Benedict Irwin edited 9090+3.tex  over 9 years ago

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Will all be divisible by $9$.  For the next strings take one from the appended 90's. We have the J-sentence \\  (-.' '=x)#x=:":>: ($$#89),88 \\  \\  Where,$$ corresponds to the number of $90$'s to prepend before the $89$. I.e $=4\to9090909089$. \\  \begin{equation}  \begin{array}{|c|c|c|} 

(90;;3)-1 & 1 & (90;;3)-1 \\  (90;;4)-1 & 1 & (90;;4)-1 \\  (90;;5)-1 & 0 & 2549×3566461 \\  (90;;6)-1 & 0 & 11^2×7513148009 11^2×7513148009==11×[[82644628099]]  \\ (90;;7)-1 & 0 & 79×203279×5660929 \\  (90;;8)-1 & 0 & 19×6679×71637804989 \\   (90;;9)-1 & 0 & 23×29×743×1834394194069 \\ 

Where [[]] signifies a non-prime, calculated factor, in a bid to reduce the undetermined representations.  We see that if $n+(0,1,2,3...\%1)=1m\in\mathbb{Z^0}$, $n+0=1m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is div by $1$. [0th primeish?] \\ We see that if $n+(5,16,27,38...\%11)=11m\in\mathbb{Z^0}$, $n+5=11m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is div by $11$. [5th prime] \\ We see that if $n+(1,10,19,28,37,...\%9)=9m\in\mathbb{Z^0}$, $n+1=9m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is div by $19$. [8th prime] \\ We see that if $n+(2,13,24,35,46,...\%11)=11m\in\mathbb{Z^0}$, $n+2=11m\in\mathbb{Z^0}$,  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:\\  If $n+5=11m\in\mathbb{Z^0}$, then $(90;;n)-1$ is div by $11$. [5th prime] \\  If $n+5=14m\in\mathbb{Z^0}$, then $(90;;n)-1$ is div by $29$. [10th prime] \\  If $n+5=21m\in\mathbb{Z^0}$, then $(90;;n)-1$ is div by $43$. [14th prime] \\  If $n+5=32m$ div $59$,$61$,$67$... ish $97$.  COULD TABULATE/CHART N+... AGAINST Xm\in Z... 

\end{array}  \end{equation}  Extrapolating forward $(90;;152)-1$, will be divisible by $19,23,61,67$. \\  Extrapolating forward $(90;;163)-1$, will be divisible by $23,29,43,79$. \\  Extrapolating forward $(90;;449)-1$, will be divisible by $19,23,67,79$. \\  Many others inbetween are divisible by $3$ known factors.  8264462809917355371900  8264462809917355371900  8264462809917355371900  8264462809917355371900  8264462809917355371900  8264462809917355371900  8264462809917355371900  $p_H$=8264462809917355371900  $q_H$=82644628099