What is special about the numbers 9801, 998001, 99980001 ..?

Solution 1:

There is actually a straightforward reason. As $99$ is $1$ less than $100$, we get a fairly simple expression for its decimal $$\frac{1}{99}=0.01010101010101\overline{01}\dots$$ Now, $$\frac{1}{9801}=\left(\frac{1}{99}\right)^2,$$ and the decimal expansion follows from the formula for general power series $$\left(\sum_{n=1}^\infty x^n\right)^2= x\sum_{n=1}^\infty nx^n.$$

Letting $x=\frac{1}{100}$, the decimal expansion for $\frac{1}{99}$ given above is exactly the same thing as writing $\frac{1}{99}=\sum_{n=1}^\infty x^n$. Applying our identity, the $x$ in front accounts for the double zero. Once $n$ is around $99$ we expect to miss a number because we are forcing things to be in decimal, and there will be carrying, which is why the number 98 is missed.

A similar pattern will occur for $\frac{1}{998001}=\left(\frac{1}{999}\right)^2,$ since as before $$\frac{1}{999}=0.001001\overline{001}.$$

Solution 2:

Since this was recently brought back to the front page, I wanted to point out a particular numberphile blog video on this exact question.

In this video, Brady Haran interviews Dr. James Grime and they discuss why this happens and how to generalize the result.

If I were to briefly summarize the key idea: in the video, they show that this boils down to considering $\dfrac{12345679}{999999999} = \dfrac{1}{81}$, or to evaluating $\dfrac{1}{(1-x)^2}$ at $x = \frac{1}{10}$ and considering a sort of generating functions.

Solution 3:

It happens in every base that $1/(n-1)^2$ is equal to the series of numbers in increasing order, dropping just the number n-2

  1/5²  = 0.0 1 2 3 5 0 1 2 3 5 0 ...    base 6
 1/15²  = 0.0 1 2 3 4 5 6 7 8 9 A B C D F 0 1 2  ... base 16

The examples quoted are for bases 10, 100, 1000 &c.

Similar series occur with eg these. These are calculated in base 1000, with leading zeros in each place suppressed. It's true for all bases and their powers.

 1/(n-1)³ =   0. 0 0  1 3 6 10,15 21 28,36 45 55,66 78 91,105 ... (trianguar numbers)
 1/(n²-n-1) = 0. 0 1 1 2 3 5 8 13 21 34 55 89 144 &c.  
 1/(n²-2n-1) = 0.0 1 2,5 12 29,70 169 408,987   (approxmates to sqrt(2)).

Solution 4:

$1/9801$ and the other numbers are all in the form of $$\frac{1}{999...9^2}.$$ You can see similar results for things like $1/9.9$ squared (only you get fewer zeros after the decimal). You can also try things like $1/9.999$ squared ($1/(9.999^2)$) and get more zeros between the magic numbers.

Now try $1/9800$ or $1/998000$ and see the magic number sequences that you get. :) I'm still not understanding that one.

Solution 5:

I was about to say: Let's not forget that 9801 is a fundamental solution for x in $x^2-29y^2=1$

When I saw that there is an error in Wolframalpha's treatment of this equation. While 9801 which the program offers, is indeed a solution for x, is it not the fundamental solution (which is x=70 and y=13). This solution combined with itself Brahmagupta style gives x=9801 and y=1920

That is a special cirumstance in itself, I guess.