I made the following conjecture

al-mahed

yes, S' infinite <==> S infinite, but to prove S' infinity seems to be as difficult as the original problem... if we prove that we could put all the primes into these two sets:

1. S is infinite ==> S' is infinite *
2. S' is infinite ==> S is infinite

* S infinity ==> S' infinity because if q = p + 2^n, q > 2^n ==> q - 2^n = p positive

CRGreathouse

Certainly there are many primes not in either S or S', so there goes that approach.

al-mahed

Sorry about my poor english but I did not follow you... you mean that in fact there are primes not in S or S'? Primes that cannot be written as p + 2^n or p - 2^n?

dodo

Post #9 shows examples of primes not of the form p + 2^n. I imagine that finding counterexamples for p - 2^n (in particular, checking those on A065381) is harder, since there is no limit to p or 2^n.

al-mahed

I made another conjecture: those primes that cannot be written as p + 2^n could be written as p - 2^n

Consider the two sets:

S = set of primes of the form p + 2^n
S' = set of primes of the form p - 2^n

If we prove that all primes should be in one of these two sets the infinity of the two sets will be proved.

In fact post #5 shows that the primes not in S may could be found in the infinity of the negative integer line as -p + 2^n, which I wonder is the same thing of find this primes in the infinity of the positive integer line as p - 2^n. We can search to infinity for them, increasing both p and 2^n until find a match.

This is why [tex]S \subset\ S'[/tex], I think. This is why trying to show that primes are in the form p + 2^n contain "gaps" = p - 2^n.

And now, by Tchebychef: for m > 1 there is at least one prime p such that m < p < 2m

if we put m = 2^n, n a natural > 0 ==> 2^n < p < 2^(n+1)

perhaps this theorem could be usefull

al-mahed

Let all positive whole number be written as a sum of power of two:

[tex]1 = 2^0[/tex]
[tex]
2 = 2^1[/tex]
[tex]
3 = 2^0 + 2^1[/tex]
[tex]
4 = 2^2[/tex]
[tex]
5 = 2^0 + 2^2[/tex]
[tex]
6 = 2^1 + 2^2 [/tex]
[tex]
7 = 2^0 + 2^1 + 2^2[/tex]
[tex]
8 = 2^3 [/tex]
[tex]
9 = 2^0 + 2^3[/tex]
[tex]
10 = 2^1 + 2^3[/tex]
[tex]
11 = 2^0 + 2^1 + 2^3[/tex]
[tex]
12 = 2^2 + 2^3[/tex]
[tex]
13 = 2^0 + 2^2 + 2^3[/tex]
[tex]
14 = 2^1 + 2^2 + 2^3[/tex]
[tex]
15 = 2^0 + 2^1 + 2^2 + 2^3[/tex]
[tex]
16 = 2^4 [/tex]
[tex]
17 = 2^0 + 2^4[/tex]
[tex]
18 = 2^1 + 2^4[/tex]
[tex]
19 = 2^0 + 2^1 + 2^4[/tex]
[tex]
20 = 2^2 + 2^4

etc...[/tex]

We know by Tchebychef: for m > 1 there is at least one prime p such that m < p < 2m

if we put m = 2^n, n a natural > 0 ==> 2^n < p < 2^(n+1)

We know by Gauss: the number of primes from 1 to x is approximately =[tex]\frac{x}{ln(x)}[/tex]

The number of primes between [tex] 2^n[/tex] and [tex] 2^{n+1}[/tex] is approximately =

[tex]\frac{2^{n+1}}{ln(2^{n+1})} - \frac{2^n}{ln(2^n)}[/tex] = [tex]\frac{2^n}{ln(2)}*(\frac{2}{n+1} - \frac{1}{n})[/tex] = [tex]\frac{2^n}{ln(2)}*(\frac{n-1}{n^2+n})[/tex]

and we know that [tex]2^n > n^2[/tex] for [tex]\geq 3[/tex] ==> [tex]\frac{2^n}{ln(2)}*(\frac{n-1}{n^2+n})[/tex] diverges for n --> [tex]\infty[/tex]

for n increasing the Tchebychef's theorem underestimates the amount of primes

CRGreathouse

I'm not sure why you're measuring the primes between [itex]2^n[/itex] and [itex]2^{n+1}[/itex]. The relevant probabilities are

[tex]1/\ln(p+2)[/tex]
[tex]1/\ln(p+4)[/tex]
[tex]1/\ln(p+8)[/tex]
. . .

Thus the relevant question is: Does [tex](1-1/\ln(p+2))(1-1/\ln(p+4))\cdots[/tex] diverge to 0?

