three consecutive numbers with exactly different four prime factors

Question

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

Answer 1

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)