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 2021-06-07, 15:32 #1 sweety439   "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 5·7·83 Posts Smallest possible prime of the form (31^n+1)/2 I want to solve the generalized Sierpinski conjectures in bases 2<=b<=128, but for base b=31, the CK is 239, and there are 10 k-values remaining with no known (probable) primes: {1, 43, 51, 73, 77, 107, 117, 149, 181, 209}, for k=1, the formula is (1*31^n+1)/2, and if this formula produce prime, then n must be power of 2 (since if n has an odd factor m>1, then (31^n+1)/2 is divisible by (31^m+1)/2, thus cannot be prime), for the status for (31^n+1)/2: (see http://factordb.com/index.php?query=...%29%2B1%29%2F2) Code: n factors 2^0 2^4 2^1 13*37 2^2 409*1129 2^3 17*P11 2^4 1889*... 2^5 4801*... 2^6 257*641*... 2^7 P58*P133 2^8 P11*P11*P361 2^9 25601*... 2^10 114689*... 2^11 composite 2^12 composite 2^13 1196033*... 2^14 4882433*... 2^15 65537*... 2^16 composite 2^17 composite (there is no n<11559 such that (n^(2^17)+1)/2 is prime, see http://www.fermatquotient.com/PrimSerien/GenFermOdd.txt) 2^18 255666946049*... 2^19 1775270625281*... 2^20 unknown Thus what is the true test limit for S31 k=1, is it 2^20-1 = 1048575? Can someone check whether (31^(2^20)+1)/2, (31^(2^21)+1)/2, etc. is probable prime or not? Also for other Sierpinski bases with GFN (b^(2^n)+1) or half GFN ((b^(2^n)+1)/2) remain, such as 15 (k=225), 18 (k=18), 22 (k=22), 37 (k=37), 38 (k=1), 40 (k=1600), 42 (k=42), 50 (k=1), 52 (k=52), 55 (k=1), 58 (k=58), 60 (k=60)? What are the true test limit for these GFNs? I know that for all even bases, this test limits must be at least 2^23-1, see http://www.primegrid.com/stats_genefer.php and http://www.primegrid.com/forum_thread.php?id=3980