, 17, 19, 31, 37, 41, 61, 73, 97, 109, 151, 241, 257, 673] P = 67587788790255388691025974985 e_i = [2, 4, 3, 10, 12, 8, 18, 5, 36, 20, 60, 9, 48, 36, 15, 24, 16, 48] e = 720 Covering set 2: C1 = [3, 5, 17, 97, 241, 257, 673] P1 = 1031043870735 e_i1 = [2, 4, 8, 48, 24, 16, 48] e1 = 48 C2 = [3, 7, 11, 13, 19, 31, 37, 41, 61, 73, 109, 331] P2 = 431085479609978193 e_i2 = [2, 3, 10, 12, 18, 5, 36, 20, 60, 9, 36, 30] e2 = 180 C = [3, 5, 7, 11, 13, 17, 19, 31, 37, 41, 61, 73, 97, 109, 241, 257, 331, 673] P = 148156013838241944746553627285 e_i = [2, 4, 3, 10, 12, 8, 18, 5, 36, 20, 60, 9, 48, 36, 24, 16, 30, 48] e = 720 The first covering set was originally used by Arkadiusz to cover k = 56284389701328043058161 (Wesołowski 5). The second one was originally used by Vantieghem to cover k=11615103277955704975673 (Vantieghem 2). Would you like to verify it? *** Emilia Gurisatti wrote agaiin on May 29, 2026: Here is another Brier number with two distinct covering sets:
k = 216644304573327922644547 (24 digits)
Covering set 1: [3, 5, 7, 11, 17, 19, 31, 37, 41, 61, 73, 97, 109, 151,
241, 257, 331, 673]
C1 = {3,5,17,97,241,257,673}
C2 = {3,7,11,19,31,37,41,61,73,109,
Covering set 2: [3, 5, 7, 11, 13, 17, 19, 31, 37, 61, 73, 97, 109, 151,
241, 257, 331, 673]
C1 = {3,5,17,97,241,257,673}
C2 = {3,7,11,13,19,31,37,61,73,109,
*** On August 9, 2026, Emilia Gurisatti sent the following link to her OEIS A395816 sequence of "Brier numbers that have at least two covering sets.But if you want to see her list of 29 Brier integers with double CS, and see these double CSs explicitly, click here. *** On June 1, 2026, Eric Brier wrote about Q3:"... My view on it is the following one: For Riesel or Sierpinski numbers, finding a covering set is "easy". It means there is sufficiently many covering prime so that: - You can find several covering sets - You can find covering set for Riesel and Sierpinski simultaneously (i.e. the so-called Brier numbers) This is a funny variation and worth a little bit of investigation. However, I would not consider it as very important. What would be really a breakthrough is to decide whether or not there can exist Riesel/Sierpinski/Brier numbers without covering set. Another important question remains to prove which number is the smallest one in each family. But these are just my view and my two cents..." |
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