Benedict Irwin edited Primality.tex  over 9 years ago

Commit id: a4fcf4b61b696c361d6afb35bb2e67441bb9df8e

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1;20 & 0 & 11×41×101×271×3541×9091×27961 \\  1;21 & 0 & 3×37×43×239×1933×4649×10838689 \\  1;22 & 0 & 11^2×23×4093×8779×21649×513239 \\  1;23 & 1 & 1;23 \\  1;24 & 0 & 3×7×11×13×37×73×101×137×9901×99990001 \\  1;25 & 0 & 41×271×21401×25601×182521213001 \\  1;26 & 0 & 11×53×79×859×265371653×1058313049 \\  1;27 & 0 & 3^3×37×757×333667×440334654777631 \\  1;28 & 0 & 11×29×101×239×281×4649×909091×121499449 \\  1;29 & 0 & 3191×16763×43037×62003×77843839397 \\  1;30 & 0 & 3×7×11×13×31×37×41×211×241×271×2161×9091×2906161 \\  \end{array}  \end{equation}  We can observe a potential pattern that evey other $1;2n\in\mathbb{Z}$ string of ones is divisible by $11$, also $1;3n\in\mathbb{Z}$ is divisible by $37$... We also note that if this trend continues there should be only be distinct primes of the form $1;p$, where $p$ must be prime, but not any prime will do. [2,19,...] [2,19,23,...]  special primes. Due to the nature of the occurence of divisors, the number of prime factors on $1;n$, should be directly proportional to the number of divisors of $n$!  Let us tabulate \begin{equation} 

\end{array}  \end{equation}  This is curious as the prime is not in ascending order. There are multiples, which together would make some other factor. We have for $3$ and $9$ $9$, [thus $27$ ...]  an interesting relationship where the number is as divisible by $3$ as the integer prefactor $m$ is...