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.

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