According to your list, a counter-example, if it exists, must have more than $60,000$ digits. So, a counterexample would be a quite gigantic prime.
Unfortunately, a proof of the conjecture will almost certainly be out of reach.
The search for a counter-example can be painful as well, it is we...
@EnzoCreti How did you get this list ? It shows that a counterexample must have more than $60\ 000$ digits. I think that the numbers of this form (no matter of the congruences) can have least prime factor very high. The probability that the least prime factor is very high is much higher than one might expect. A large "random" number has no prime factor less than $10^{20}$ with probability about $1.2$%. I do not think there is a special reason. Moreover, my list of candidates has become obsolete.
@EnzoCreti No, this is only known for arithmetic progressions of the form $an+b$ with coprime $a$ and $b$ (Dirichlet). The numbers here will behave like "random" numbers , hence we won't be able to rule out a large prime.
@EnzoCreti Not really realistic. Sometimes, we only have a very small prime factor and a prime cofactor and sometimes the smallest prime factor has more than $15$ digits. This does not indicate hidden factors. So, it seems to be a pure coincidence that $6$ does not appear early. If factordb would allow concatenating numbers, that would be fine! But it seems that factordb does not support this yet. I have passed $120\ 000$ now without success.
Primes are of limited importance. They are used for cryptography, but there are better methods and the safety of the RSA is debatable. Such primes as the one you search are purely for curiousity. Nevertheless, many such primes were calculated. It is a bit sad that this kind of primes is not considered to be interesting by most of the people here. Of course, special primes, as the Mersenne primes, are more important because they answer important mathematical questions.
@EnzoCreti Concerning the structure of the factors of those numbers, I search a prime factor of $f(530)$ and in a new project I search $f(n)$ of the form $7k+6$ such that $\frac{f(n)}{5}$ is prime which turns out to be tough as well.
@EnzoCreti No, seems that the smallest prime factor has more than $25$ , probably even more than $30$ digits. Of course, trial division does not work anymore to find such factors. The elliptic curve method is the key. With factordb , the concatenation for numbers works, if the numbers are concrete or depend linear on $n$ , for example $c(123,456)$ gives $123456$ and $c(n,n+2)$ correctly concatenates $n$ and $n+2$. Strangely, $c(M(n+1),M(n))$ does NOT cancatenate $2^{n+1}-1$ and $2^n-1$ correctly, otherwise this would be the perfect way to search further primes.
For clarification, $M(n)=2^n-1$ , the $n-th$ Mersenne number is a valid syntax in factordb. I also tried "c(2^(n+1)-1,2^n-1)" , but without success either.
My post have 870 views but only 13 votes...so I think that there is skepticism about that conjecture or nobody believes it has some importance...
@Peter between 10^5 and 2*10^5 there are about 2000 candidates (or even less)...if the probability of a random number to be prime is 1 out of 4200, perhaps it will be still difficult to find a prime?
@EnzoCreti The logarithm is roughly $n\cdot \ln(4)$ , hence the probability that a candidate is prime is about $1:\frac{e^{-\gamma}}{\ln(10^6)}\cdot \ln(4)\cdot n\approx 1:0.0563\cdot n$
A large number has no prime factor less than $x$ with probability about $\frac{e^{-\gamma}}{\ln(x)}$, if $x$ is large. The approximation is already good for $x=10^6$
Didn't continue, probably no factor below $10^{30}$
@VishaalSelvaraj A nice observation, but I don't think that is considered to be of great importance. Otherwise we would already have found helpers. r.e.s was a helper, but quit at $10^5$, now we are only two despite of 1k+ views. And without some luck or more helpers, we won't be able to decide the conjecture.
@Peter this morning I arrived at 248.000, the search has slowed down. I was hoping that on Saturday morning we would have arrived at k=300.000 but now I think that this is not the case...
@Vishaal Selvaraj you told me about a library...you said that it is necessary high power of the computer. How much do you believe?
@Vishaal Selvaraj Ok thank you very much, I have half a mind to buy a new computer (mine at home is broken I'm using the computer office). So I could help also Peter...
@Peter f(n)/5 conjecture is another good question...
@Peter with my computer office now the search is almost 3 times slower than before...
@Peter This morning I arrived only at 260.000 now the search is awfully slow...i can't imagine if we have to arrive to k=500.000 how long and consuming will be the search!
Look here : primes.utm.edu/glossary/page.php?sort=WallSunSunPrime to see a kind of prime numbers for which it is expected that there are infinite many of them although none is known. Noone knows a reason why this should not be the case, and it is very likely that we are in the same situation.
@Peter yes it's true...it would be interesting to know how many primes with other residues are up to k=260.000...if I remember up to k=100.000 there were 30 or 31...the chance that random numbers with residue 6 mod 7 do not occur for 30 times shouldn't be (8/9)^30?
The growth rate of our sequence is exponential and even if it produces infinite many primes (I would still guess it does), the smallest can be huge. We didn't search for primes with other residues, but I would not be surprised if there were none as well in the range [10^5,260000]
You mean (5/6)^30. Yes, but consider that you didn't fix this residue at the beginning. The coupon-collector problem emphasis this detail. To get the last coupon, you need n trials in the average, no matter how many trials you already have made.
However, here chances are not equally distributed because of the frequencies.
But for our problem we need the probability that a random chosen prime of this seqeucne has residue 6. What is the distribution of the residues ? I think it is 1/18 for 0 and 1 , 1/9 for 2,4,5,6 and 4/9 for 3 , right ?
If we would have found, lets say , 200 primes and none were of residue 6, this would be a strong indication that this has a reason. But 31 is not enough to draw any conclusion.
At least, the conjecture holds upto very large numbers. Are you actually interested in other primes in this range ? If yes, I can calculate the candidates without considering the residues and post them on github.
But of course, we will have much more candidates ...
This was surprising in the case of your conjecture and it would be surprising in my case as well because the frequencies should be approximately the same.
Doublechecking my conjecture, I passed 104 000 , pfgw displayed resuming a second time , I don't know why.Seems that the program sometimes saves some checkpoints. Concerning the Wall Sun Sun primes, obviously the property that has to be checked is not trivial.
Heuristic indicates infinite many primes, yes. Similar situation with the Wieferich-primes.
My conjecture is doublechecked upto 10^5 if we can assume that displaying "resume" does not indicate some problems. Hopefully not a sign of an unstable software.
The BPSW-test is an efficient primality-test for which no counterexample is known. It combines a fermat-test and a lucas-test and if a number fails this test is must be composite. If it passes, the number is very likely prime. The numbers upto 2^64 have been checked.
Did you reach 270 000 in the mean-time ?
Do I have to prepare for calculating more candidates ?
Discussion between Enzo Creti and Peter
Discussion between Enzo Creti and Peter