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I made this observation and it seems reasonable to me to ask :if $n$ is a natural number then the number of the primes less than or equal to $n$ is denoted by $π(n)$ . is that true that in any interval of length $n$ there are at most $π(n)+1$ primes?(the $+1$ is needed for the trivial occasion where $n=p-1$ and the interval of length $n$ is $[2,p]$) Alternative we can say that in any interval of length $n-1$ there are at most $π(n)$ primes.

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-1: What is your evidence? Up to where have you checked? What have you tried? –  t.b. Feb 12 '11 at 13:20
    
This may be true, I will think about a proof. Theo I don't know why you are so negative, it is a good question. –  Eric Naslund Feb 12 '11 at 14:26
    
Fair enough. I maybe should of looked at the edit history a bit closer. –  Eric Naslund Feb 12 '11 at 15:26
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3 Answers

up vote 6 down vote accepted

This is a well-known conjecture. It even has a name: the Second Hardy-Littlewood Conjecture, in the form: $\pi(x+y) \le \pi(x)+\pi(y)$ for $x, y \ge 2$.

For a long time, this was generally thought to be true. Then in 1974, Ian Richards showed that it was incompatible with the First Hardy-Littlewood Conjecture! He did this by constructing explicitly an admissible prime constellation of length $x$ and size larger than $\pi(x)$. Computers were involved. See here for more details.

The First H-L Conjecture is considered a sure thing, which has led most mathematicions to abandon the Second H-L Conjecture (although any counterexamples are likely to be extremely large).

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why we believe that the first Hardy -Littlewood conjecture is the right –  minasteris Feb 20 '11 at 22:03
    
I'd really like to know where the names First and Second H-L Theorems came from. Neither one appears in their famous 1923 paper. –  Charles Feb 24 '11 at 15:50
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A combination of the Brun-Titchmarsh inequality and the Prime Number Theorem will yield the following: For every $\epsilon>0$ there exists $N$ such that for $y>N$ and for every $x>0$ we have $$\pi(x+y)-\pi(x)<(2+\epsilon)\pi(y).$$

However, this is not quite as good as what you are asking, since you want for every $M,N$ $$\pi(M+N)-\pi(N)\leq \pi(M)$$ Is this true or not? According to my analytic number theory text (Montgomery and Vaughn):

It was once conjectured that $$\pi (M+N)\leq \pi (M)+\pi (N)$$ for $M>1$, $N>1$, but there is now serious doubt as to the validity of this inequality. Indeed, it seems likely that $\rho(y)>\pi (y)$ for all large $y$.

Here, $\rho(y)$ is defined as $$\limsup_{x\rightarrow\infty}(\pi (x+y)-\pi (x)).$$

Hope that helps.

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so do you think that it is a well known open problem? proposed an open problem again? we do not have any counterexample?? –  minasteris Feb 12 '11 at 14:40
    
It is one of Hardy and Littlewood's conjecture. Counterexamples for this kind of thing are likely to occur far far beyond the realm of numbers we can check up to. Intuition tells me that if there are counter examples, the first one would be enormous. (Similar to the first time $\pi(x)$ passes $\text{li} (x)$, at around $10^{316}$) –  Eric Naslund Feb 12 '11 at 15:24
    
Some preliminary calculations suggest that the first counterexample almost surely has more than 1250 digits but fewer than 1700 digits. The bounds could be sharpened with more work (bringing them closer by 50-100 digits). –  Charles Feb 24 '11 at 16:20
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This is indeed a known open problem, the Hardy-Littlewood conjecture:

$$\pi(x+y) - \pi(x) \le \pi(y)$$

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