Concantenations of non-zero squares in two (or more) different ways.
164 is concatenation of 16 and 4, and of 1 and 64.
1441 is concatenation of 144 and 1, and of 1 and 441.
Are there more examples?
How many numbers with exactly $N$ digits are the concatenation of two squares? Suppose the number is $g(N)$.
If $f(k)$ is the number of squares with exactly $k$ digits, then the answer is
\[ g(N)= \sum_{k=1}^{N-1} f(k)f(N-k) \]
Clearly
\[ f(1)=3\]
\[ f(2) = 9-3=6\]
\[ f(3) = 31- 9 =22\]
\[ f(4) = 99-31 = 68 \]
We can now count the number of concatenations of squares having total length 5:
\[ f(1)f(4) + f(2)f(3) + f(3)f(2) + f(4)f(1) = 672 \]
We see that the probability that a random 5 digit integer chosen uniformly will be a concatenation of two squares is about 0.75 $\%$
The probability that two integers chosen independently and uniformly will be concatenations of two squares is about $(.0075)^2=0.000056$, that is $.0056\%$
This suggests that the probability that a number is a concatention of squares in two different ways ought to be about the same.
672 672 |
90000 90000 |
0.00746666666666667 0.00746666666666667 |
0.0000557511111111112 0.0000557511111111112 |
5.01760000000000 5.01760000000000 |
Project: Let $g(N)$ be the number of concatenations of squares of total length $N$.
\begin{enumerate}
\item Compute $f(k)$ for $k< 20$.
\item Compute $g(N)$ for $N\leq 20$.
\item Compute the estimate for the probability that an $N$ digit number is a concatenation of two squares in two different ways for $N\leq 20$.
\item Estimate the number of $N$ digit numbers that are concatenations of two squares in two different ways for $N\leq 20$.
\end{enumerate}
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