20220806, 14:22  #12  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2×5^{2}×73 Posts 
Quote:
Last fiddled with by sweety439 on 20220806 at 14:23 

20220808, 13:27  #13  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2·5^{2}·73 Posts 
Quote:


20220809, 13:45  #14 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2·5^{2}·73 Posts 
For odd prime q, 2^q1 is Phi(q,2), and (2^q+1)/3 is Phi(2*q,2), and p divides Phi(n,2) if and only if order(2,p) = n, and order(2,p) = 2 if and only if (p1)/n is the largest m dividing p1 such that 2 is mth power residue mod p

20221016, 09:50  #15 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
111001000010_{2} Posts 
The numbers in https://oeis.org/A121719 are composite in every base, and assuming the Bunyakovsky conjecture, all numbers not in https://oeis.org/A121719 (except 0 and 1, of course) are primes in infinitely many bases (e.g. the bases such that these numbers are primes: 12 21 101 111)

20221022, 05:33  #16 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2·5^{2}·73 Posts 
(71^16384+1)/2, 30331 decimal digits
this PRP is generalized unique prime in base 71, and also the only one unproven generalized half Fermat PRP (of the form (b^(2^n)+1)/2 with odd base b<1000 (recently (799^2048+1)/2 was proven prime), thus if the primality of this PRP is proven, then maybe all generalized half Fermat primes (of the form (b^(2^n)+1)/2 with odd b, see http://www.fermatquotient.com/PrimSerien/GenFermOdd.txt) for odd bases b<1000 are known and proven (and hence (heuristically) we can complete the classification of the generalized Fermat primes of the form (b^(2^n)+1)/gcd(b+1,2) for bases b<1000), since heuristically there are only finitely many generalized half Fermat primes to every odd base b, e.g. there may be no prime of the form (3^(2^n)+1)/2 with n>6, see https://oeis.org/A171381 Last fiddled with by sweety439 on 20221022 at 05:44 
20221108, 12:59  #17  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2×5^{2}×73 Posts 
Quote:


20221120, 04:06  #18 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
7102_{8} Posts 
This number is very important, since this is the last number 2^n+1 with n <= the smallest Wieferich prime (1093), thus when this number is fully factored, all 2^n+1 with n <= the smallest Wieferich prime (1093) will be fully factored (note that the most important number 2^10931 was factored in 2014, the SNFS difficulty of this number is higher than that of 2^1091+1, but this number was already factored 8 years ago)

20221124, 06:33  #19 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
7102_{8} Posts 
Since 2,1091+ is C307, and it has been sieved many days with no result (as long as 7,889M), I think that it may be the worst case (i.e. P154*P154), and if so, then it will broke 7,889M's P151 record (for the penultimate prime factor, i.e. secondlargest prime factor), but only God knows that this C307 is a product of 3 prime factors!!! (the same holds for 10,323, in April I think that its C271 may be a semiprime and its smaller prime factor may have > 125 digits, since this number seems to be very hard, but the fact is that it is a product of 3 prime factors and the smaller two prime factors both have only 70 digits (however, their P1 and P+1 are not (10^25)smooth, so ECM cannot find them), and I noted that the only Phi(n,10) with n<=352 with penultimate prime factor > googol (10^100) are Phi(223,10) and Phi(337,10), whose penultimate prime factor have 105 digits and 101 digits (a little (1.004 times) > googol), respectively, while for Phi(n,2) with n<=1169 (2^1169 also has 352 digits, thus < 10^352), there are many Phi(n,2) with penultimate prime factor > googol)

20221124, 12:06  #20  
"Bob Silverman"
Nov 2003
North of Boston
1110101001000_{2} Posts 
Quote:


20221124, 20:27  #21 
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
2^{6}×181 Posts 

20221127, 14:03  #22 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2×5^{2}×73 Posts 
this number is the 133549th minimal prime in base 25 if we only consider the primes > base, i.e. primes > 25 when written as base 25 strings have no proper subsequence which (again, read in base 25) represents primes > 25, see https://github.com/xayahrainie4793/m...primenumbers, for this PRP, N1 is 27.686% factored, I know that for 27.686% factored, CHG proving may be possible, but unfortunately, factordb lacks the ability to verify CHG proofs, so are there any interest to factor the composite cofactors of N1 (which is equivalent to factor 5^159601) to enable N1 proof for this PRP?
Last fiddled with by sweety439 on 20221127 at 14:05 
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