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2022-08-06, 14:22   #12
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

365010 Posts

Quote:
 Originally Posted by charybdis The indices of the first 17 partition primes are 2, 3, 4, 5, 6, 13, 36, 77, 132, 157, 168, 186, 188, 212, 216, 302, 366... I'm sure that if we were in the 1950s and this was the full known extent of the sequence, you'd have been speculating that there are no more odd terms. When the next term was discovered to be 417, you'd have said that maybe there are no terms that are 3 mod 4 apart from 3. And then it turns out 491 is in the sequence. It's easy to go looking for patterns. But if it's not outrageously unlikely for them to have appeared by chance, and there's no good reason for them to exist, you shouldn't expect them to continue forever.
See https://oeis.org/A005849, the numbers n such that n*2^n+1 is prime (Cullen prime), there is no single known prime term, even worse, all known terms except 1 and 6611 are divisible by either 2 or 3 (or both), also there is no single known term == 2 mod 4

Last fiddled with by sweety439 on 2022-08-06 at 14:23

2022-08-08, 13:27   #13
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

2×52×73 Posts

Quote:
 Originally Posted by paulunderwood FactorDB has verified it. Now The Prime Pages awaits your submission under: Generalized Lucas number, Wagstaff, ECPP
this page for New Mersenne Conjecture needs to be updated

 2022-08-09, 13:45 #14 sweety439     "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 365010 Posts For odd prime q, 2^q-1 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 (p-1)/n is the largest m dividing p-1 such that 2 is m-th power residue mod p
 2022-10-16, 09:50 #15 sweety439     "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 71028 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)
 2022-10-22, 05:33 #16 sweety439     "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 2·52·73 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 2022-10-22 at 05:44
2022-11-08, 12:59   #17
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

2×52×73 Posts

Quote:
 Originally Posted by unconnected 2744952 terminates! My first termination for a while.
Its termination is unusual, terminate at a relative larger prime 347, most terminations of a number which reaches to 100 digits or more are a prime < 100

2022-11-20, 04:06   #18
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

365010 Posts

Quote:
 Originally Posted by frmky 2,1091+ is in LA now. Code: linear algebra completed 226672 of 115553837 dimensions (0.2%, ETA 78h56m)
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 2^1093-1 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)

2022-11-24, 06:33   #19
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

2×52×73 Posts

Quote:
 Originally Posted by frmky And it's done. Another surprisingly smooth run.
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. second-largest 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 P-1 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)

2022-11-24, 12:06   #20
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

23·937 Posts

Quote:
 Originally Posted by sweety439 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. second-largest 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 P-1 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)
Complete gibberish. Devoid of content. Please move this to misc. math.

2022-11-24, 20:27   #21
xilman
Bamboozled!

"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across

26×181 Posts

Quote:
 Originally Posted by R.D. Silverman Complete gibberish. Devoid of content. Please move this to misc. math.
+1

Moved somewhere even more appropriate by a mod with whose actions I agree.

 2022-11-27, 14:03 #22 sweety439     "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 2·52·73 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...-prime-numbers, for this PRP, N-1 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 N-1 (which is equivalent to factor 5^15960-1) to enable N-1 proof for this PRP? Last fiddled with by sweety439 on 2022-11-27 at 14:05

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