- #1
Rudy Toody
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If this series https://www.physicsforums.com/showthread.php?t=485665 is proved to be infinite, then proofs of these two conjectures can be done as simple corollaries.
Legendre's Conjecture states that for every $n\ge 1,$ there is always at least one prime \textit{p} such that $n^2 < p < (n+1)^2$.
Our stronger conjecture states that for every $n\ge 1,$ there are always at least \textbf{two} primes \textit{p} such that $n^2 < p_{m},p_{m+1} < (n+1)^2$.
Brocard's Conjecture states that for every $n\ge 2,$ the inequality $\pi((p_{n+1})^2)-\pi((p_n)^2) \ge 4$ holds where $\pi(n)$ is the prime counting function.
Our stronger conjecture states that for every $n\ge 2,$ the inequality $\pi((p_{n+1})^2)-\pi((p_n)^2) \ge 2(p_{n+1}-p_n)$ holds where $\pi(n)$ is the prime counting function.
Sorry, I couldn't get the [tex] stuff to work.
Legendre's Conjecture states that for every $n\ge 1,$ there is always at least one prime \textit{p} such that $n^2 < p < (n+1)^2$.
Our stronger conjecture states that for every $n\ge 1,$ there are always at least \textbf{two} primes \textit{p} such that $n^2 < p_{m},p_{m+1} < (n+1)^2$.
Brocard's Conjecture states that for every $n\ge 2,$ the inequality $\pi((p_{n+1})^2)-\pi((p_n)^2) \ge 4$ holds where $\pi(n)$ is the prime counting function.
Our stronger conjecture states that for every $n\ge 2,$ the inequality $\pi((p_{n+1})^2)-\pi((p_n)^2) \ge 2(p_{n+1}-p_n)$ holds where $\pi(n)$ is the prime counting function.
Sorry, I couldn't get the [tex] stuff to work.