!       ------------------    3N+1 Problem.    ---------------------
!        We have an integer number N>1 and we apply the following procedure:
!       If N is even we divide N with 2. If N is odd we multiply N with 3 and we add 1,
!       so we have 3N+1. In the new number in each case (if it's not 1) we make again the 
!       same. Until now, for every number humans have tested the final number in which
!       the procedure will terminate, is 1! This can't be proved until now. 

!        This program finds, how many times for a number N, we have to 
!       apply the procedure from which the 3N+1 problem is defined,
!       to find number 1. ++++++Example for number 5: We have 5->16->8->4->2->1.
!       So the procedure is being used 5 times. This program writes to a file (3N1_DATA)
!       the number 5 and the number of times the above procedure occurs, plus 1, that means
!       writes 5  6. And you choose the interval between two numbers. 
!       For example use A=1 , B=10000 and then import the file in Origin and make a plot.

        integer i,a,b,s,d
 1      continue
        print *,'Choose A>0:  (A is for the start-number)'
        read*,a
        if (a.lt.1) then
        goto 1
        end if
 2      continue
        print *,'Choose B>A:  (B is the end-number)'
        read*,b
        if (b.lt.a) then 
        goto 2
        end if
        open(1,file='3N1_DATA',status='unknown')
        
        do i=a,b
         s=0
         d=i
4        continue
          if (d.eq.1) then 
          goto 3
          end if
          s=s+1
          if (mod(d,2).eq.0) then
          d=d/2
          goto 4
          else 
          d=3*d+1
          goto 4
          end if
 3        continue
          write(1,*) i , s+1
        end do
        close=(1)
        end