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Is it really true that, for any n > 0 , there is a prime between 2 ^ n

by goldust
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Office_Shredder
#19
Nov11-13, 10:42 AM
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So is your new claim something like there is a prime number between 2k and 2k+2kn for some n? (changing it from a linear range in k to a polynomial in k)
goldust
#20
Nov11-13, 10:53 AM
P: 85
Quote Quote by Office_Shredder View Post
So is your new claim something like there is a prime number between 2k and 2k+2kn for some n? (changing it from a linear range in k to a polynomial in k)
That's exactly right. n must be more than 1, of course.

This time, as the exponent n in 2 ^ n doubles, the probability that any randomly picked number under 2 ^ n is prime halves, and the range more than doubles. Therefore, the statement becomes more probable to being true as n increases.

So far, 2 ^ 1040 checks out. Would be cool if I can publish this result in a journal as an interesting little project I done, especially considering I'm only 18 years old.
D H
#21
Nov11-13, 01:13 PM
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Checking with small numbers such as all integers between 2 and 21040 is no guarantee that your conjecture is true for all numbers. If your new conjecture is that there is a prime between 2k and 2k+2*k2, that's essentially a variant on Cramer's conjecture. This conjecture has been tested for against much larger numbers than your 21040 and it still stands as a conjecture. Checking against a finite subset will never prove a universal statement about an infinite set. Checking can disprove the statement, but it can't prove it.
arildno
#22
Nov11-13, 01:22 PM
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Quote Quote by D H View Post
Checking with small numbers such as all integers between 2 and 21040 is no guarantee that your conjecture is true for all numbers. If your new conjecture is that there is a prime between 2k and 2k+2*k2, that's essentially a variant on Cramer's conjecture. This conjecture has been tested for against much larger numbers than your 21040 and it still stands as a conjecture. Checking against a finite subset will never prove a universal statement about an infinite set. Checking can disprove the statement, but it can't prove it.
Cramér's conjecture, not Cramer's conjecture.
goldust
#23
Nov11-13, 05:43 PM
P: 85
My conjecture is that, for any n > 1, there is a prime between n and n + how many prime factors n has [itex]\times[/itex] the sum of n's prime factors. When n is prime, the range would be between n and 2n, which has been proven. When n is of the form 2k, the range would be between 2k and 2k + 2 k2, which, as D H pointed out, is a variant on Cramér's conjecture.

So far, I've checked from 2 to 700,000. It seems more likely to be true as n gets bigger, because the range seems to be growing much faster than the growth of prime gap and seems to exceed the biggest prime gap as a function of n. Not sure if my conjecture could be considered original? I'll have to check with my prof on this one. It's essentially of the same nature as Bertrand's postulate.

It seems to me that the range between 2k and 2k + 2 k2 is sufficiently large to be true, because the range should be able to exceed the biggest prime gap as a function of n by the time n gets big enough. At n = 21000, the range is 1,999,999. Anyone by any chance know how big the biggest known prime gap is?

If the range as a function of n grows faster than and always exceeds the biggest prime gap as a function of n, wouldn't it be proof that the statement is true?
goldust
#24
Nov12-13, 02:45 PM
P: 85
Checked with my prof on this. He says it could be considered an original conjecture. However, because I'm still in high school, I can only publish it in a journal. He said my conjecture is fairly interesting in that it sort of unifies Bertrand's postulate (actually a theorem), Cramér's conjecture, and prime factorization. Under the special cases of n being prime and n being of the form 2k, my conjecture reduces to Bertrand's postulate (actually a theorem) and a variation on Cramér's conjecture, respectively. Because my prof is a relative of mine, he says he could pull some strings and there's a good chance it can get published in a particular university journal as an interesting home project.

This morning, while thinking over my first conjecture, I came up with a second conjecture, which is closely related to my first conjecture and which could lend more substance to my first conjecture. My second conjecture is that, for any k > 2, there is always a composite number in the range (2k , 2k + 1) whose range size is less than that of 2k. The range size of a number n is how many prime factors n has [itex]\times[/itex] the sum of n's prime factors. The range size of 2k is therefore 2k2.

