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Just saw this post, and realized that

1/9801 = 0.00(010203040506070809101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979900)(repeat)

Similar properties are also exhibited by numbers 998001, 99980001, .. and so on.

It is not very obvious to me why this happens. Is there some simple explanation to this property?

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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 $$\frac{1}{999}=0.001001\overline{001}.$$

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    $\begingroup$ Is there a related formula that wouldn't result in the '98' element being skipped? $\endgroup$ – Dan Neely Jan 26 '12 at 20:16
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    $\begingroup$ @Dan: The simplest case of $0.\overline{1234567890}$ equals $\frac{1234567890}{9999999999}$ which in lowest terms equals $\frac{137174210}{1111111111}$. So my guess would be no. $\endgroup$ – Dejan Govc Jan 26 '12 at 20:42
  • $\begingroup$ How do you get $$\left(\sum_{n=1}^\infty x^n\right)^2= x\sum_{n=1}^\infty nx^n.$$ $\endgroup$ – user4951 Jan 27 '12 at 8:22
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    $\begingroup$ @Jim multiply the series term-by-term and gather terms of like degree: $$(x+x^2+x^3+\ldots)\cdot(x+x^2+x^3+\ldots)=\left[ (x\cdot x) + (x\cdot x^2 + x^2 \cdot x) + (x\cdot x^3+x^2\cdot x^2 + x^3\cdot x) + \ldots \right]$$ $$= x \left[ 1x^1 + 2x^2 + 3x^3 + \ldots \right]$$ $\endgroup$ – Philip Jan 27 '12 at 17:59
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    $\begingroup$ oh we got binomial when se do (x+y)^n never mind. Even with limited series the pattern emerge. $\endgroup$ – user4951 Jan 29 '12 at 9:25
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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.

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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)).
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$1/9801$ 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.

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  • $\begingroup$ this isn't an answer because it doesn't explain why it happens $\endgroup$ – phuclv Apr 2 '16 at 16:26
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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.

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This number “998001” is very interesting but it has 2997 repeating digits and “99980001” has 39996 repeating digits they are too large. The three numbers above and the next next one all have a factor of 3^4 which you will see next in my two examples also have powere of 3’s. The number 81 is 3^4. 1/ 81 = 0.012345679012345679} { 18 repeating digits in brackets} 1/ 8181 = 0.000122234445666788901112333455567779} { 36 repeating digits in brackets} 1/ 818181 = 0.000001222223444445666667888890111112333334555556777779} { 54 repeating digits in brackets} The number 243 is 3^5. 1/243=0.004115226337448559670781893} { 27 repeating digits in brackets} 1/243243=0.000004111115222226333337444448555559666670777781888893} { 54 repeating digits in brackets} 1/243243243=0.000000004111111115222222226333333337444444448555555559666666670777777781888888893} { 81 repeating digits in brackets} Ok the number 729 is look a lot like your number 998001 not as systematic as these two numbers. Of all the other poweres of 3’s I did not find any other numbers that worked like these two. If you would like to see other examples go to “engert.us/erwin/miscellaneous/Four%20families%20of%20numbers.pdf”.

Erwin

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  • $\begingroup$ Welcome to Math SE! Your answer is pretty hard to read because of how cramped everything is. Take a look at MathJax for more information about formatting/typesetting. $\endgroup$ – DMcMor Apr 5 '17 at 16:38

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