Fermat 28 Output

277 days ago by b0yscout

num = 28 ind_count =[25,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1] count = 0 fermat = (2^(2^num))+1 # With apologies, file names are alphabetized, but output is RULE BASED as denoted # at end. # From right to left, hack off 10m digit parts of the number, and save them to files in decreasing order. # Not sure why python doesn't use 25, but the intermediate files number 24, and # decrease by one from right to left. # Hack off LAST 10m n = fermat%10^10000000 fermat = fermat - n fermat = fermat / 10^10000000 o = open('f_dat_end','w') o.write(str(n)) o.close() # Hack off 10m digits at a time, and creatively name data files in descending order # until less than 10m digits remain for initial portion. while (fermat > 10^10000000): n=fermat%10^10000000 count = count + 1 n_count = ind_count[count] chomp = str(n_count) nomdeplume = ('f_dat' + chomp) o = open(nomdeplume,'w') o.write(str(n)) o.close() fermat = fermat - n fermat = fermat / 10^10000000 # put the final(LEFTMOST) digits in a "first part" file. o = open ('first_part', 'w') o.write(str(fermat)) o.close() print print "done" # To concatenate, use datafiles increasing/ascending Left to Right, starting with "first_part," # then f_dat low to f_dat high, and ending with "f_dat_end." 
       
done
done
f_dat18   f_dat19   f_dat20   f_dat21   f_dat22   f_dat23   f_dat24   f_dat_end   first_part