Can a sequence like $\{n!\alpha\}$ or $\{(n!+1)\alpha\}$ (fractional parts), where $\alpha$ is irrational, be not uniformly distributed in $(0,1)$? I know about Weyl's result for polynomials $p(n)\cdot\alpha$. David
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Use latex. What do you get from the sequence $p_n/q_n$ in my comment (it exists for example thanks to continued fractions) and from Dirichlet's theorem in Diophantine approximation ? – reuns Oct 21 '17 at 02:47
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@reuns I don't quite follow: the coefficients of continued fractions are very sparse, so how can they tell you much about equidistribution which concerns statistics across all $n$? – Erick Wong Oct 21 '17 at 05:15
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Yes. The standard proof that $e$ is irrational rests on proving that $\{n!e\}$ lies strictly between $0$ and $1$, and in fact the estimate we get shows that it converges to $0$ (since it's strictly less than $\frac1n$), which makes it highly non-equidistributed.
Consequently, $\{(n!+1)e\} = \{n!e + e\}$ will converge to $\{e\} \approx 0.71828$ since $n!e$ is so close to an integer, so again it will be non-equidistributed.
Erick Wong
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@Dave I would point out that your question only asks whether it can be non-equidistributed, not how often it is :). – Erick Wong Oct 23 '17 at 19:32
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@Dave Well, it shouldn't be hard to show that set of values that are as well-approximable as $e$ in this sense is uncountable and dense in $\mathbb R$. But, I don't know whether it's first or second category. – Erick Wong Oct 23 '17 at 19:42
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@Dave See Gerry Myerson's answer here https://math.stackexchange.com/a/1944292/30402. I lack the context to be sure of what the quoted section is referring to, but it does seem that Weyl already considered this question and proved that the non-equidistributed set has measure zero, so it's not typical. – Erick Wong Oct 23 '17 at 19:54