It's more common than you think

There are two infinite sequences of numbers $\{a_n\}$ and $\{b_n\}$ where $a_k^2\equiv a_k$ (mod $10^k$ ) and $b_k^2\equiv b_k$ (mod $10^k$ ). For every value of $n\geq1$ , the satisfy the following equation:

$\displaystyle \sum_{k=1}^{n} 10^{k-1}(a_k+b_k)=10^k+1$

By analyzing this equation we see that $a_k+b_k=9$ for $n>1$ . If $a_k$ or $b_k$ equal $0$ , then $M_n$ will only have one solution, and therefore will not satisfy they equation $L_k=10^k+1$ . We need to find an algorithm that will find count the number of times that $a_k$ or $b_k$ equals $0$ . There is a fairly simple method to find subsequent values of $a_k$ where $a_1=5$ . Using $a_k+b_k=9$ , we must now count when $a_k\in\{0,9\}$ .

We define $A_n$ as the following, relate it to our sequence ${a_n}$ , and use the following equations and congruences to build our algorithm.

$A_n=\displaystyle \sum_{k=1}^{n} a_k10^{k-1}$

$(A_n^2-A_n)10^{-k}\equiv a_{n+1} \text{ (mod }10)$

$A_{n+1}=a_{n+1}10^n+A_n$

Using these formulas I was able determine each value of $k$ such that $a_k\in\{0,9\}$ . Here's an example of how it works:

$a_1=5$ , $a_2=2$ , $A_2=5(10^0)+2(10^1)=25$

$\frac{25^2-25}{10^2} = 6\equiv 6 \text{ (mod }10) \Longrightarrow a_3=6$

$A_3=6(10^2)+25=625$

Save the value of $A_n$ and check values of $a_{n+1}$ at every iteration.

Here is the list I came up with using this algorithm:

$4,5,12,13,20,24,25,29,32,33,34,35,38,40$

Notice that $k=1$ is also a non-solution as $M_n = {1,5,6} \Longrightarrow L_k=1+5+6=12>10^1+1$

Assuming that Pi Han Goh doesn't count $k=1$ as a solution, the $14$ th value of $k$ in this list is $\boxed{40}$

All I can say is

this

and

this

and for $a(n) = 0, n+1$ is your $k$ . Count till $14$ .

Still I would love if anybody gives a number theoretical approach. How did mathematicians came up with that table and that for others, there will only $2$ automorphic numbers and that too adding to ${10}^{k} +1$