Ignoring p, this is [tex](1-k)(1-k/2)(1-k/3)\cdots[/tex] for [tex]k=1/\ln2[/tex].

[tex]\prod_n(1-k/n)=\exp\sum_n\ln(1-k/n)=\exp\left(-\sum_n k/n+k^2/2n^2+\cdots\right)\ge\exp\left(-\sum_n k/n\right)=e^{-\infty}=0[/tex]

so the product is 0 and so we do expect that all odds are of the form [itex]p-2^n[/itex] for prime p and some positive n.

al-mahed

This is the point: seems to me that you already are considering that there are such primes of the form [tex]p + 2^n[/tex], and we don't know yet. Or did I not understand some particular point?

My idea on measuring the primes between [itex]2^n[/itex] and [itex]2^{n+1}[/itex]:

Lets discriminate the numbers in classes according to the highest exponent, such that the nth-class has [itex]2^n[/itex] numbers, etc. Considering some prime number in nth-class like [itex]2^0 + 2^a + 2^b + 2^n[/itex], we should have [itex]2^0 + 2^b + 2^n[/itex], [itex]2^0 + 2^a + 2^n[/itex] or [itex]2^0 + 2^a + 2^b[/itex] = prime, or if for some reason this particular prime couldn't be written of the form [itex]p + 2^n[/itex], we should have [itex]2^0 + 2^a + 2^b + 2^n + 2^k[/itex] a prime number for some [itex]2^k[/itex] exponent, writting this prime in the [itex]p - 2^n[/itex] form.

My idea: [tex]\frac{2^n}{ln(2)}*(\frac{n-1}{n^2+n})[/tex] should give us the number of primes expected in the nth-class, and then we'll be able do find the probability that some particular prime number have a permutation of the exponents that represents another prime number. By doing these calculations I hope that the probability increases when n increases.

Honestly I don't feel that these statistical procedures could give us an absolute proof, only heuristics evidences.

al-mahed

* because obviously [tex]2^0 + 2^a + 2^b + 2^n + 2^k - 2^k[/tex] is the same prime, so we can keep adding exponents until find some k such that [tex]2^0 + 2^a + 2^b + 2^n + 2^k[/tex] represents a prime; and as we know that for every prime of the form [tex]p - 2^n[/tex] there is a prime of the form [tex]p + 2^n[/tex], this should give us our proof.

ps: I had to write this sentence here because TEX become crazy and change the expressions

CompuChip

Sure I will do that, if it makes you happy

This gives for the first 100 primes

where a line like
{491,{8,7,6,5,3}}
should be read as
[tex]491 - 3 = 2^8 + 2^7 + 2^6 + 2^5 + 2^3[/tex].

In the attachment I have plotted the occurring powers for the first 1000 primes (except 2 and 3). Have fun

PS the Mathematica code, in case anyone wants to reproduce this

Attached Thumbnails

 

al-mahed

Thank you CompuChip!!! You did a very good job !!

CRGreathouse, nevermind what I said, I do understand now reading more carefully what is your idea. I was just confused because:

1. using numbers like [tex]p+3, p+5, p+9, p+17,..., p+2^n+1[/tex], or [tex]p+1, p+3, p+7, p+15,..., p+2^n-1[/tex] to calculate the convergence to zero, seems to return almost the same result, so I think that this approach may not work

2. this seems to work only with p fixed, so we cannot search for [tex]p - 2^n[/tex] because we cannot calc the log of negative numbers (in case of to exclude a fixed p), and specially because we should increase p when n increases, so p cannot be a fixed constant such that it can be excluded

3. to exclude p should give us the same result of calculate the probabilities of [tex] 2, 2^2, 2^3, ..., 2^n, ... [/tex] be prime numbers, when only 2 is a prime number ==> both results cannot be the same

CRGreathouse

For information on the Cramer model of the primes, see (for example):
http://www.dartmouth.edu/~chance/cha...ann/cramer.pdf

The magnitude of the number is the only thing that matters for the Gauss-Tchebychev expectation 1/log(x). Obviously if you know about the divisibility of the number by small primes that will tell you more -- in fact this can be stated explicitly in what I call the Cramer-Maier model of the primes -- but that's only changing a multiplicative constant; the magnitude remains only the exponential part 2^k.

al-mahed

I was thinking about this approach (I don't know the difference between odds, chances and probability in english, so I used the word probability) :


The expression Q = [tex]\frac{2^n}{ln(2)}*(\frac{n-1}{n^2+n})[/tex] give us an estimate of the amount of primes between [tex]2^{n+1} and \ 2^n[/tex]

The formula [tex]\frac{\frac{2^n}{ln(2)}*(\frac{n-1}{n^2+n})}{2^{n+1} - 2^n} = \frac{\frac{2^n}{ln(2)}*(\frac{n-1}{n^2+n})}{2^n*(2 - 1)} = \frac{1}{ln(2)}*(\frac{n-1}{n^2+n})[/tex] should give us the probability of a particular sum of powers of two (inside the " nth-class [tex]2^n[/tex] " whose amount of numbers is obviously = [tex]2^n[/tex] ) represent a prime.

