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Old 2021-08-08, 05:35   #1
sweety439
 
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"99(4^34019)99 palind"
Nov 2016
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Default A related conjecture about twin primes

Are there infinitely many triples of three consecutive numbers whose product has exactly four different prime factors, i.e. A001221((k-1)*k*(k+1)) = 4? Reference: https://oeis.org/A325204, https://math.stackexchange.com/quest...886521_3345481

There are six situations: k == 0 mod 6, k == 1 mod 6, k == 2 mod 6, k == 3 mod 6, k == 4 mod 6, k == 5 mod 6

* If k == 0 mod 6, 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 == 1 mod 6, then k-1 is of the form 2^r*3^s, and both k and the odd part of k+1 (i.e. 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, besides, A000265(k+1) is equal to (k+1)/2 unless r=1 (since if r>1, then k is == 1 mod 4, and the only known such k-values with r=1 are 19, 163, 487, 86093443, and it is conjectured that all other such k-values with r=1), 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 == 5 mod 6, then k+1 is of the form 2^r*3^s, and both k and the odd part of k-1 (i.e. 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, besides, A000265(k-1) is equal to (k-1)/2 unless r=1 (since if r>1, then k is == 3 mod 4, and the only known such k-values with r=1 are 53 and 4373, and it is conjectured that all other such k-values with r=1), 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 == 2 mod 6, then k+1 is power of 3, k is of the form 2^r*p^s with p prime, k-1 is prime power, by the generalization of Pillai conjecture, all sufficient large such k have p and k-1 both primes, the only known such k-values with r=1 are 26, 242, 1326168790943636873463383702999509006710931275809481594345135568419247032683280476801020577006926016883473704238442000000602205815896338796816029291628752316502980283213233056177518129990821225531587921003213821170980172679786117182128182482511664415807616402, and it is conjectured that all other such k-values with r=1, but it is conjectured that there are infinitely many such numbers.

* If k == 4 mod 6, then k-1 is power of 3, k is of the form 2^r*p^s with p prime, k+1 is prime power, by the generalization of Pillai conjecture, all sufficient large such k have p and k+1 both primes, the only known such k-values with r=1 are 10 and 82, and it is conjectured that all other such k-values with r=1, but it is conjectured that there are infinitely many such numbers.

* If k == 3 mod 6, 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, however, if the r-value for k-1 is >1, then (k-1)/(2^r) has algebra factors (like (9^n-1)/8 = (3^n-1)*(3^n+1)/8, etc.) and cannot be prime, thus the r-value for k-1 is 1, and if (k-1)/2 is prime then log_3(k) must be prime, and since k+1 cannot be divisible by 8, thus the r-value for k+1 is 2, therefore, log_k(3) must be in sequences A028491 and A007658 in common, the only known such k-values are 27, 243, 2187, 1594323, and it is conjectured that all other such k-values

Thus, such k-values (i.e. k-values with omega(k*(k-1)*(k+1)) = 4 are conjectured to exist infinitely many with all residue class mod 6, except 3 mod 6.

Last fiddled with by sweety439 on 2021-08-08 at 06:23
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