The three consecutive numbers 127, 128, 129 have exactly four different prime factors, namely, 2, 3, 43, and 127. Are these numbers infinite?
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– Oscar LanziSep 5, 2019 at 17:32
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2– Erick WongSep 5, 2019 at 17:36
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– luluSep 5, 2019 at 17:41
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– Erick WongSep 5, 2019 at 17:42
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2– Keith BackmanSep 5, 2019 at 19:36
1 Answer
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)