So far, I've tested my second conjecture up to k = 30. I would need more than unsigned int to handle powers of 2 over 231.
goldust
#25
Nov12-13, 07:10 PM
P: 85
There is also a Firoozbakht's conjecture, which is of the same nature as Cramér's conjecture.

http://en.wikipedia.org/wiki/Firoozb...99s_conjecture

Legendre's conjecture and Oppermann's conjecture are of the same nature.

http://en.wikipedia.org/wiki/Legendre%27s_conjecture

http://en.wikipedia.org/wiki/Oppermann%27s_conjecture

I suppose my first conjecture is of the same nature as Bertrand's postulate.

http://en.wikipedia.org/wiki/Bertrand%27s_postulate
goldust
#26
Nov13-13, 03:25 PM
P: 85
I formulated my original first conjecture into a weak conjecture and a strong conjecture. The weak one is the same as my original first conjecture. The strong one is that, for any n > 1, the number of primes in the range (n , n + how many prime factors n has × the sum of n's prime factors) [itex]\geq[/itex] how many prime factors n has.

Now it should be good enough for publication in a journal as an interesting amateur home project.
goldust
#27
Nov14-13, 08:12 PM
P: 85
Interestingly, it appears that, for any k > 0, the number of primes in the range (2k , 2k + 2 [itex]\times[/itex] k2) [itex]\geq[/itex] k. The difference between 2k + 2 [itex]\times[/itex] k2 and the kth prime above 2k appears to be a non monotonically increasing function of k and that only when k = 1 do these two values equal each other. So far, I've checked up to k = 400 using Mathematica. This is a special case of my strong conjecture when n takes the form of 2k.
Robert1986
#28
Nov15-13, 07:27 AM
P: 828
I don't understand why you would want to publish your problem so that other people can solve it? You came up with the problem; don't let someone else solve your problem for you. Publishing a conjecture in some kind of journal might be nice, but working hard to solve it and then publishing the proof is even better. And, so far, you haven't even worked very long on it (since you only came up with it a few days age.)
goldust
#29
Nov16-13, 08:12 PM
P: 85
Quote Quote by Robert1986 View Post
I don't understand why you would want to publish your problem so that other people can solve it? You came up with the problem; don't let someone else solve your problem for you. Publishing a conjecture in some kind of journal might be nice, but working hard to solve it and then publishing the proof is even better. And, so far, you haven't even worked very long on it (since you only came up with it a few days age.)
If it's so easy to prove, I wouldn't have considered it worthy of being a conjecture. When n is prime, my conjecture reduces to Bertrand's postulate when n is prime. It's what happens when n is composite that makes my conjecture interesting. I won't publish anything unless I'm quite certain it is true.

In increments of a million, I look at how the smallest of the (n + how many prime factors n has [itex]\times[/itex] the sum of n's prime factors) - the (how many prime factors n has)th prime above n values change. I see a non monotonic increasing trend in these values as n gets larger. I haven't tested too many numbers so far, only up to about 28 million, where these values have increased from 1 to about 400. When n is a power of 2, I have tested to 610 as the exponent, and by now I'm quite certain of the non monotonic increasing nature of the difference when n is a power of 2, since the difference has increased from 1 to nearly 500,000.
goldust
#30
Nov19-13, 05:02 PM
P: 85
So far, tested my strong conjecture, which is a strengthening of Bertrand's postulate, to 655 million less 1, where the smallest difference has increased from 1 to 527. As for n being powers of 2, so far, tested using Mathematica to the exponent being 700 where the difference has increased from 1 to over 600,000. I have recorded all of my experimental results.
mfb
#31
Nov30-13, 04:10 PM
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Quote Quote by goldust View Post
This morning, while thinking over my first conjecture, I came up with a second conjecture, which is closely related to my first conjecture and which could lend more substance to my first conjecture. My second conjecture is that, for any k > 2, there is always a composite number in the range (2k , 2k + 1) whose range size is less than that of 2k. The range size of a number n is how many prime factors n has [itex]\times[/itex] the sum of n's prime factors. The range size of 2k is therefore 2k2.

So far, I've tested my second conjecture up to k = 30. I would need more than unsigned int to handle powers of 2 over 231.
Proof: For k>2, 2k is divisible by 8 and the "range size" (never heard of that before, but I have no idea about number theory) of 9/8 * 2k is 2k(k-1) < 2k2. The number is in your range, as 1<9/8<2.



Between 2k and 2k + 2 k2, we expect 2 k/ln(2) primes. Neglecting correlations, the probability of finding a prime is ##\displaystyle p(k) \approx 1-\exp\left(-\frac{2k}{ln(2)}\right)##. The product over all k should converge to some finite value, so the non-existence of a small counterexample is a good argument that could be no such gap. It is not a proof, of course.


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