Let $s$ be the solution of the equation $e^{-x^2}=x$
The first $1000$ digits are :
0.6529186404192047155350807673531963699201168811029977306249214940750472761980389255118225716068055968688851940429563511542639972035162540546184014314736515563495845288693035814478587853842857780658616279193692084130958642023333796276411296868531689474262892215608745009598152369180178149105626587066739951550269510748069902717760287395134889049475223218060931472513069981211016428190656317364055693355450003146948306013507970744265217412215016132587649430680044056787884029408663952102966652334903872704616645841575454513552731162180548149393628681582632188137740104386997009737873013156541916137229839360201360375345885525247598732173530922269150166716076862085406497386415145782103562063958906901301976160463902960472417318997321462573730838234493446339156402727582278943274237402310315758136645964565044814291326069247530654509787921402283476101814488620715302088429768179782396908060337455967125160567056011574933240619841127046321975145916237832717973201545089797553631095887493551466234611735605
Let $z_k=floor(s\times 10^k)$
For which $k$ is $z_k$ prime ? Due to my calculation ,$z_k$ is prime for $k=4,11,27$ and for no other $k\le 10000$. Please check this.
I am also interested in the factorization of
$z_{153}=$
65291864041920471553508076735319636992011688110299773062492149407504
72761980389255118225716068055968688851940429563511542639972035162540
54618401431473651
I only found the prime factor $P24=$
972147411124463059337939
- What is the next prime $z_k$ ?
- What is the complete factorization of $z_{153}$ ?