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The three consecutive numbers 127, 128, 129 have exactly four different prime factors, namely, 2, 3, 43, and 127. Are these numbers infinite?

J. M. Bergot
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  • Whether the set is infinite or not, the claimed triple is far from minimal. We have 5, 6 7; then 9, 10, 11; etc. – Oscar Lanzi Sep 05 '19 at 17:32
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    @OscarLanzi I would think the triple becomes more remarkable as the numbers get larger, so minimal examples are far from the OP’s concern. – Erick Wong Sep 05 '19 at 17:36
  • I'd have thought Mersenne primes were a good place to start...after all, if $n$ is a Mersenne prime then there are only two primes dividing $n, n+1$ so you just need $n+2$ to be a semiprime. – lulu Sep 05 '19 at 17:41
  • Just to clarify, do you require the 3 numbers to be pairwise relatively prime (aka, does the first number need to be odd)? Would you count 10,11,12 as having this property? (There are four distinct prime divisors, but the number of prime factors in each is 2,1,2 totaling to 5). – Erick Wong Sep 05 '19 at 17:42
  • @lulu Unfortunately $2^p +1$ is always (save for one case) going to be divisible by $3$, so semi-primeness will be pretty rare, given the already extreme rarity of Mersenne primes. – Erick Wong Sep 05 '19 at 17:49
  • Also 12, 13, 14 with exactly four distinct prime factors 2, 3, 7, and 13. An obvious place to look is for prime, 2^a*3^b, prime, as in 17, 18, 19. Remember that it is not the total number of all prime factors to be found in these three consecutive numbers but merely that there are exactly FOUR different ones. – J. M. Bergot Sep 05 '19 at 17:59
  • So, basically the question is whether infinite many positive integers $n$ satisfy $$\omega(n^3+3n^2+2n)=4$$ – Peter Sep 05 '19 at 18:09
  • Terms $<10^6$ are $${5, 9, 10, 11, 12, 14, 15, 17, 18, 22, 23, 24, 25, 26, 27, 30, 31, 32, 36, 46, 47, 48, 52, 62, 71, 72, 79, 80, 81, 96, 106, 107, 126, 127, 162, 191, 192, 241, 242, 256, 382, 431, 486, 512, 576, 862, 1151, 1152, 2186, 2591, 2592, 2916, 4372, 8191, 8746, 131071, 131072, 139967, 472391, 524287, 786431, 995326, 995327}$$ – Robert Israel Sep 05 '19 at 19:04
  • I expect that there are infinitely many 3-smooth terms, i.e. $2^i \cdot 3^j$ where $2^i \cdot 3^j+1$ and $(2^i\cdot 3^j+2)/2 = 2^{i-1} \cdot 3^j+1$ are prime. – Robert Israel Sep 05 '19 at 19:07
  • True, as long as there are an infinite number of twin primes. – J. M. Bergot Sep 05 '19 at 19:10
  • $$2^{i-2}3^j\neq 2ab+a+b$$ would be a requirement for that to occur. –  Sep 05 '19 at 19:19
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    I refer you to this question concerning twin primes of the form $2^a3^b\pm 1$. All such examples of three consecutive numbers will feature the desired property of having only four prime factors: the primes themselves, and the median number having only $2$ and $3$ as factors. – Keith Backman Sep 05 '19 at 19:36
  • What about prime, 2^n, semiprime or semiprime, 2^n, prime? – J. M. Bergot Sep 05 '19 at 20:00
  • @J.M.Bergot first forces a potentially infinite number of Mersenne primes, and the Second an infinite supply of Fermat primes requiring n to be a power of 2. –  Sep 05 '19 at 21:03
  • 12,13,14 is of neither form. –  Sep 06 '19 at 14:25
  • A deceptively simple question. – Robert Soupe Sep 06 '19 at 17:01
  • See OEIS sequence A325204. – Robert Israel Sep 08 '19 at 16:12
  • I don't think that anyone will be able to actually solve this problem. But maybe, someone can give a good evidence for infinite many such triples. My guess is that there are infinite many. – Peter Sep 10 '19 at 13:12

