Half way decent fortunate conjecture attack?

[CRGreathouse]

Originally Posted by Joshua2

So your saying that you don't think that proof is valid. You haven't found any errors with my q-kPrimorial(n) is prime (or the first couple on a few can be 9) have you?

I haven't tried to check it. Can you give a precise statement of that, please? "For each positive integer k, ..."

[Joshua2]

Let q(x) be the next prime after x. q(x) - k Primorial(n) is prime for all positive integer n and k. For some k, the sequence will have a 9 within the first few terms, afterwhich all will be prime. That second sentence needs some refining probably by some additional testing or math analysis. I have checked k up to 2500 for the first 50 terms, and sometimes the 1st, 2nd, or 3rd produced 9, after which all were primes. Testing out randomly at large k being several million, or trillion, I came across two values which produced 9 on the 4th iteration.
EDIT:It seems my value of always 9 is not true, but I think that maybe the function needs to stabilze for a few values to produce primes.

[CRGreathouse]

I'm still having a little trouble with your statement. What's x in "q(x)"? And the value 1 is very common; I assume you meant to include it. Here's what I assume your conjecture to be:
"Let q(x) be the next prime after x and Q(x) = q(x) - x.
1. For all positive integers n and k, Q(k * primorial(n)) is prime, 1, or 9.
2. For all positive integers k, there is some N such that for all n > N, Q(k * primorial(n)) is prime or 1."


Part 2 can have no finite counterexample. The counterexamples I found to part 1 are
2# * {263, 446, 536, 565, 568, ...}
3# * {413, 523, 717, 806, 922, ...}
5# * {2018, 2736, 3679, 4988, ...}
7# * {48907, 60997, 76102, 95354, ...}
11# * {100284, 169695, 188919, ...}
13# * {869700, 9452087, ...}

I expect that for each value of n there are infinitely many counterexamples.

[fivemack]

Another small problem is that pari's implementation of APRCL takes really quite a lot of memory; about 600M for a 2000-digit number, and a little over 4G for a 3000-digit one.

I think this is a rather uninteresting problem, since by construction K = nextprime(p#)-p# must be prime if it's less than p^2, and for it to be greater than p^2 would mean you had a very unusually large gap between primes (p# is roughly e^p, so it'd be a gap of O(log(N)^2) between primes near a number of size N; this is about what one of the conjectures mentioned in http://www.ieeta.pt/~tos/gaps.html gives as the size of the largest-ever gap between primes up to N, but it would be very odd for such a largest-ever gap to be of the very special form of fortunate numbers.

[CRGreathouse]

Originally Posted by fivemack

this is about what one of the conjectures mentioned in http://www.ieeta.pt/~tos/gaps.html gives as the size of the largest-ever gap between primes up to N, but it would be very odd for such a largest-ever gap to be of the very special form of fortunate numbers.

Right. If the maximum prime gap ~ (log x)^2 then almost certainly all fortunate numbers are prime. On the other hand, there seems to be at least a shred of a possibility that there are composite fortunate numbers is the maximum prime gap ~ 2/e^gamma (log x)^2.

[Joshua2]

Originally Posted by fivemack

I think this is a rather uninteresting problem, since by construction K = nextprime(p#)-p# must be prime if it's less than p^2, and for it to be greater than p^2 would mean you had a very unusually large gap between primes.

I follow the rest of the argument, but I don't see why it must be prime or why greater than p^2 would mean a large prime gap.

As well, I believe you have both my conjecture and the fortunate conjecture misinterpreted, because neither one can produce 1 as the output (see below post). It is nextprime(p#+1) for the fortunate conjecture or nextprime(Kp#+1) for mine.

Edit: because the next number would be even after the +1 term, it could be stated as +2 instead to make it odd as in nextprime(p#+2).

[Joshua2]

Originally Posted by CRGreathouse

I'm still having a little trouble with your statement. What's x in "q(x)"?

q is the nextprime function, so x is any integer.

Originally Posted by CRGreathouse

And the value 1 is very common; I assume you meant to include it. Here's what I assume your conjecture to be:
"Let q(x) be the next prime after x and Q(x) = q(x) - x.
1. For all positive integers n and k, Q(k * primorial(n)) is prime, 1, or 9.

There shouldn't be any value of 1, because q(x) is next prime after x+1.

So, the Jordan generalized Fortune conjecture. Let q(x) be the next prime after x+1, and Q(x) = q(x) - x. For all positive integers k and n positive integers ten or greater (i'm thinking better is n integer greater than ceiling(log k)), Q(k*primorial(n)) is prime.

k=1 would be the fortune conjecture, with n any positive integer.

I discovered k = 9.99999 x 10 ^ 87 + 1 took nine terms to stabilize to primes.

[CRGreathouse]

Originally Posted by Joshua2

q is the nextprime function, so x is any integer.

You write:
"Let q(x) be the next prime after x. q(x) - k Primorial(n) is prime for all positive integer n and k."

I think you mean:
"Let q(x) be the next prime after x. q(k Primorial(n)) - k Primorial(n) is prime for all positive integer n and k."

That's what I meant when I wrote "What's x in 'q(x)'?", and why I defined Q(x) to avoid the issue.

Originally Posted by Joshua2

There shouldn't be any value of 1, because q(x) is next prime after x+1.

Once again, this comes from not defining terms carefully enough. I'm using Pari, and in Pari nextprime(x) is the smallest prime greater than or equal to x. Thus nextprime(6) = nextprime(7) = 7.

[Joshua2]

If you look at my pari, I have nextprime(x+2 or x+3) (pari's notation is somehow equivalent to actually x+1 or x+2). nextprime(6+1) = 11. The smallest possible value is 3, which is the first term of the fortunate series. So I guess I need to change to:

Let q(x) be the next prime after x, and Q(x) = q(x+2) - x. The rest is good as is.

k = 44 took two terms to stabilize (first term was 9).
k= 221 took three.
k=2018 took four
k = 47867 took five.
k= 100284 took six
k = 9.99999 x 10 ^ 87 + 1 took 9 terms to stabilize to primes.
k = 9.999 x 10 ^ 118 + 234 took 10 terms to stabilize.
k= 9 x 10 ^ 173 + 81 took 12 terms

So it seems that I need as k tends to infinity it would take more and more terms. However, 99.9% at least seem to work for all n and not need to stabilize. So I need something like "For all positive integers k and n integer greater than log k, Q(k*primorial(n)) is prime, where Q(x) = q(x+2) - x.

[CRGreathouse]

Sorry, I missed your program.

In any case, I don't think I'll be doing any more checking of this conjecture. I think when you say Primorial(9) you mean what I mean by 23# = 223092870, and checking for k that cause problems there would take a lot of crunching (in the petatick range, or several weeks on my computer).

[Joshua2]

Yes, I by primorial 9 i meant the first 9 primes multiplied together. You don't necessarily need to check there at 9 now that I changed my conjecture to not use nine. Instead of the first saying n starting at 9, see previous post it is log k. So if you are trying k's with a ten digits you might need ones that large, but there are plenty of k's and n's to check that are fast. I am doing some right now and it is churning. Do you still believe my conjecture is related to the size of prime gaps? And if so could you help me understand how?