==> arithmetic/digits/prime/101.s <==
Note that the sequence
101 , 10101, 1010101, ....
can be viewed as 
100**1 +1, 100**2 + 100**1 + 1, 100**3 + 100**2 + 100**1 +1 ....
that is, 
the k-th term in the sequence is 
100**k + 100**(k-1) + 100**(k-2) + ...+ 100**(1) + 1
= (100)**(k+1) - 1
  ----------------
    11 * 9
= (10)**(2k+2) - 1
  ----------------
    11 * 9
= ((10)**(k+1) - 1)*((10)**(k+1) +1)
   ---------------------------------
       11*9
thus either 11 and 9 divide the numerator. Either they both divide the
same factor in the numerator or different factors in the numerator. In
any case, after dividing, they leave the numerators as a product of two
integers.  Only in the case of k = 1, one of the integers is 1. Thus
there is exactly one prime in the above sequence: 101.