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 2021-08-08, 05:35 #1 sweety439     "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 2×52×61 Posts 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
 2021-08-08, 05:59 #2 sweety439     "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 2×52×61 Posts If k == 2 mod 6, then: ---either--- * k+1 is power of 3, k is of the form 2^r*p^s with p prime, k-1 is prime power, the r-value of k must be 1 or the k will have algebra factors and cannot be prime, the only known such k-values are 26 and 1326168790943636873463383702999509006710931275809481594345135568419247032683280476801020577006926016883473704238442000000602205815896338796816029291628752316502980283213233056177518129990821225531587921003213821170980172679786117182128182482511664415807616402, and it is conjectured that all other such k-values ---or--- * 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 all other such k-values (related to New Mersenne Conjecture) If k == 4 mod 6, then: ---either--- * k-1 is power of 3, k is of the form 2^r*p^s with p prime, k+1 is prime power, the r-value of k must be 1 or 2 or the k will have algebra factors and cannot be prime, the only known such k-values with r=1 are 10 and 82, and it is conjectured that all other such k-values, the only known such k-values with r=2 are 28, and it is conjectured that all other such k-values ---or--- * 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
 2021-08-08, 06:03 #3 sweety439     "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 2·52·61 Posts Code: 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 Thus it is conjectured to exist infinitely many such k-values, however, all but finitely many such k-values are in A027856, A325255+1, or A327240-1 (thus, all but finitely many such k-values have one of k, k-1, k+1 divisible by 12) Last fiddled with by sweety439 on 2021-08-08 at 06:23
 2021-08-08, 06:11 #4 sweety439     "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 305010 Posts The two numbers which must be prime powers simultaneously are: Code: 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 (only for k=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 (only for k=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 Last fiddled with by sweety439 on 2021-08-08 at 06:14
 2021-08-08, 06:13 #5 sweety439     "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 57528 Posts It can be easily proven that there are eight k-values which have A001221((k-1)*k*(k+1)) < 4: 2, 3, 4, 5, 7, 8, 9, 17

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