One-Thirteenth

(This post is a cleaned up and expanded version of this thread .) A cool fact I've seen shared around the internet   a few times : The decimal expansion of $1/998001$ starts with $$\frac{1}{998001}=0.000001002003\dots 996997999\dots$$ That is, it begins with three-digit strings from $000$ to $999$, in order, except that it skips $998$ for some reason. The first thing to observe is that $998001={999}^2$. Recall the formula for the infinite geometric series: $$\sum _{n=0}^{\mathrm{\infty }}{r}^n=\frac{1}{1-r}.$$ If we differentiate both sides with respect to $r$, we get $$\sum _{n=1}^{\mathrm{\infty }}n{r}^{n-1}=\frac{1}{(1-r{)}^2},$$ and multiplying by $r$ gives $$\sum _{n=1}^{\mathrm{\infty }}n{r}^n=\frac{r}{(1-r{)}^2}.$$ (This can also be obtained by some series manipulations.) Now, take $r=0.001$. We have $$\frac{0.001}{(1-0.001{)}^2}=\frac{1000}{998001}=0.001+0.000002+0.000000003+\dots$$ From here, the appearance of the numbers from $001$ to \( 997...

The OEIS article on the number $100$  opens with an interesting factoid:  "$100$ is the square of $10$, and the smallest square that is the sum of four [positive] consecutive cubes: $1^3+2^3+3^3+4^3=100$." In fact, it is the only one. To see this, let's look at the equation $$\begin{array}{rl}{y}^2& =x^3+(x+1{)}^3+(x+2{)}^3+(x+3{)}^3\\ & =4x^3+18x^2+42x+36.\end{array}$$ Let $X=4x+6,Y=4y$; the above equation then reduces to $$Y^2=X^3+60X.$$ Note that a positive integer solution in $(x,y)$ will give a positive integer solution in $(X,Y)$, though the converse is not true. Now, generally speaking, whenever one sees an equation of the form $Y^2=X^3+aX+b$, one has an elliptic curve . Well, we have the extra condition $4a^3+27b^2\ne 0$ to get rid of problematic cases like $y^2=x^3$, which are referred to as singular curves ; we'll see the logic of this later on. I will not go too...

(This is a cleaned-up and somewhat expanded version of this Twitter thread .) What follows is a silly little proof that $$\zeta (2)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{6}$$ where $\zeta$ is the Riemann zeta function. Consider the integral $$I:={\int }_0^1\frac{\log (1-x+x^2)}{x(x-1)}dx.$$ We have, by using partial fractions and performing some other algebraic manipulations, $$\begin{array}{rl}I& =-{\int }_0^1\frac{\log (1-x+x^2)}{x}dx-{\int }_0^1\frac{\log (1-x+x^2)}{1-x}dx\\ & =-2{\int }_0^1\frac{\log (1-x+x^2)}{x}& (x\mapsto 1-x)\\ & =2({\int }_0^1\frac{\log (1+x)}{x}dx-{\int }_0^1\frac{\log (1+x^3)}{x}dx)\\ & =\frac{4}{3}{\int }_0^1\frac{\log (1+x)}{x}dx& (x\mapsto x^{1/3}).\end{array}$$ To evaluate this integral, we take the Maclaurin series: $${\int }_0^1\frac{\log (1+x)}{x}dx={\int }_0^1\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n{x}^{n-1}}{n}dx$$ ...

(This is a cleaned-up, somewhat revised/expanded version of my Twitter thread here .) Among math enthusiasts, the number $142857$ is pretty cool. Move its leftmost digit to the right, and you get $428571$, which is three times the original: $428571=142857\times 3$. Do this again, and you get $285714$, which is two times the original: $285714=142857\times 2$. We can keep doing this until we return to $142857$, as follows: $$\begin{array}{rlrl}142857& =142857\times 1& 142857\times 1& =142857\\ 428571& =142857\times 3& 142857\times 2& =285714\\ 285714& =142857\times 2& 142857\times 3& =428571\\ 857142& =142857\times 6& 142857\times 4& =571428\\ 571428& =142857\times 4& 142857\times 5& =714285\\ 714285& =142857\times 5& 142857\times 6& =857142\end{array}$$ Numbers that give you consecutive...