1 Answers1

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Let the product of three consecutive numbers (k and k+-1) has exactly four different prime factors, i.e. A001221((k-1)*k*(k+1)) = 4

There are twelve situations: k == 0 mod 12, k == 1 mod 12, k == 2 mod 12, k == 3 mod 12, k == 4 mod 12, k == 5 mod 12, k == 6 mod 12, k == 7 mod 12, k == 8 mod 12, k == 9 mod 12, k == 10 mod 12, k == 11 mod 12

  • If k == 0 mod 12, then k is of the form 2^r*3^s, and both k+-1 must be prime powers, by the generalization of Pillai conjecture, all sufficient large such k have k+-1 both primes, such k's are the Dan numbers, and it is conjectured that there are infinitely many such numbers (however, for any fixed m values, it is conjectured that there is only finitely many r values such that m*2^r+-1 are both primes), k+1 is Pierpont prime, k-1 can be called "Pierpont prime of the second kind".

  • If k == 6 mod 12, then k is of the form 2*3^r, and both k+-1 must be prime powers, by the generalization of Pillai conjecture, all sufficient large such k have k+-1 both primes, the only known such k-values are 6 and 18, and it is conjectured that there are no other such k-values.

  • If k == 1 mod 12, then k-1 is of the form 2^r*3^s, and both k and (k+1)/2 must be prime powers, by the generalization of Pillai conjecture, all sufficient large such k have k and (k+1)/2 both primes, the known such k-1 values are listed in https://oeis.org/A325255, and it is conjectured that there are infinitely many such numbers.

  • If k == 11 mod 12, then k+1 is of the form 2^r*3^s, and both k and (k-1)/2 must be prime powers, by the generalization of Pillai conjecture, all sufficient large such k have k and (k-1)/2 both primes, the known such k+1 values are listed in https://oeis.org/A327240, and it is conjectured that there are infinitely many such numbers.

  • If k == 7 mod 12, then k-1 is of the form 2*3^r, and both k and A000265(k+1) must be prime powers, by the generalization of Pillai conjecture, all sufficient large such k have k and A000265(k+1) both primes, the only known such k-values are 19, 163, 487, 86093443, and it is conjectured that there are no other such k-values.

  • If k == 5 mod 12, then k+1 is of the form 2*3^r, and both k and A000265(k-1) must be prime powers, by the generalization of Pillai conjecture, all sufficient large such k have k and A000265(k-1) both primes, the only known such k-values are 53 and 4373, and it is conjectured that there are no other such k-values.

  • If k == 2 mod 12, then k+1 is power of 3, k is of the form 2*p^r with p prime, k-1 is prime power, by the generalization of Pillai conjecture, all sufficient large such k have r=1 and k-1 prime, the only known such k-values are 26, 242, 1326168790943636873463383702999509006710931275809481594345135568419247032683280476801020577006926016883473704238442000000602205815896338796816029291628752316502980283213233056177518129990821225531587921003213821170980172679786117182128182482511664415807616402, and it is conjectured that there are no other such k-values.

  • If k == 10 mod 12, then k-1 is power of 3, k is of the form 2*p^r with p prime, k+1 is prime power, by the generalization of Pillai conjecture, all sufficient large such k have r=1 and k+1 prime, the only known such k-values are 10 and 82, and it is conjectured that there are no other such k-values.

  • If k == 8 mod 12, then k is power of 2, k-1 and (k+1)/3 are both prime powers (k+1 cannot be divisible by 9 or the (k+1)/(3^r) will have algebra factors and cannot be prime), the only known such k-values are 8, 32, 128, 8192, 131072, 524288, 2147483648, 2305843009213693952, 170141183460469231731687303715884105728, and it is conjectured that there are no other such k-values (related to New Mersenne Conjecture); or k+1 is power of 3, k/4 and k-1 are both primes, such k-values do not exist since k/4 has algebra factors.

