A square number that is a concatenation of at least 25 squares

Question

A pen and paper challenge:
Write down a number whose square (repetition allowed) is the concatenation of at least 25 nonzero squares.

examples
7, 13, 12: Their squares are the concatenation of 2 squares. (4,9 and 16,9 and 1,4,4 resp.)
10: Its square is the concatenation of 3 squares (1,0,0) but only 1 nonzero square (i.e. no concatenation at all) for the purpose of this question! The two 0s do not count.

Answer 1

Here is a simple solution

4020400000201020000040204² = 16163616161616361616363681363616163616161616361616

which is the concatenation of

81, 8 x 36 and 16 x 16

and entirely built on

212² = 44944

Answer 2

Consider the sequence $a_n = \prod_{k=0}^{n-1}(10^{3^k}+2)$. The sequence begins like this:
$$a_0 = 1, a_1 = 12, a_2 = 12024, a_3 = 12024000024048, \ldots$$

By induction, we observe the following: $a_n^2$ has exactly $3^n$ digits, the first of which is $1$, so $4a_n^2$ also has exactly $3^n$ digits.

Since $a_n^2 = \prod_{k=0}^{n-1}(100^{3^k}+4\cdot 10^{3^k}+4) = (100^{3^{n-1}}+4\cdot 10^{3^{n-1}}+4)a_{n-1}^2$, this implies that $a_n^2$ is formed by concatenating $a_{n-1}^2$, $4a_{n-1}^2$, and $4a_{n-1}^2$ without zeros in between. This means that $a_n^2$ is always the concatenation of two more squares than $a_{n-1}^2$, and since $a_0^2 = 1$ is the concatenation of one square, the number $a_{12}^2$ is the concatenation of $25$ squares.

Solution:

$a_{12} = \prod_{k=0}^{11}(10^{3^k}+2)$ satisfies the requirements. Its square begins as follows:
$a_{12}^2 = \color{red}{1}4\color{red}{4}576\color{red}{576}578306304\color{red}{578306304}578306306313225218313225216\ldots$

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Better solution:

Consider the sequence $b_n = 12\cdot \prod_{k=0}^{n-1}(10^{6 \cdot 5^k}+10^{3 \cdot 5^k}/2 + 1)$. With the same reasoning as before, you see that $b_n$ has $3 \cdot 5^n$ digits, but now each $b_n^2$ is a concatenation of $b_{n-1}^2, b_{n-1}^2, \frac 94 b_{n-1}^2, b_{n-1}^2, b_{n-1}^2$, so it is made of four times as many squares as $b_{n-1}^2$, plus one. So if $b_0^2$ is made of three squares ($1, 4, 4$), $b_1^2$ is made of $13$ squares and $b_2^2$ is made of $53$ squares.

Solution: $b_2 = 12006012000000006003006000000012006012$ and $b_2^2 = 144144\color{red}{324}144144144144\color{red}{324}144144\color{red}{324324729324324}$ $144144\color{red}{324}144144144144\color{red}{324}144144$. Observe that we can coincidentally separate $324324729324324$ into five squares so we can actually separate this square into $57$ squares.