Conjecture:if $i

Conjecture:if $i
2011-07-30T06:15:12Z

p[i] is the i-th prime. $\pi(x)$ is prime counting function.

Firstly, I think that this Prime gap inequality holds true,

$ p[i+1] - p[i] <= i $

Prove:for any i>0, we can always find distinct prime factors for {p[i], p[i]+1,...,p[i+1]}. For example, i=11, p[11]=31, p[12]=37, {31,32,33,34,35,36,37} have distinct prime factors {31,2,11,17,5,3,37}. Pigeonhole principle shows this simple inequality!

My question is the title,

Conjecture(Prime counting inequality) :

if $ i\lt j $, then $\pi(p[i]+i)-i<=\pi(p[j]+j)-j+1$

Edit: sorry, this conjecture is false! But another question is arising: if $ i\lt j $, what is the max value of $g(i,j) = (\pi(p[i]+i)-i) - (\pi(p[j]+j)-j) /; i\lt j$

I find g(i,j) may be 12. g(150065,150090)=12.

and more, what is the max value of h(i,j)=j-i /; $(\pi(p[i]+i)-i) > (\pi(p[j]+j)-j) $

Answer by Gerhard Paseman for Conjecture:if $i
2011-07-30T07:03:45Z

I believe your inequality $p(i+1) \leq p(i) + i$ is true, but that there is no short and elementary proof. It follows from inspection and some known results on the length of gaps between primes, cf. Dusart or Harman.

Your titled inequality I believe fails for some $i$ with $j=i+2$. I don't have a specific value for $i$, but I suspect that there is a sequence such that your arguments to $\pi$ could fall into a large gap, especially when $p(i+2)= p(i)+6$. That is where I would start looking for counterexamples.

Gerhard "But It Looks Quite Plausible" Paseman, 2011.07.30