An
emirp (prime spelled backwards) is a prime number that results in a different prime when its decimal digits are reversed.
This definition excludes the related palindromic primes.
13, 17, 31, 37, 71... are
emirps.
List values of q(n) (the number of emirps of length n) for n=2 to 6.
digits 2 3 4 5 6
emips 4 14 102 703 4769
pals 1 15 0 93 0
The code here is far cruder than Charlie's. But, this does make it a bit easier to follow. :-)
program pp
implicit none
integer i,j,flag1,flag2,e2,e3,e4,e5,e6,
1 p1,p2,p3,p4,p5,p6
data e2,e3,e4,e5,e6,p2,p3,p4,p5,p6/10*0/
do i=3,999999,2
call isprime(i,flag1)
if(flag1.eq.1)then
call rev(i,j)
call isprime(j,flag2)
if(flag2.eq.1.and.i.ne.j)then
if(i.ge.10.and.i.le.99)e2=e2+1
if(i.ge.100.and.i.le.999)e3=e3+1
if(i.ge.1000.and.i.le.9999)e4=e4+1
if(i.ge.10000.and.i.le.99999)e5=e5+1
if(i.ge.100000.and.i.le.999999)e6=e6+1
endif
if(flag2.eq.1.and.i.eq.j)then
if(i.ge.10.and.i.le.99)p2=p2+1
if(i.ge.100.and.i.le.999)p3=p3+1
if(i.ge.1000.and.i.le.9999)p4=p4+1
if(i.ge.10000.and.i.le.99999)p5=p5+1
if(i.ge.100000.and.i.le.999999)p6=p6+1
endif
endif
enddo
print*,'emips ',e2/2,e3/2,e4/2,e5/2,e6/2
print*,'pals ',p2,p3,p4,p5,p6
end
subroutine rev(i,j)
implicit none
integer i,j,k,i1,i2,i3,i4,i5,i6
k=i
i1=k/10**5
k=k-i1*10**5
i2=k/10**4
k=k-i2*10**4
i3=k/10**3
k=k-i3*10**3
i4=k/10**2
k=k-i4*10**2
i5=k/10
i6=k-i5*10
j=((((i6*10+i5)*10+i4)*10+i3)*10+i2)*10+i1
do 1 k=1,5
1 if((j/10)*10.eq.j)j=j/10
return
end
subroutine isprime(i,n)
implicit none
integer i,j,k,l,m,n
n=0
k=sqrt(1.*i)
do j=2,k
m=(1.*i)/(1.*j)
l=m*j
if(l.eq.i)go to 1
enddo
n=1
1 return
end
Edited on October 12, 2018, 10:02 pm
different language, same answer