The formula [tex]\frac{1}{2^n}[/tex] should give us the probability of a particular number be this particular sum of powers of two + some power of two, because every number has a unique sum of powers of two representation.

The probability of both events is = [tex]\frac{\frac{1}{ln(2)}*(\frac{n-1}{n^2+n})}{2^n}[/tex]

Let this particular sum be = q, so we ask if there are infinite prime numbers p such that [tex] p = q + 2^n[/tex].

The questions are:

1) what is the probability of the same prime represent another primes taking increasing exponents, for example: [tex]p_1 = q + 2^{n_1}, p_2 = q + 2^{n_2}, p_3 = q + 2^{n_3}, ..., p_k = q + 2^{n_k}[/tex]; or

2) what is the probability of "there are infinite classes with at least one prime of the form [tex] p = q + 2^n [/tex] inside the class"?[tex]/[/tex]

The first questions I think may be answered by the product:

[tex]\prod^{n}_{k=2} \frac{\frac{1}{ln(2)}*(\frac{k-1}{k^2+k})}{2^k}[/tex] = [tex]\prod^{n}_{k=2} \frac{\frac{1}{ln(2)}*(\frac{k-1}{k*(k+1)})}{2^k}^{[1]}[/tex]

This product converges to zero because it represents the probability of each of the existing classes as having at least a prime number such like that ([tex]p_k = q + 2^{n_k}[/tex] with q fixed), and I think that this is because the fixed "q".

But this is not the "right question to ask" since this result cannot be interpreted as a proof of the "non infinity" (I don't know if exists such word in english...) of the set of [tex]p_1 = q + 2^{n_1}; p_2 = q + 2^{n_2}; p_3 = q + 2^{n_3}; ...; p_k = q + 2^{n_k}[/tex], because even if there is not a prime such that [tex]p_k = q + 2^{n_k}[/tex] in each of the existing classes we still could have infinite number primes of that form in another classes (like only in odd classes, etc).

Then I think that perhaps could be more prolific to work with the more general question:

" to calculate the probability of an any prime of the nth-class, considering all primes into this class, be combined with an exponent -2^n such that will be equal to a prime of the (n-1)th-class

OR the probability of an any prime of the (n-1)th-class, considering all primes into this class, be combined with an exponent +2^n such that will be equal to a prime of the (n-1)th-class

OR the probability of an any prime of the nth-class, considering all primes into this class, be combined with an exponent -2^m (m<n) such that will be equal to a prime of the same nth-class

OR the probability of an any prime of the nth-class, considering all primes into this class, be combined with an exponent +2^m (m<n) such that will be equal to a prime of the same nth-class"

This should increase our chances, because the probabilities will be added. But supose that we find a result that diverge to infinite, or that is convergent to a value > 1, in this cases how to interpret such results in terms of probabilities?

[1] k=1 doesn't matter because [tex]2^1[/tex] is the first class such that among the elements there is a prime (in fact, the only element is a prime)

CRGreathouse

Chance and probability have almost exactly the same meaning in English. I'd shy away from odds because it's sometimes used differently: 2:1 odds against means 1/3 probability, for example.

Good so far. It might be worth simplifying to [itex]1/n\log2[/itex] since if n is large the minor terms have little effect.

This is the probability that a number is a given number in the range, say 2^n - 78352785. I'm not sure this is what you want to do.

Now you're just being silly. You know there's just one number in the range which is your chosen number, so the probability is 0 for all other numbers in the range and
[tex]\frac{n-1}{n(n+1)\log2}\approx\frac{1}{n\log2}[/tex], the probability of being prime, for the chosen number.

This can't be what you mean. q is strictly between 0 and 1, so there are no prime numbers of this form.

If you mean "p = q + 2^n, where p and q are prime" then we're back to your original question.

I agree that it's the wrong question. I'm not sure at all about your form, though: the expected number of primes 2^k + q for a fixed odd q is
[tex]\sum_{k=1}^\infty\frac{1}{(2^k + q)\log(2^k+q)}\approx C+\sum_{k=2}^\infty\frac{1}{k\cdot2^k}[/tex]
for some (small) constant C.

But if you accept that each such class (for given prime q) is finite, then the problem simplifies to "are there infinitely many primes q such that q+2^k is prime for some integer k?".

