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

Commit id: 424f9d63ceb98f029a8579f383a277605cd2d747

deletions | additions      

       

(90;;47)-1& 0 & \\  (90;;48)-1& 0 & \\  (90;;49)-1& 0 & \\  (90;;50)-1& 0 & 11×...  \\ (90;;51)-1& 0 & \\  (90;;52)-1& 0 & 6983×187171×1076213×64629155095902444263963806439089201090498271510681620879393026187936269908369318103338921 \\  (90;;53)-1& 0 & 19×23×67×310492472110013692718020051603848864684275729734932576559681310526626281945794224219033809525288059391 \\ 

(90;;58)-1& 0 & \\  (90;;59)-1& 0 & \\  (90;;60)-1& 0 & \\  (90;;61)-1& 0 & \\ 11×...\\  (90;;62)-1& 0 & \\  (90;;63)-1& 0 & \\  (90;;64)-1& 0 & \\ 

(90;;69)-1& 0 & \\  (90;;70)-1& 0 & \\  (90;;71)-1& 0 & 19×86399748109×5537850630280835793385158924798585959711630523655733039869444968470991361168447402811525949960092140267365793582245494756146065959 \\  (90;;72)-1& 0 & \\ 11×...\\  (90;;73)-1& 0 & \\  (90;;74)-1& 1 & (90;;74)-1\\  (90;;75)-1& 0 & \\ 

\end{array}  \end{equation}  We see that if $n+0=m\in\mathbb{Z^0}$, then $(90;;n)-1$ is divisible div  by $1$. [0th primeish?]\\ primeish?] \\  We see that if $n+5=11m\in\mathbb{Z^0}$, then $(90;;n)-1$ is divisible div  by $11$. [5th prime]\\ prime] \\  We see that if $n-8=9m\in\mathbb{Z^0}$, $n+1=9m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is divisible div  by $19$. [8th prime] \\ Pot_{lly} We see that  if $n-9=11m\in\mathbb{Z^0}$, $n+2=11m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is divisible div  by $23$. [9th prime] \\ We see that if $n-9=14m\in\mathbb{Z^0}$, $n+5=14m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is divisible div  by $29$. [10th prime] \\ Pot_{lly} We see that  if $n+14=30m\in\mathbb{Z^0}$, then $(90;;n)-1$ is divisible div  by $43$. [14th prime]\\  Pot_{lly} prime] \\  We see that  if $n-2=30m\in\mathbb{Z^0}$, $n+28=30m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is divisible div  by $61$. [18th prime] \\ Pot_{lly} We see that  if $n-20=33m\in\mathbb{Z^0}$, $n+13=33m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is divisible div  by $67$. [19th prime] \\ Pot_{lly} We see that  if $n-7=13m\in\mathbb{Z^0}$, $n+6=13m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is divisible div  by $79$. [22nd prime] \\ We see that if $n-1=22m\in\mathbb{Z^0}$, $n+21=22m\in\mathbb{Z^0}$,  then $(90;;n)-1$ is divisible div  by $89$. [24th prime] \\Try n-10 = 7m  9,23,37 10th prime 29...  7,20, 22nd prime 79...  n-7=13m  COULD TABULATE/CHART N+... AGAINST Xm\inZ...  primes of form $(90;;n)+1$: 2, 3, 9, 15, 26, 33, 146, 320, 1068, 1505 \\  primes of form $(90;;n)-1$: 1, 3, 4, 11, 15, 21, 36, 74 \\  Twin primes of form $(90;;n)\pm1$ 3, 15, ...