Return to Question

Let $\ p\ $ be an arbitrary prime. Then an integer $\ s\ $ is called $p$-simple $\ \Leftarrow:\Rightarrow\ s\ $ is not divisible by any prime $\ q<p.\ $

Could you prove my conjecture (or is it known one way or another?):

For every integer $\ n\ $ there exists a $p$-simple integer $\ s\ $ such that $\ n\le s < n+p$.

• There is a $p$-simple integer such that $\ n\le s<n+p\ $ for every $\ n\ $ such that $\ -2\cdot p < n \le p$.

• If the conjecture holds restricted to all positive integers $\ n\ $ then it holds in full, for all integers $\ n$.

• There is a $p$-simple integer $\ s\ $ such that $\ p+1 \le s\le 2\cdot p\ $ (Chebyshev--actually, $\ s\ $ can be a prime).

• The conjecture holds for $\ n=2\ $ (trivial); and for $\ n=3\ $ (at least one of the three consecutive integers is odd); and for $\ p=5\ $ (among any $5$ consecutive integers, at the most $3$ are even, at the most $2$ are divisible by $3$, whole one of the even ones is divisible by $3$ when there are there three even numbers--thus one of these consecutive $5$ integers is not divisible by $2$ nor by $3$).

Let $\ p\ $ be an arbitrary prime. Then an integer $\ s\ $ is called $p$-simple $\ \Leftarrow:\Rightarrow\ s\ $ is not divisible by any prime $\ q<p.\ $

Could you prove my conjecture (or is it known one way or another?):

For every prime $\ p\ $ and for every every integer $\ n\ $ there exists a $p$-simple integer $\ s\ $ such that $\ n\le s < n+p$.

• There is a $p$-simple integer such that $\ n\le s<n+p\ $ for every $\ n\ $ such that $\ -2\cdot p < n \le p$.

• If the conjecture holds restricted to all positive integers $\ n\ $ then it holds in full, for all integers $\ n$.

• There is a $p$-simple integer $\ s\ $ such that $\ p+1 \le s\le 2\cdot p\ $ (Chebyshev--actually, $\ s\ $ can be a prime).

• The conjecture holds for $\ n=2\ $ (trivial); and for $\ n=3\ $ (at least one of the three consecutive integers is odd); and for $\ p=5\ $ (among any $5$ consecutive integers, at the most $3$ are even, at the most $2$ are divisible by $3$, whole one of the even ones is divisible by $3$ when there are there three even numbers--thus one of these consecutive $5$ integers is not divisible by $2$ nor by $3$).

Let $\ p\ $ be an arbitrary prime. Then an integer $\ s\ $ is called $p$-simple $\ \Leftarrow:\Rightarrow\ s\ $ is not divisible by any prime $\ q<p.\ $

Could you prove my conjecture (or is it known one way or another?):

For every integer $\ n\ $ there exists a $p$-simple integer $\ q\ $ such that $\ n\le s < n+p$.

• There is a $p$-simple integer such that $\ n\le s<n+p\ $ for every $\ n\ $ such that $\ -2\cdot p < n \le p$.

• If the conjecture holds restricted to all positive integers $\ n\ $ then it holds in full, for all integers $\ n$.

• There is a $p$-simple integer $\ s\ $ such that $\ p+1 \le s\le 2\cdot p\ $ (Chebyshev--actually, $\ s\ $ can be a prime).

• The conjecture holds for $\ n=2\ $ (trivial); and for $\ n=3\ $ (at least one of the three consecutive integers is odd); and for $\ p=5\ $ (among any $5$ consecutive integers, at the most $3$ are even, at the most $2$ are divisible by $3$, whole one of the even ones is divisible by $3$ when there are there three even numbers--thus one of these consecutive $5$ integers is not divisible by $2$ nor by $3$).

Let $\ p\ $ be an arbitrary prime. Then an integer $\ s\ $ is called $p$-simple $\ \Leftarrow:\Rightarrow\ s\ $ is not divisible by any prime $\ q<p.\ $

Could you prove my conjecture (or is it known one way or another?):

For every integer $\ n\ $ there exists a $p$-simple integer $\ s\ $ such that $\ n\le s < n+p$.

• There is a $p$-simple integer such that $\ n\le s<n+p\ $ for every $\ n\ $ such that $\ -2\cdot p < n \le p$.

• If the conjecture holds restricted to all positive integers $\ n\ $ then it holds in full, for all integers $\ n$.

• There is a $p$-simple integer $\ s\ $ such that $\ p+1 \le s\le 2\cdot p\ $ (Chebyshev--actually, $\ s\ $ can be a prime).

• The conjecture holds for $\ n=2\ $ (trivial); and for $\ n=3\ $ (at least one of the three consecutive integers is odd); and for $\ p=5\ $ (among any $5$ consecutive integers, at the most $3$ are even, at the most $2$ are divisible by $3$, whole one of the even ones is divisible by $3$ when there are there three even numbers--thus one of these consecutive $5$ integers is not divisible by $2$ nor by $3$).

Let $\ p\ $ be a prime. Then an integer $\ s\ $ is called $p$-simple $\ \Leftarrow:\Rightarrow\ s\ $ is not divisible by any prime $\ q<p.\ $

Could you prove my conjecture (or is it known one way or another?):

For every integer $\ n\ $ there exists a $p$-simple integer $\ q\ $ such that $\ n\le s < n+p$.

• There is a $p$-simple integer such that $\ n\le s<n+p\ $ for every $\ n\ $ such that $\ -2\cdot p < n \le p$.

• If the conjecture holds restricted to all positive integers $\ n\ $ then it holds in full, for all integers $\ n$.

• There is a $p$-simple integer $\ s\ $ such that $\ p+1 \le s\le 2\cdot p\ $ (Chebyshev--actually, $\ s\ $ can be a prime).

• The conjecture holds for $\ n=2\ $ (trivial); and for $\ n=3\ $ (at least one of the three consecutive integers is odd); and for $\ p=5\ $ (among any $5$ consecutive integers,at the most $3$ are even, at the most $2$ are divisible by $3$, whole one of the even ones is divisible by $3$ when there are there three even numbers--thus one of the $5$ is not divisible by $2$ nor by $3$).

Let $\ p\ $ be an arbitrary prime. Then an integer $\ s\ $ is called $p$-simple $\ \Leftarrow:\Rightarrow\ s\ $ is not divisible by any prime $\ q<p.\ $

Could you prove my conjecture (or is it known one way or another?):

For every integer $\ n\ $ there exists a $p$-simple integer $\ q\ $ such that $\ n\le s < n+p$.

• There is a $p$-simple integer such that $\ n\le s<n+p\ $ for every $\ n\ $ such that $\ -2\cdot p < n \le p$.

• If the conjecture holds restricted to all positive integers $\ n\ $ then it holds in full, for all integers $\ n$.

• There is a $p$-simple integer $\ s\ $ such that $\ p+1 \le s\le 2\cdot p\ $ (Chebyshev--actually, $\ s\ $ can be a prime).

• The conjecture holds for $\ n=2\ $ (trivial); and for $\ n=3\ $ (at least one of the three consecutive integers is odd); and for $\ p=5\ $ (among any $5$ consecutive integers, at the most $3$ are even, at the most $2$ are divisible by $3$, whole one of the even ones is divisible by $3$ when there are there three even numbers--thus one of these consecutive $5$ integers is not divisible by $2$ nor by $3$).