CRGreathouse

Even better, since we agree that the expected number in each "class" is finite, consider showing that there is an infinite set of primes q such that q+2^k is prime for some integer k.

A094076 seems to strongly suggest that such an infinite set exists.

al-mahed

As a first "look" seems to be an strong suggestion indeed! I'll think about something!

Perhaps the fact [tex]2^{n+1} - 2^n = 2^n[/tex] could be usefull to infer adjacent classes relations, or even not adjacent classes relations.

Got it !

I was trying to calculate the probability of "a given number be prime" and "be a paticular sum of powers of two plus another power of two". What I meant saying "particular sum of powers of two" is "a fixed sum", for example, [tex]5 = 2^0 + 2^2[/tex] is the only sum of powers of two for 5, called "particular sum of powers of two" or "q".

Then in the nth-class what is the probability of [tex]x = 2^0 + 2^2 + 2^n[/tex] be a prime number? The probability of x be prime "times" the probability of [tex]2^0 + 2^2 = 5[/tex] be combined with an exponent [tex]2^n[/tex]. But you are right, this last probability is wrong.

This is not what I meant, q is not strictly between 0 and 1 because q = a particular sum of powers of two, I did not make myself clear.

My mistake was the second probability calculation, let me try to make myself clear and correct the mistake:

Lets take a look on:

[tex]1 = 2^0[/tex]
[tex]
2 = 2^1[/tex]
[tex]
3 = 2^0 + 2^1[/tex]
[tex]
4 = 2^2[/tex]
[tex]
5 = 2^0 + 2^2[/tex]
[tex]
6 = 2^1 + 2^2 [/tex]
[tex]
7 = 2^0 + 2^1 + 2^2[/tex]
[tex]
8 = 2^3 [/tex]
[tex]
9 = 2^0 + 2^3[/tex]
[tex]
10 = 2^1 + 2^3[/tex]
[tex]
11 = 2^0 + 2^1 + 2^3[/tex]
[tex]
12 = 2^2 + 2^3[/tex]
[tex]
13 = 2^0 + 2^2 + 2^3[/tex]
[tex]
14 = 2^1 + 2^2 + 2^3[/tex]
[tex]
15 = 2^0 + 2^1 + 2^2 + 2^3[/tex]
[tex]
16 = 2^4 [/tex]
[tex]
17 = 2^0 + 2^4[/tex]
[tex]
18 = 2^1 + 2^4[/tex]
[tex]
19 = 2^0 + 2^1 + 2^4[/tex]
[tex]
20 = 2^2 + 2^4

etc...[/tex]

This should let us conclude, without any formal proof, that all the elements of a "class" (considering a "class" the set of all elements with the same highest exponent) are a sum of the highest exponent of the class and each of the elements of all the precedents "classes" (but not the first element 2^n itself[tex]^{[1]}[/tex]). For example, 3th-class (exponent 2^3):

[1] this is because [itex]1+2^1+2^2+...+2^n=2^{n+1}-1[/itex],i.e, the total of the precedent exponents (which I call "classes") is the the total of elements of the following class minus one

[tex]8 = 2^3 [/tex] the first element is the exception

[tex]9 = 2^0 + 2^3[/tex] where [tex]2^0[/tex] is an element of the zero-class

[tex]10 = 2^1 + 2^3[/tex] where [tex]2^1[/tex] is an element of the first class

[tex]11 = 2^0 + 2^1 + 2^3[/tex] where [tex]2^0+2^1[/tex] is an element of the first class

[tex]12 = 2^2 + 2^3[/tex] where [tex]2^2[/tex] is an element of the second class

[tex]13 = 2^0 + 2^2 + 2^3[/tex] where [tex]2^0+2^2[/tex] is an element of the second class

[tex]14 = 2^1 + 2^2 + 2^3[/tex] where [tex]2^2+2^3[/tex] is an element of the second class

[tex]15 = 2^0 + 2^1 + 2^2 + 2^3[/tex] where [tex]2^1+2^2+2^3[/tex] is an element of the second class

This lead us to ask for the probability of an arrangement of 2-exponents of a prime in the last class, already taking off the greater 2-exponent, be exacly a prime of one of the precedent classes.

More generally, we could ask for the probability of an arrangement of 2-exponents of a prime in the last class be exacly a prime of one of the precedent classes, tanking off ONE of the any 2-exponents (but not 2^0 obviously). This should consider an arrangement inside the class itself.

And of course we can extend this to the primes of the form p - 2^n.

Hope I make myself clear now, please consider the english barrier.

Why did you use sum instead product?

al-mahed

We know that [tex]\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... \ =1[/tex], but your sum above seems to converges to 0.16667 or something like that