More off-topic

[sweety439]

Originally Posted by Batalov

- p(n) and p(n+1) are prime. E.g. n = 2, 1085, <next term?>

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,b-1) (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^n-1 are twin primes
* For two different k-values, there are only finitely many n such that k*2^n+1 is prime for both of the k
* For two different k-values, there are only finitely many n such that k*2^n-1 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^n-1 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^n-1 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^n-1 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 3^541-1 (this is equivalent to: there are only finitely many n such that both of these formulas generate primes: (3^n-2,(3^n-1)/2), ((3^n+1)/2,3^n+2), (2*3^n-1,2*3^n+1))
etc.

[sweety439]

Originally Posted by charybdis

How about you look up the asymptotics of those sequences yourself and use the Prime Number Theorem to work out whether they ought to contain infinitely many primes?

I have a research of minimal primes, i.e. finding the minimal set of the “primes > b written in base b” strings for the subsequence ordering, e.g. in decimal (base b = 10), there are 77 such primes: 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 227, 251, 257, 277, 281, 349, 409, 449, 499, 521, 557, 577, 587, 727, 757, 787, 821, 827, 857, 877, 881, 887, 991, 2087, 2221, 5051, 5081, 5501, 5581, 5801, 5851, 6469, 6949, 8501, 9001, 9049, 9221, 9551, 9649, 9851, 9949, 20021, 20201, 50207, 60649, 80051, 666649, 946669, 5200007, 22000001, 60000049, 66000049, 66600049, 80555551, 555555555551, 5000000000000000000000000000027, since there is no infinite antichain for the subsequence ordering, there are only finitely many such primes in every base b, however, determine the minimal set is hard when the base b is large, and this is an interesting problem, also see my data of the minimal primes and the unsolved families (sorry, I lost the “left17” file), I doubt whether an exponential sequence (a*b^n+c)/gcd(a+c,b-1) (with fixed integers a>=1, b>=2, c != 0, gcd(a,c) = 1, gcd(b,c) = 1, and variable integer n>=1) contains infinitely many primes, if it cannot be proven to only contain composites or only contain finitely many primes, by using covering congruence, algebraic factorization, or combine of them.

[paulunderwood]

Originally Posted by sweety439

I have a research of minimal primes

<big snip>

Yet another thread hijacked with the same off-topic ramblings from sweety. Sweety, please confine your "studies" to your blog

[charybdis]

Originally Posted by sweety439

I have a research of minimal primes...

Sweety.

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?

[Batalov]

Originally Posted by sweety439

This sequence should be 2, 3, 4, 5, 1085, … rather than E.g. 2, 1085, …

What part of "E.g." was too difficult to understand?

[sweety439]

Originally Posted by williamsheikspa

Hello its william sheiksapre, I'm new to this forum.

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^n-1)/(b-1) 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*206-11)/15 (38815 digits), (22^22002*251-335)/21 (29538 digits), 30^24609*18+13 (36352 digits).

I am interesting in number theory, this is my article about primes

[sweety439]

Originally Posted by ryanp

W117239 = \((2^{117239}+1)/3\) has been proven prime with ecpp-mpi, and the certificate is processing on factordb.com.

Wow!!!

Can you also run this number? This number is just few digits longer.

[paulunderwood]

Originally Posted by sweety439

Wow!!!

Can you also run this number? This number is just few digits longer.

Can't you stop asking all and sundry "Can you run...." or "Please factor...". If you persist, we mods will create a repository of such requests. You have hereby been warned.

[Uncwilly]

Originally Posted by sweety439

Can you also run the PRP of the form Phi(n,2)?

Originally Posted by Batalov

Fair warning: one more time we will hear "But I... But I... Win$ows doesn't work! There is no binary!" - you will receive an educational ban for 7 days. Then for 30 days, then for a year. It's better that you reply nothing (that's what we advise), than one more time on top of dozens that you have already done: "Run this for me!" [/Warning]

Originally Posted by sweety439

Can you also run

You were warned. You continued to ask people to run numbers for you. Enjoy your ban.

[sweety439]

In the top generalized repunit primes page, the condition of the generalized repunit primes (b^n-1)/(b-1) 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^n-1, they have a condition n >= b-1, 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^n-1)/(b-1) 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

[sweety439]

Are there any interest to extend the sequence https://oeis.org/A275530?