Does there exist prime number of the form $1010101010101....$ after the trivial one $101$? I have checked numbers of the form $101010101...$ up to $1(01)_{2500}$ and the only prime I found is 101.  I found that numbers of such form are quite rich in number of distinct prime factors. And $1(01)_{18}$ is the only semiprime I found So far(!!). Are there anymore primes of such form ?
 A: No, there is no other such prime.  The number of this form with $n$ 1's is equal to $a_n = \frac{10^{2n}-1}{99}$.  Now:
If $n$ is even, then $a_n$ is divisible by 101, since $10^{2n} = 100^n \equiv (-1)^n = 1 \pmod{101}$ and 99 is relatively prime to 101.  So if $n>2$ also, then $a_n$ is composite.
If $n$ is odd, then $a_n$ is divisible by $\frac{10^n-1}{9}$: the quotient is $\frac{10^n+1}{11}$ and $10^n \equiv (-1)^n = -1 \pmod{11}$.  If $n>1$, then $\frac{10^n-1}{9} > 1$ and $\frac{10^n+1}{11} > 1$ so $a_n$ is composite; if $n=1$, then $a_n=1$ which is not prime.
A: The number $1(01)_n$ is $(100^{n+1}-1)/(100-1)$ by thinking of the number as a geometric series.  The numerator factors as $(10^{n+1}+1)(10^{n+1}-1)$.  Once $n$ is larger than $1$, one of these two factors will have factors outside of $99$.  The $n=1$ case is $101$, which is prime.
If this were base-2, your observations stem from this being nearly a product of a Fermat and a Mersenne number, so perhaps this is the base-10 version of those numbers.
A: The number is of the final form $(10^{2n} -1)/99$. Where $n \geq 2$.
Lets assume that to be a prime $p$, $p\geq 1$. Then we have $99p = (10^{2n} -1) = (10^n-1)\times (10^n+1)$ with $n\geq 2$.
Now we observe that in R.H.S. we have 2 numbers separated by 2.
$(10^n+1)-(10^n-1) = 2$
Hence, we can factorize L.H.S. = $99p = 3^2\times 11\times p$ into 2 factors $a\times b$ where $(a-b)=\pm 2$.
All possible pairs (without the obvious symmetric pairs) are listed as follows.
$(99,p),(11,9p),(9,11p)$ and $(33,3p),(99p,1)(33p,3)$.
$33p-3 =\pm2 $ and $(33-3p) =\pm2$ can be discarded as $ 3\not| 2$.
$(99p-1)=2$ gives $99p =3$ not possible for integer $p$.
$99-p = \pm2$ gives $p=101$ or 97. $p=101$ $\implies$ n=2 which gives the prime 101.
$p=97$ which is clearly not possible as the number is '1 $mod(100)$' (by definition).
Now we are left with other 2 choices.
$(11-9p)= \pm2$ means $9p= 13 $ or $9$. So only possibility is the trivial $p=1$. We discard $9p =13$  as not possible.
Similarly,$(9-11p) =\pm2$ gives $11p = 11$ or $7$. $7$ can be discarded as $11 \not| $  7. For $11p=11$ we get trivial answer p=1.
Finally, 101 is the only prime of this form.
