20220715, 07:04  #1 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2·5^{2}·73 Posts 
More offtopic
This sequence should be 2, 3, 4, 5, 1085, … rather than 2, 1085, …
For p(n) and p(n+2), the known n are 2, 3, 4, 186, … For p(n) and p(n+3), the known n are 2, 3, … For p(n) and p(n+4), the known n are 2, 212, … For p(n) and p(n+5), the known n are 2727, … For p(n) and p(n+6), the known n are 2502, … For p(n) and p(n+7), the known n are 6, 491, … For p(n) and p(n+8), the known n are 5, 1578, … For p(n) and p(n+9), the known n are 4, 2493, … For p(n) and p(n+10), the known n are 3, … Also, I conjectured that for all integer k>=1, there are only finitely many n such that p(n) and p(n+k) are both primes (like that I conjectured that for all pairs of two exponential sequences (a*b^n+c)/gcd(a+c,b1) (with fixed integers a>=1, b>=2, c != 0, gcd(a,c) = 1, gcd(b,c) = 1), there are only finitely many n >= 1 such that both sequences generate primes, e.g. * For any k divisible by 3, there are only finitely many n such that k*2^n+1 and k*2^n1 are twin primes * For two different kvalues, there are only finitely many n such that k*2^n+1 is prime for both of the k * For two different kvalues, there are only finitely many n such that k*2^n1 is prime for both of the k * For any k divisible by 3, there are only finitely many n such that k*2^n+1 and k*2^(n+1)+1 are both primes * For any k divisible by 3, there are only finitely many n such that k*2^n1 and k*2^(n+1)1 are both primes * For any base and any even k, there are only finitely many n such that the repunits with length n and n+k are both primes * For two different bases, there are only finitely many n (n must be primes) such that the repunit with length n in these two bases are both primes * 127 is the largest n such that 2^n1 and (2^n+1)/3 are both primes (although (2^n+1)/3 does not satisfy the condition of the sequences, but we can change them to 2*4^n1 and (2*4^n+1)/3) * There are only finitely many n such that all of n and n+1 have primitive roots, and the largest such n is 3^5411 (this is equivalent to: there are only finitely many n such that both of these formulas generate primes: (3^n2,(3^n1)/2), ((3^n+1)/2,3^n+2), (2*3^n1,2*3^n+1)) etc. Last fiddled with by sweety439 on 20220715 at 07:12 
20220715, 09:29  #2  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2×5^{2}×73 Posts 
Quote:
Last fiddled with by sweety439 on 20220715 at 09:41 

20220715, 10:27  #3  
Sep 2002
Database er0rr
29×151 Posts 
Quote:


20220715, 11:05  #4  
Apr 2020
36B_{16} Posts 
Quote:
You clearly put a lot of time into your posts here, and you must also have done a fair amount of reading about your preferred topics. And yet you still ask people on here to do things for you all the time. Usually those are computational tasks, and maybe there's a reason you can't do those yourself, e.g. perhaps you're using your parents' computer and they don't let you use it for that. (If that is the case, there's no need to keep it secret, just let us know.) But in this case all you need to do is a little bit of reading and a couple of small calculations. You've done that many times before for your own "research" purposes. So why not do it now, rather than making a reply that has absolutely nothing to do with the thread or the task that I asked you to do? 

20220722, 09:32  #6 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2·5^{2}·73 Posts 
Hello!!! Nice to meet you! Welcome of this forum!!!
This forum is about finding large primes (e.g. top proven primes and top probable primes) and factoring integers (see this: https://en.wikipedia.org/wiki/Integer_factorization), for example, you can try to find the smallest prime of the form (b^n1)/(b1) for b = {185, 269, 281, 380, 384, 385, 394, 396, 452, 465, 511, 574, 598, 601, 631, 632, 636, 711, 713, 759, 771, 795, 861, 866, 881, 938, 948, 951, 956, 963, 1005, 1015} or factor the numbers 13^282+1 and 13^288+1 (both of them have known prime factors, but not completely factored, you can try to completely factor them!!) Also, some people in this forum also prove the primality of large PRPs (by use elliptic curves primality proving programs such as PRIMO), currently the record elliptic curves primality proving number is 10^50000+65859 with 50001 decimal digits, see this page, you can try these large PRPs: (13^23756*149+79)/12 (26464 digits), 13^32020*8+183 (35670 digits), (16^32234*20611)/15 (38815 digits), (22^22002*251335)/21 (29538 digits), 30^24609*18+13 (36352 digits). I am interesting in number theory, this is my article about primes 
20220722, 21:21  #7  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2·5^{2}·73 Posts 
Quote:
Can you also run this number? This number is just few digits longer. 

20220722, 21:39  #8  
Sep 2002
Database er0rr
1000100011011_{2} Posts 
Quote:


20220723, 00:56  #9  
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
25067_{8} Posts 
Quote:


20220806, 11:43  #10 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2×5^{2}×73 Posts 
Limit of generalized repunits
In the top generalized repunit primes page, the condition of the generalized repunit primes (b^n1)/(b1) is n >= b/5, otherwise every prime can be called “generalized repunit prime”, like the generalized Cullen primes page n*b^n+1 and the generalized Woodall primes page n*b^n1, they have a condition n >= b1, otherwise every prime can be called “generalized Cullen prime” or “generalized Woodall prime”, thus I think the more reasonable condition of generalized repunit primes (b^n1)/(b1) is n >= b, under my definition, the “generalized repunit primes” are 3, 7, 13, 31, 127, 1093, 8191, 19531, 55987, 131071, 524287, 797161, 12207031, 305175781, 2147483647, 16148168401, 50544702849929377, 1111111111111111111, 2305843009213693951, 6115909044841454629, 29043636306420266077, 459715689149916492091, 7369130657357778596659, 11111111111111111111111, 109912203092239643840221, 618970019642690137449562111, 162259276829213363391578010288127, 177635683940025046467781066894531, 3754733257489862401973357979128773, 26063080998214179685167270877966651, 170141183460469231731687303715884105727, 243270318891483838103593381595151809701, 568972471024107865287021434301977158534824481, 7538867501749984216983927242653776257689563451, 6957596529882152968992225251835887181478451547013, 26656068987980386414408582952871386493955339704241, …, but under my definition, what are the top 20 generalized repunit primes (to bases other than 2 and 10)? Of course, the top generalized repunit prime (7176^24691  1)/7175 is still generalized repunit prime, since 24691 >= 7176

20220806, 13:04  #11 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2·5^{2}·73 Posts 
Are there any interest to extend the sequence https://oeis.org/A275530?
Last fiddled with by sweety439 on 20220806 at 13:04 
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