View Single Post
Old 2021-06-07, 15:32   #1
sweety439
 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

5·7·83 Posts
Default 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
sweety439 is offline   Reply With Quote