Interesting series

First, we have that

$\frac { 1 }{ 9 } =0,111111...\\ \frac { 1 }{ 99 } =0,010101...\\ \frac { 1 }{ 999 } =0,001001...\\ \frac { 1 }{ 9999 } =0,00010001...\\ .\\ ..\\ ...\\$

So clearly the digit in the place will be equal to the number of factors of that number.

Now to calculate what is the number of factors of $2014$ , before we need the prime representation $2014=2x19x53$ .

So we use the function $d(2014)=d(2\cdot 19\cdot 53)=8$ .

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$Note:$

If we have a number $n={ { a }_{ 1 } }^{ { k }_{ 1 } }{ { a }_{ 2 } }^{ { k }_{ 2 } }{ { a }_{ 3 } }^{ { k }_{ 3 } }...{ { a }_{ n } }^{ { n }_{ n } }$ , the number of factor of n is $({ k }_{ 1 }+1)({ k }_{ 2 }+1)({ k }_{ 3 }+1)...({ k }_{ n }+1)$ .