Consider the following polynomials:
$$
f\_1(n\_1, m\_1) = 30n\_1m\_1 + 23n\_1 + 7m\_1 + 5\\
f\_2(n\_2, m\_2) = 30n\_2m\_2 + 17n\_2 + 13m\_2 + 7\\
f\_3(n\_3, m\_3) = 30n\_3m\_3 + 23n\_3 + 11m\_3 + 8\\
f\_4(n\_4, m\_4) = 30n\_4m\_4 + 11n\_4 + 29m\_4 + 11\\
f\_5(n\_5, m\_5) = 30n\_5m\_5 + 29n\_5 + 17m\_5 + 16\\
f\_6(n\_6, m\_6) = 30n\_6m\_6 + 19n\_6 + 7m\_6 + 4\\
f\_7(n\_7, m\_7) = 30n\_7m\_7 + 31n\_7 + 13m\_7 + 13\\
$$
where each $n\_1, m\_1,..., n\_7, m\_7 \in \mathbb{N}$


How can I prove that $\left\vert{\:\mathbb{N} \setminus (f\_1 \cup f\_2\: \cup\: ... \cup \:f\_7)\:}\right\vert = \infty$?



Are there infinitely many natural numbers not covered by one of these 7 polynomials?
Notice that
$$30\cdot f\_1(n\_1,m\_1) = (30\cdot m\_1+23)\cdot (30\cdot n\_1+7) - 11\\
30\cdot f\_2(n\_2,m\_2) = (30\cdot m\_2+17)\cdot (30\cdot n\_2+13) - 11\\
30\cdot f\_3(n\_3,m\_3) = (30\cdot m\_3+23)\cdot (30\cdot n\_3+11) - 13\\
30\cdot f\_4(n\_4,m\_4) = (30\cdot m\_4+11)\cdot (30\cdot n\_4+29) + 11\\
30\cdot f\_5(n\_5,m\_5) = (30\cdot m\_5+29)\cdot (30\cdot n\_5+17) - 13\\
30\cdot f\_6(n\_6,m\_6) = (30\cdot m\_6+19)\cdot (30\cdot n\_6+7) - 13\\
30\cdot f\_7(n\_7,m\_7) = (30\cdot m\_7+31)\cdot (30\cdot n\_7+13) - 13
$$


These representations imply that the polynomials cannot represent integers $k>0$ such that $30k \pm 11$ and $30k + 13$ are primes. There are likely infinitely many such prime constellations. See <http://mathworld.wolfram.com/PrimeConstellation.html> 


P.S. The polynomial $f\_4(n\_4,m\_4)$ is the only one, whose representation has $+11$ as a free term. If it were $-11$ or $-13$ as in the other representations, then we would need existence of infinitely many twin primes of the form $(30k+11,30k+13)$.


