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Consider some even integer number $n$.

Let: $$C = \{c_i|i - n \mod i\}_{i=2..n}$$

For example for $n = 50$:

$c_2 = 2 - 50 \mod 2 = 2$

$c_3 = 3 - 50 \mod 3 = 1$

$c_4 = 4 - 50 \mod 4 = 2$

$c_5 = 5 - 50 \mod 5 = 5$

...

Let $C_{\text{odd}} = \{c \mid c \in C,\text{c is odd}\}$ and $C_{\text{even}} = \{c\mid c \in C,\text{c is even}\}$

Question: Is it true that number $p = n + x$ where:

  • $x$ - minimal odd integer number that is not contained in the set $C_{odd}$
  • $n+x < n^2$

is minimum prime number greater than $n$?

Numerical results shows that it is true up to $10^6$.

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  • $\begingroup$ Defining $n$ mod $i$ as the least non-negative remainder when $n$ is divided by $i$ means $n+c_i$ is the least multiple of $i$ greater than $n$, so if $x$ is contained in $C$ then $n+x$ is not prime, but the converse is not true as you showed with $n=50$, $x=15$ $\endgroup$ – J. W. Tanner Feb 20 at 15:18
  • $\begingroup$ Why do you separate $C$ into odd and even subsets? $\endgroup$ – J. W. Tanner Feb 20 at 15:19
  • $\begingroup$ I tried to give an estimate of the gap between consequitive primes as an estimate of cardinality of sets $C_{odd}$ or a $C_{even}$. In proving this, I made the same mistake as you and I realized that it is too difficult for me. I decided to ask a question, suddenly this is some kind of well-known problem, just in a different formulation. $\endgroup$ – Dmitry Pyatin Feb 20 at 16:56
  • $\begingroup$ You have written $c_2 = 2 - 50 \pmod 2 = 2$ (or it can be $ = 0$) $c_5 = 5 - 50 \pmod 5 = 5$ (or it can be $ = 0$) If you follow the general rule for the remainder (less than the divisor), the resulting value of $C_{odd}$ and $C_{even}$ will change. I am just curious whether the result still holds or not? $\endgroup$ – SARTHAK GUPTA Jun 20 at 5:39

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