2.71828. And then another 1828.

Question

This may qualify as the silliest math.SE question ever, but am I really the first person ever to worry about this? The decimal expansion of $e$ has a 2. And then a 7. And then a 1828. And...well, then another 1828. Bernoulli proved that $e$ was irrational, and I'm sure it was known even before Bernoulli that $e$ wasn't a repeating decimal consisting of 2.7182818281828... ad infinitum. But this really bothers me. I'm not sure it makes sense to ask for some intellectually stimulating reason why the 137th digit of $\pi$ is this, or why the 42nd digit of $e$ is that. Nor do I think that it makes sense to talk about these things in terms of probabilities. Obviously the first digit of $e$ is 2, with unit probability. But ... it kind of seems like the probability that the 7th through 10th digits of $e$ would be the same as the 3rd through 6th is kind of small, like maybe $10^{-4}$. Is there anything interesting to be said about this "coincidence?" Is there any sense in which we can say that there is not likely to be any interesting reason for this "coincidence?"

Answer 1

The sequence you're so worked up about is generated by its first twelve terms, which also adds yet another $28$ to it:

$$\displaystyle\sum_{n=0}^{12}\frac1{n!}=2.7~1828~1828~28~6\ldots\in\mathbb Q$$

So unless you think there's something spectacularly interesting about $\dfrac{260412269}{95800320}$, I honestly doubt that there's any deep and meaningful signification to it.

Answer 2

There is a somehow similar pattern in $\sqrt{2}$, which starts $1.41421...$

Write $\dfrac{\sqrt{2}}{10}$ and note that this is $7·0.020203$, or $7·(\dfrac{2}{100}+\dfrac{2}{100^2}+\dfrac{3}{100^3}+...)$

Although we may (also) attribute to base $10$ bias the perception of this pattern, there happens to be a series that explains it.

$$\sqrt{2}=\frac75\sum_{k=0}^\infty\binom{2k}{k}\frac1{200^k}$$

This was taken from an answer by robjohn.

I spent some time worried about $e$ being close to the eighth harmonic number, until the integral

$$\frac{1}{14} \int_0^1 x^2(1-x)^2(e^x-1-x)dx = e-\frac{761}{280}=e-H_8\approx 0$$

gave me some peace of mind.

Would you consider that a similar specimen evaluating to $e-\dfrac{271801}{99990}$ explains your observation?

This fraction can be written with smaller numbers as $$\dfrac{271801}{99990}=\frac{1}{10}\left(3^3+\frac{4!-1}{10^2-1}-\frac{3!-1}{10^2+1}\right)$$

although this can be misleading.

Answer 3

"Google at one time posed a "challenge" problem whose solution amounted to recognizing it was about finding the next place in the decimal expansion of e where a group of four digits repeats. – hardmath Apr 22 '14 at 11:08"

"Google at one time posed a "challenge" problem whose solution amounted to recognizing it was about finding the next place in the decimal expansion of e where a group of four digits repeats. – hardmath Apr 22 '14 at 11:08"

 I found another 4 digits that repeat and they're none other than 99999999!  They might not be the next 4 digits in e to repeat.  

 I copy and pasted 2 million+ digits of e from this webpage into a .rtf file to save. https://apod.nasa.gov/htmltest/gifcity/e.2mil

 I then went searching for 00000000, 11111111, 22222222 ..... up to 99999999 and believe it or not it found 99999999 one time.

 I then copied and pasted all the digits down to the 9s and pasted them in a .rtf file and the size was 414 KB.  So, the 9s are around 414*1024 = 423,936 digits in.  

 There might be more - 4 number repeats - although the odds are 1 in 10^8 or 100 million for 8 stated consecutive numbers to repeat.  :-)

Answer 4

Mostly a coincidence, but not entirely.

Fun fact #1. Some of the convergents of the continued fraction of $e$ are given by the integrals $$ I(n)=\int_{0}^{1} P_n(2x-1)e^{-x}\,dx. $$ For instance, as mentioned in the comments, $I(4)=1001-\frac{2721}{e}$ leads to $e\approx \frac{2721}{1001}$. The approximation is quite good since by the properties of Legendre polynomials (Rodrigues' formula) we have $|I(n)|\leq \frac{1}{n! 4^n}$. The recurrence relation of the $\{I(n)\}_{n\geq 0}$ sequence then gives the continued fraction of $e$ $$ e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,\ldots] $$ clearly showing that $e\not\in\mathbb{Q}$, since all the elements of $\mathbb{Q}^+$ have finite continued fractions.

Fun fact #2.

$$ 2.7\overline{1828} = \frac{271801}{99990} = [2;1,2,1,1,4,1,1,6,1,1,8,1,1,\color{red}{5}] $$ since the continued fraction of $e$ and the continued fraction of $2.7\overline{1828}$ are the same for the first $14$ terms, the decimal representation of $e$ has to contain at least two copies of $1828$.

Fun fact #3. 1828 is also the year of birth of Green's functions.