20220806, 14:22  #12  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2·7·263 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
2×7×263 Posts 
Quote:


20220809, 13:45  #14 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2·7·263 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
3682_{10} 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
2×7×263 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 
20221028, 03:38  #17 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
7142_{8} Posts 
398*23^9703+1 is prime, it is the first prime for S23 k=398 (while S23 has only CK=182, thus this prime is for a k>CK)
(this prime is found in the process of solving the generalized minimal prime problem in base b=23, this prime (when written in base 23) is H7000...0001 with 9702 0's (using AโZ to represent digit values 10 to 35)) Last fiddled with by sweety439 on 20221028 at 04:01 
20221108, 12:59  #18  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
111001100010_{2} Posts 
Quote:


20221120, 04:06  #19 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2×7×263 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  #20 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2·7·263 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  #21  
"Bob Silverman"
Nov 2003
North of Boston
2^{2}×1,877 Posts 
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20221124, 20:27  #22 
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
2×3×29×67 Posts 

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