2020-09-09 Math 3600

440 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}