a_1 = 101=10^{2\times1}+10^0=10^{2\times1}+1

a_2 =10101= 10^{2\times2}+ 10^{2\times1}+1

a_3 = 1010101=10^{2\times3}+ 10^{2\times2}+ 10^{2\times1}+1

\dots

a_n = \underbrace{10101\dots 01}_{\text{n 01}}= 10^{2n}+10^{2n-2}+\dots+10^2+1

If n =1, a_1 = 101, a_1 is a prime number.

If n>1

Using the Sum formula for a geometric sequence

S_n=\dfrac{a_1(1-r^n)}{1-r}

If n = 2k+1 (k\geq 1)

10^{n+1}-1 = 10^{2k+2}-1=99\times(10^k+10^{k-1}+10+1) = 99\dfrac{10^{k+1}-1}{9}

a_n = (10^{n+1}+1) \dfrac{10^{k+1}-1}{9}

If n = 2k (k\geq 1)

10^{n+1}-1 = 10^{2k+1}-1 = 9 (10^{2k}+10^{2k-1}+\dots+10+1) = 9\dfrac{10^{2k+1}-1}{9}

10^{n+1}-1 is divsible by 9 when n is an even number.

10^{n+1}+1 = 10^{2k+1}+1, which is divisible by 11 according to the rule of divisibility of 11.

The difference of the sum of all digits on even places of the number and the sum of all digits on odd places is zero or divisible by 11

a_n is a composite number for both even and odd numbers for n

In summary, there's only 1 prime number in the sequence