Fortran code to compute prime numbers.

prime_numbers.f95

!****************************************************
! 素数一覧
! : 入力値以下の素数を全て出力する
!
! date          name            version
! 2017.04.20    mk-mode.com     1.00 新規作成
!
! Copyright(C) 2017 mk-mode.com All Rights Reserved.
!****************************************************
!
program prime_numbers
  implicit none
  integer i, j, n
  logical is_prime
  print *, "自然数を入力してください N:"
  read *, n
  do i = 2, n
    is_prime = .true.
    do j = 2, int(sqrt(dble(i)))
      if (mod(i,j) == 0) then
        is_prime = .false.     ! 割り切れるので素数ではない
        exit
      end if
    end do
    if (is_prime) print *, i   ! もし素数ならば出力
  end do
end program

This is awesome! Thank you!

What change do we have to make in the code if we want to distinguish prime numbers from the numbers we enter?

program prime number
implicit none
integer::n,s,i
read*,n
s=0
do i=1,(n/2)+1
if (mod(n,i)==0)then
s=s+i
end if
end do
if (s==n)then
print*,"yes"
else
print*,"no"
end if
end program prime number

Thanks for your comment.
I'm sorry.
I'm not sure what you want to do with this code.
(Is this the code to determine if it is a prime number?)

yes

If you ask me what you should do,
I'm very sorry, but I can only say that you should refer to this 'prime_numbers.f95' code.

It could be a little better/more efficient/faster. The program doesn't have to check all numbers from 2 to n , only the odd ones.

                ...
                ! *** New lines block. ***
		If (n .GE. 2) Then 
			Write(*,*) 2                          ! Two is a prime.
		Else
			Return
		End If
                ! *** End of new lines block. ***

		Do i = 3, n, 2                                !<--- Change here.
			is_prime = .TRUE.
			Do j = 3, Int(Sqrt(Dble(i))), 2       !<--- Change here.
                        ...