# 2020-09-09 Math 3600

## 621 days ago by calkin

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.

2*(3*68+6*22)
 672 672
99999-9999
 90000 90000
672/90000.
 0.00746666666666667 0.00746666666666667
0.00746666666666667^2
 0.0000557511111111112 0.0000557511111111112
90000*0.00746666666666667^2
 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}