Hi,

Welcome to my little math blog.

This is where I will dump thoughts on recreational mathematics, fun puzzles, and other stuff every so often.

Thanks for stopping by! Hopefully you find something to enjoy here.

Hi,

Welcome to my little math blog.

This is where I will dump thoughts on recreational mathematics, fun puzzles, and other stuff every so often.

Thanks for stopping by! Hopefully you find something to enjoy here.

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(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: \[ \sum_{n=0}^{\infty} r^n = \frac{1}{1 - r}. \] If we differentiate both sides with respect to \( r\), we get \[ \sum_{n=1}^{\infty} nr^{n-1} = \frac{1}{(1 - r)^2}, \] and multiplying by \( r \) gives \[ \sum_{n=1}^{\infty} nr^n = \frac{r}{(1 - r)^2}. \] (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

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(This is a cleaned-up and somewhat expanded version of this Twitter thread .) What follows is a silly little proof that \[ \zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \] 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, \[ \begin{align*} I &= -\int_0^1 \! \frac{\log(1 - x + x^2)}{x} \, dx - \int_0^1 \! \frac{\log(1 - x + x^2)}{1 - x} \, dx \\ &= -2\int_0^1 \! \frac{\log(1 - x + x^2)}{x} & (x \mapsto 1 - x ) \\ &= 2\left( \int_0^1 \! \frac{\log(1 + x)}{x} \, dx - \int_0^1 \! \frac{\log(1 + x^3)}{x} \, dx \right) \\ &= \frac{4}{3}\int_0^1 \frac{\log(1 + x)}{x} \, dx & (x \mapsto x^{1/3}). \end{align*} \] To evaluate this integral, we take the Maclaurin series: \[ \int_0^1 \! \frac{\log(1 + x)}{x} \, dx = \int_0^1 \! \sum_{n=1}^{\infty} \frac{(-1)^nx^{n-1}}{n} \, dx \] Since for

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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 \[ \begin{align*} y^2 &= x^3 + (x + 1)^3 + (x + 2)^3 + (x + 3)^3 \\ &= 4x^3 + 18x^2 + 42x + 36. \end{align*} \] 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

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