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\section{(90;;n)+3}  Perhaps if one categorises all of the sequences 9090+1, 9090+3, 9090+7, 9090+9, there is added value. These likely already exist as OEIS numbers for primality. However it is tables of the divisors that reveal the realtionship.  \begin{equation}  \begin{array}{|c|c|c|}  \hline  (90;;n)+3 & p? & rep \\  \hline  (90;;1)+3 & 0 & 3×31 \\  (90;;2)+3 & 0 & 3×7×433 \\  (90;;3)+3 & 0 & 3×19×41×389 \\  (90;;4)+3 & 0 & 3×11×61×45161 \\  (90;;5)+3 & 0 & 3×7^2×61842919 \\  (90;;6)+3 & 0 & 3×17×103×173061281 \\  (90;;7)+3 & 0 & 3×30303030303031 \\  (90;;8)+3 & 0 & 3×7×41×10558547143913 \\  (90;;9)+3 & 0 & 3×409×740905386382159 \\  \hline  \end{array}  \end{equation}  These appear always divisible by 3. We may have unearted a form, $3030... +1$. As expected.  The nest thing to check would be all numbers $1$ through $9$ for a similar +1 form. If such numbers are always divisible by $3$ thenn we know if any prime of the form $(90;;n)+1$, were a double prime, it must be the upper of the two with a pair of the form $(90;;n)-2$.  \begin{equation}  \begin{array}{|c|c|c|}  \hline  (90;;n)+7 & p? & rep \\  \hline  (90;;1)+7 & 1 & 97 \\  (90;;2)+7 & 0 & 11×827 \\  (90;;3)+7 & 0 & 7^2×18553 \\  (90;;4)+7 & 0 & 151×602047 \\  (90;;5)+7 & 0 & 229×39698293 \\  (90;;6)+7 & 0 & 7×569×228242759 \\  (90;;7)+7 & 0 & 1607×56570685071 \\  (90;;8)+7 & 0 & 9282839×979324223 \\  (90;;9)+7 & 0 & 7×129870129870129871 \\  (90;;10)+7& 0 & 5449×44953×363563×1020827 \\  (90;;11)+7& 0 & 71×643×2232323×89203307863 \\  (90;;12)+7& 0 & 7×353×33961×10833133612779287 \\  (90;;13)+7& 0 & 11×34693×1080818041×220404363479 \\  (90;;14)+7& 0 & 3761×2417152111382369292499577 \\   (90;;15)+7& 0 & 7×2777×172507×271098224007250850789 \\  \hline  \end{array}  \end{equation}  Apart fromt the first no early primes in this sequence. Oscillating factors of $7$ are visible.  \begin{equation}  \begin{array}{|c|c|c|}  \hline  (90;;n)+3 & p? & rep \\  \hline  (90;;1)+3 & 0 & 3×31 \\  (90;;2)+3 & 0 & 3×7×433 \\  (90;;3)+3 & 0 & 3×19×41×389 \\  (90;;4)+3 & 0 & 3×11×61×45161 \\  (90;;5)+3 & 0 & 3×7^2×61842919 \\  (90;;6)+3 & 0 & 3×17×103×173061281 \\  (90;;7)+3 & 0 & 3×30303030303031 \\  (90;;8)+3 & 0 & 3×7×41×10558547143913 \\  (90;;9)+3 & 0 & 3×409×740905386382159 \\  \hline  \end{array}  \end{equation}  These appear always divisible by 3. We may have unearted a form, $3030... +1$. As expected.  The nest thing to check would be all numbers $1$ through $9$ for a similar +1 form. If such numbers are always divisible by $3$ thenn we know if any prime of the form $(90;;n)+1$, were a double prime, it must be the upper of the two with a pair of the form $(90;;n)-2$.  By the digit sum we probably know the next sequence $9090+9...$ cannot be prime.  \begin{equation}  \begin{array}{|c|c|c|}  \hline  (90;;n)+9 & p? & rep \\  \hline  (90;;1)+9 & 0 & 3^2×11 \\  (90;;2)+9 & 0 & 3^3×337 \\  (90;;3)+9 & 0 & 3^2×83×1217 \\  (90;;4)+9 & 0 & 3^2×541×18671 \\  (90;;5)+9 & 0 & 3^3×113×2979649 \\  (90;;6)+9 & 0 & 3^2×101010101011 \\  (90;;7)+9 & 0 & 3^2×10101010101011 \\  (90;;8)+9 & 0 & 3^4×47×223×431×4079×6091 \\  (90;;9)+9 & 0 & 3^2×43×49669×47294532133 \\  \hline  \end{array}  \end{equation}  Will all be divisible by $9$.