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Benedict Irwin edited 9090+3.tex
over 9 years ago
Commit id: b332367ebe96e8b3ba1178ae63b373586e3d57e2
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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