  • If k == 4 mod 12, then k is power of 2, k+1 and (k-1)/3 are both prime powers, the only such k is 16, since (k-1)/3 has algebra factors; or k-1 is power of 3, k/4 and k+1 are both primes, the only known such k-values are 28, and it is conjectured that there are no other such k-values.

  • If k == 3 mod 12, then k is power of 3, and both k+-1 must be of the form 2^r*p^s with p prime, by the generalization of Pillai conjecture, all sufficient large such k have both p primes, thus (k-1)/2 and (k+1)/4 are both primes or prime powers, the only known such k-values are 27, 243, 2187, 1594323, and it is conjectured that there are no other such k-values.

  • If k == 9 mod 12, then k is power of 3, and both k+-1 must be of the form 2^r*p^s with p prime, by the generalization of Pillai conjecture, all sufficient large such k have both p primes, thus (k+1)/2 and (k-1)/(2^r) are both primes or prime powers, the only such k is 81, since (k-1)/(2^r) has algebra factors.

Thus it is conjectured that although there are infinitely many such k, all but finitely many k are == 0 or +-1 mod 12

k mod 12: such k-values

0: A027856 except 6 and 18 (i.e. A027856 except the numbers not divisible by 4)

1: A325255+1 except 3 and 5

2: 26, 242, 1326168790943636873463383702999509006710931275809481594345135568419247032683280476801020577006926016883473704238442000000602205815896338796816029291628752316502980283213233056177518129990821225531587921003213821170980172679786117182128182482511664415807616402 (conjectured no others)

3: 27, 243, 2187, 1594323 (conjectured no others)

4: 16 (power-of-2-related numbers are proven no others because of algebra factors), 28 (power-of-3-related numbers are conjectured no others)

5: 53, 4373 (conjectured no others)

6: 6, 18 (conjectured no others)

7: 19, 163, 487, 86093443 (conjectured no others)

8: 32, 128, 8192, 131072, 524288, 2147483648, 2305843009213693952, 170141183460469231731687303715884105728 (conjectured no others)

9: 81 (proven no others because of algebra factors)

10: 10, 82 (conjectured no others)

11: A327240-1 except 5 and 7

The two numbers which must be prime powers simultaneously are:

k mod 12: the forms for the two numbers

0: 2^r*3^s+-1

1: 2^r*3^s+1 and (2^r*3^s+2)/2

2: (3^r-1)/2 and 3^r-2

3: (3^r-1)/2 and (3^r+1)/4

4: 2^r+1 and (2^r-1)/3 (the only such k-value is 16, since (2^r-1)/3 has algebra factors), or (3^r+1)/4 and 3^r+2

5: 2*3^r-1 and A000265(2*3^r-2)

6: 2*3^r+-1

7: 2*3^r+1 and A000265(2*3^r+2)

8: 2^r-1 and (2^r+1)/3

9: (3^r-1)/(2^s) and (3^r+1)/2 (the only such k-value is 81, since (3^r-1)/(2^s) has algebra factors)

10: (3^r+1)/2 and 3^r+2

11: 2^r*3^s-1 and (2^r*3^s-2)/2

It is notable that there are only eight k-values which have A001221((k-1)*k*(k+1)) < 4:

2, 3, 4, 5, 7, 8, 9, 17

Also, this problem is related to three consecutive numbers all have primitive roots (see https://oeis.org/A305237), the only known k such that k and k+-1 all have primitive roots are 2, 3, 4, 5, 6, 10, 18, 26, 82, 242, 1326168790943636873463383702999509006710931275809481594345135568419247032683280476801020577006926016883473704238442000000602205815896338796816029291628752316502980283213233056177518129990821225531587921003213821170980172679786117182128182482511664415807616402, and it is conjectured that there are no other such k-values, since such k-values are exactly the k-values in this problem (i.e. three consecutive numbers with exactly different four prime factors) which are == 2, 6, 10 mod 12 (i.e. == 2 mod 4) plus the k-values 2, 3, 4, 5 (i.e. the only k-values which have A001221((k-1)*k*(k+1)) < 4 and k and k+-1 all have primitive roots)

Richard Chen
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