This is my first post here... My algorithm is about the 3x+1 problem. I want to test if some types of numbers have a stoping time. Until now i managed to do it by just testing one type of number each time... So i would like to automate the proccess... Here is an example: > k := 16 n + 3 > while modp(k, 2) = 0 or modp(k, 2) = 1 do if modp(k, 2) = 1 then k:= (3n+1)/2 elif modp(k,2)=0 then k:=k/2 end do k Well this example will produce k= 9n+2, which is smaller than the original 16n+3 so I know that numbers of the form 16n+3 have a finite stoping time...So it passes my test. If for example i start with k:=16n+7 i will get k=27n+3 which is bigger than 16n+7 so i do not know if this number has a stoping time... so it ''fails'' my test. I hope this is clear to you.