Conjecture: There are infinitely many $N$ such that $0\equiv N\pmod2$, $1\equiv N\pmod3$, and $0\leq N\pmod P\leq P-4$ for primes $P$ with $5\leq PHere's a conjecture,
There are infinitely many numbers $N$, such that for all prime numbers $P<N$
$0 \equiv N \pmod 2$ and
$1 \equiv N \pmod 3$ and
$0 \leq N \pmod P \leq (P-4)$, for $P \geq 5$

For eg: 10
$10$ mod $2$ $\equiv 0$
$10$ mod $3$ $\equiv 1$
$10$ mod $5$ $\equiv 0$
$10$ mod $7$ $\equiv 3$

Is this a known conjecture? Does it have a proof?
If not, is there a way to prove or disprove this?
 A: I changed the question slightly to an equivalent question: Are there infinitely many N = 4 modulo 6, so that N+1, N+2 and N+3 are not divisibly by any prime 5 <= p < N?
N+2 must be of the form $2^n 3^m$: If N+2 = 6k is not of this form, then k has a prime factor p < N. So we restrict N to the form $N = 2^n 3^m - 2$, and since N+2 has no prime factor ≥ 5, the question is: Does N+1 or N+3 have such a prime factor?
N+1 = 5 modulo 6, and N+3 = 7 modulo 6. N+1 and N+3 are both greater than N, so the question is: Is either N+1 or N+3 composite? So solutions are $N = 2^n 3^m - 2$ where N+1 and N+3 are twin primes.
For random x, the chance that (x, x+2) are twin primes is about $0.66 / log^2 x$. But with x = 5 (modulo 6), neither is divisible by 2 or 3, where for the other cases at least one is divisible by 2 or 3, so the chances are 6 times hight, about $4 / log^2 x$. There are about $n / log_2(3)$ candidates $2^{n-1} < N < 2^n$, each is a solution with probability $4 / n^2$, so I expect about $4 / log_2(3) / n$ solutions between $2^{n-1} < N < 2^n$. That's the harmonic series, so about $4 log n / log_2(3)$ solutions N < 2^n. This would be a very, very slowly growing function.
So this heuristic seems to show a very, very slowly growing function going towards infinity. But since its growing so slowly, any heuristics that reduces the number just a tiny bit may be enough to make this false.
To find examples for N: Generate numbers $N = 2^n 3^m - 2$, n, m ≥ 1, then check whether N+1 or N+2 have small prime factors, and if need perform a non-deterministic primality test for N+1 and N+3.
