deletions | additions       diff --git a/9090+3.tex b/9090+3.tex  index a578ad7..bea5b7b 100644  --- a/9090+3.tex  +++ b/9090+3.tex 
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\end{array}  \end{equation}  Where [[]] signifies a non-prime, calculated factor, in a bid to reduce the undeterminaed undetermined  representations. We see that if $n+0=m\in\mathbb{Z^0}$, $n+(0,1,2,3...\%1)=1m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is div by $1$. [0th primeish?] \\ We see that if $n+5=11m\in\mathbb{Z^0}$, $n+(5,16,27,38...\%11)=11m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is div by $11$. [5th prime] \\ We see that if $n+1=9m\in\mathbb{Z^0}$, $n+(1,10,19,28,37,...\%9)=9m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is div by $19$. [8th prime] \\ We see that if $n+2=11m\in\mathbb{Z^0}$, $n+(2,13,24,35,46,...\%11)=11m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is div by $23$. [9th prime] \\ We see that if $n+5=14m\in\mathbb{Z^0}$, $n+(5=14m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is div by $29$. [10th prime] \\ We see that if $n+5=21m\in\mathbb{Z^0}$, $n+(5=21m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is div by $43$. [14th prime] \\ We see that if $n+28=30m\in\mathbb{Z^0}$, $n+(28=30m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is div by $61$. [18th prime] \\ We see that if $n+13=33m\in\mathbb{Z^0}$, $n+(13=33m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is div by $67$. [19th prime] \\ We see that if $n+6=13m\in\mathbb{Z^0}$, $n+(6=13m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is div by $79$. [22nd prime] \\ We see that if $n+21=22m\in\mathbb{Z^0}$, $n+(21=22m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is div by $89$. [24th prime] \\ \\  Consider grand-rule:\\ 
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primes of form $(90;;n)-1$: 1, 3, 4, 11, 15, 21, 36, 74, 81, \\  Twin primes of form $(90;;n)\pm1$ 3, 15, ...  Note the second sequence contains $1^2,2^2,6^2,9^2$  $11+5=4^2$  $15+10=5^2$  $21+15=6^2$  $74-25=7^2$  $74-10=8^2$...  Means almost nothing.  Now the second sequence doesn't appear to be recoginsed by OEIS. We require a way of finding the two sequences, then we may identify very large double primes.  However this being said $((90;;76)-1)/43 \nin \mathbb{Z}$ according to WA. But all other cases have been... apparently has remainder 13, Does this frown upon one of the ruels above?? [Check it out.] 
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(p_H;;3)|q_H & 0 & & 101 & 0\\  (p_H;;4)|q_H & 0 & & 129 & 0 \\  (p_H;;5)|q_H & 0 & 43×7611282473×9578180596676119497063479901434206663440373037131421633305032513428028044431106610444794385191741080502880035410580923888423332883194147047017319 & 157 & 0 \\  (p_H;;6)|q_H & 0  & & 185 & 1 0  \\ \hline  \end{array}  \end{equation}  Extrapolating forward $(90;;152)-1$, will be divisible by $19,23,61,67$.