Benedict Irwin edited Primality.tex  over 9 years ago

Commit id: bd3ccb3da607b70eaeaf41941af2cacb0aeaa020

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1;mn & \mathrm{div} \; \mathrm{by} \\  \hline  1;n\in\mathbb{Z} & 1 & 1\\  1;2n\in\mathbb{Z} & 11 & 11\\ 1;2\\  1;3n\in\mathbb{Z} & 3×37 & 111\\ 1;3\\  1;4n\in\mathbb{Z} & 101 & 101\\  1;5n\in\mathbb{Z} & 41×271 & 1;5\\  1;6n\in\mathbb{Z} & 13 & 1;6/8547 \\ 

We note that $n\in\mathbb{P}$ that are not $2$ appear to have two, larger primes introduced as divisible factors. (sometimes more? $13$. [semiprime?].  It is noted that $\forany \; p\in\mathbb{P} \; n\in\mathbb{Z}:$ $1;pn$ is divisible by $1;p$.  $10$,$12$,$14$, may or may not have some curious relationship as thier numbers are $9091,9901,909091$...$99990001$, Think [False concatenation of $90$ -> $(90;;3)+1$], [think added permutations], does this apply to $333667$ as $(3;3|6;3)+1$. This sounds ridiculous which is why it's exciting.  Addition to this idea: Try "Any number $n$ whose $1;n$ spare representation is of the form (9,0,+1), multiplied to another $m$ whose $1;m$ spare representation is of the form $1;a$, gives that $1;nm$ has spare represntation the form (9,0,+1).