PARI/GP for High Pecision Arithmetic Calculation

Question

I recently became aware that PARI/GP may be useful for high-precision calculations. Has anyone had experience with this software; if so, I would be grateful if you might provide me a couple of simple examples of its code so that I may get started. I have no colleages who can help me. I would like to practice at first some simple things like computing $log_{2}k$ (large $k$) accurate to many $k$ digits and then use this in further computation. Also, can PARI/GP convert the calculation to binary with the same accuracy as it computes, say $log_{2}n$? Any guidance is appreciated.

Answer 1

Real logarithm can calculate any CAS. But pari/gp can smallest discrete logarithm by modulo. Example:

? n=2^77+7
%1 = 151115727451828646838279
?
? m=Mod(2,n)
%2 = Mod(2, 151115727451828646838279)
?
? h=znorder(m)
%3 = 96869056058864517204
?
? z=znlog(89,m,h)
%4 = 52540840786720769887
?
? m^z
%5 = Mod(89, 151115727451828646838279)
?
? m^(z+h)
%6 = Mod(89, 151115727451828646838279)
?
? m^(z+2*h)
%7 = Mod(89, 151115727451828646838279)
?
? m^(z+3*h)
%8 = Mod(89, 151115727451828646838279)
?
? m^(z+33*h)
%9 = Mod(89, 151115727451828646838279)
?
? m^(z+100*h)
%10 = Mod(89, 151115727451828646838279)

Convert to binary:

? n=2^77+7
%11 = 151115727451828646838279
?
? binary(n)
%12 = [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1]

Setup 4th bit (5th, if counting from zero-bit):

? n2=bitor(n,2^4)
%13 = 151115727451828646838295
?
? binary(n2)
%14 = [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1]

Setdown second bit:

? n3=bitnegimply(n2,2^2)
%15 = 151115727451828646838291
?
? binary(n3)
%16 = [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1]

Count of bits:

? vecsum(binary(n3))
%17 = 4