View Single Post 2021-08-08, 05:59 #2 sweety439   "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 57608 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  