Judith Gatz on Tue, 15 Jan 2002 10:05:16 +0100

 easy precision problem ( I hope ;)

```Hi!

I am a real pari-gp-beginner, so I hope it´s easy for you:
I want to make a number of exact factorisations like:

? for(x=65,75,print(factor(truncate(precision(exp(x),1000)),0)))

the results cant be exact, there are too many 2^x:
[2, 1; 3, 2; 941605135783518730078768673, 1]
[2, 2; 19, 2; 43, 1; 1217, 1; 119723, 1; 5092511140946699, 1]
[2, 3; 401, 1; 1213, 1; 20929, 1; 1537753234015292017, 1]
[2, 3; 42553450624146756517211339837, 1]
[2, 4; 3, 1; 7, 1; 13, 1; 211853977234152646525695923, 1]
[2, 5; 5, 1; 7, 1; 19, 1; 23, 1; 433, 1; 677, 1; 17532252168972131581, 1]
[2, 9; 17, 1; 43, 1; 18269256662969027516768267, 1]
[2, 8; 5, 1; 43, 1; 239, 1; 7927, 1; 39181, 1; 4549272969690959, 1]
[2, 11; 3, 1; 4637, 1; 1773408868724932787900621, 1]
[2, 11; 67059715797860165419149830557, 1]
[2, 14; 7, 1; 181, 1; 17984136432019657849799891, 1]
^^^^

before that, I tried to increase the precision with:  (I don´t know if this
makes sence...)
? primelimit=1000000000000000000000000000000000000000000000000000000000000
%1 = 1000000000000000000000000000000000000000000000000000000000000
? parisize=100000000000000
%2 = 100000000000000
? realprecision= 500
%3 = 500

Can I do this in a better way?
I am mainly interested in the set of resulted primenumbers of the series
[exp(n)], n= 1,2,3,...

Thank You very much,
regards, Judith
```