Just a curiosity of mine.
If I define the $n^{th}$ concatenation number (denoted $Q_n$) to be the the concatenation of digits from the $1^{st}$ square number to the $n^{th}$, can $Q_n$ ever be square itself apart from $n=1$?
So for instance, $Q_1=1$, $Q_2=14$, $Q_3=149$, $Q_4=14916$. I have tested up to $Q_{14}$ and they were all not square numbers.
The only thing I can think of is that we could express $Q_n$ as a summation:
$$Q_n=\sum_{i=1}^{i=n}i^2\times10^{\epsilon_{i.n}}$$ where $\epsilon_{i,n}$ depends on $i$ and $n$ and they are various integers.
How could I proceed? Thanks in advance!