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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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- All are divible by 3,9. - They are exactly 99 less than the next rounded,I dont know what we call it. - Add 9 to each and they become divisible by 10 But is there a way to get this series? – user23738 Jan 26 '12 at 19:30
1 – wim Jan 27 '12 at 5:17
@Martin: That link appears to be dead. – The Chaz 2.0 Apr 8 '12 at 16:48
up vote 79 down vote accepted

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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Is there a related formula that wouldn't result in the '98' element being skipped? – Dan Neely Jan 26 '12 at 20:16
@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. – Dejan Govc Jan 26 '12 at 20:42
How do you get $$\left(\sum_{n=1}^\infty x^n\right)^2= x\sum_{n=1}^\infty nx^n.$$ – Jim Thio Jan 27 '12 at 8:22
@Jim multiply the series term-by-term and gather terms of like degree: $$(x+x^2+x^3+\ldots)\cdot(x+x^2+x^3+\ldots)=\left[ (x\cdot x) + (x\cdot x^2 + x^2 \cdot x) + (x\cdot x^3+x^2\cdot x^2 + x^3\cdot x) + \ldots \right]$$ $$= x \left[ 1x^1 + 2x^2 + 3x^3 + \ldots \right]$$ – Philip Jan 27 '12 at 17:59
Ah I see. Put them in a matrix and see the length of diagonal. I was expecting some form of binomial result though. Why we don't get binomials? – Jim Thio Jan 29 '12 at 9:24

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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this isn't an answer because it doesn't explain why it happens – Lưu Vĩnh Phúc Apr 2 at 16:26

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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