What is the longest possible length $k$ of a prime $k$-tuplet (smallest groupings of $k$ primes) such that each prime $p$ in the $k$-tuplet,$\gcd(p-1, 420) > 2$. In other words, each prime $p$ in this $k$-tuplet is either $1$ $\pmod 3$, $1$ $\pmod 4$, $1$ $\pmod 5$, or $1$ $\pmod 7$. Also, given the longest possible following these properties, what is the smallest prime $p$ starting this chain? Thanks for help and explanation.
For prime twins primes $p$, $p+2$, such that ${p^2+2*p}
$p$ $=$ $1$ $\pmod {4*5*7}$, $p+2$ $=$ $1$ $\pmod 3$
$p$ $=$ $1$ $\pmod 4$, $p+2$ $=$ $1$ $\pmod {3*5*7}$
$p$ $=$ $1$ $\pmod 5$, $p+2$ $=$ $1$ $\pmod {3*4*7}$
$p$ $=$ $1$ $\pmod 7$, $p+2$ $=$ $1$ $\pmod {3*4*5}$
$p$ $=$ $1$ $\pmod {5*7}$, $p+2$ $=$ $1$ $\pmod {3*4}$
$p$ $=$ $1$ $\pmod {4*7}$, $p+2$ $=$ $1$ $\pmod {3*5}$
$p$ $=$ $1$ $\pmod {4*5}$, $p+2$ $=$ $1$ $\pmod {3*7}$
The smallest twin primes yielding these properties are:
$281, 283$
$1049, 1051$
$1931, 1933$
$659, 661$
$71, 73$
$29, 31$
$41, 43$
Would The longest possible length mentioned $k$-tuplet start on one of these forms?