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Let's consider a sequence of natural numbers $a_n$, represented in binary, with the following properties:

  • $\forall n \in \mathbb{N}$ the number $a_n$ is represented with $n$ binary digits
  • $\forall n \in \mathbb{N}$ the first $n$ digits (counting from the first to the right) of $a_{n+1}$ are the same as those of $a_n$

We will say that such a sequence $a_n$ is "thin" if, by defining $\alpha_n$ as the number of "ones" in the binary representation of $a_n$, then: $$\lim_{n\rightarrow \infty} \frac{\alpha_n}{n}=0$$ Now, let's define $(a\cdot b)_n$ as the number with binary representation corresponding to that of the first n digits of the binary representation of $a_n\cdot b_n$, (I have already proven this is a valid definition). Is it true that, if $a_n$ and $b_n$ are thin sequences, then so is $(a\cdot b)_n$?

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    You mean multiplication by $a \cdot b$, correct?2011-10-26
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    yes: where i say $a_n \cdot b_n$ I mean the ordinary multiplication between natural numbers.2011-10-26
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    By "first $n$ digits", you mean "most significant $n$ digits"? I.e., $a_{n+1} = 2a_{n} + \sigma_{n}$, where $\sigma_{n}\in\{0,1\}$?2011-10-26
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    Your sequences (as well as the product) are [2-adic numbers](http://en.wikipedia.org/wiki/P-adic_number). It does not feel intuitively likely that multiplication would preserve "thinness". Do you have any particular reason to expect that it would?2011-10-26
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    @ mjqxxxx By "the first $n$ digits" I mean the less significant ones. I.e. $a_{n+1}=a_n+\sigma_n\cdot 2^n$ where $\sigma_n \in \{0,1\}$2011-10-26
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    @HenningMakholm if both $a_n$ and $b_n$ are convergent, so is $(a \cdot b )_n$. So I thought the notion of thinnes could be a useful generalization, since a convergent sequence of this kind is always "thin".2011-10-26
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    @Lorenzo, what does "convergent" mean for you here? Your second condition guarantees that $a_n$ and $b_n$ will converge in the 2-adic metric whether or not it they are thin.2011-10-26
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    ... sorry, I just meant $a_n$ seen as a sequence of natural numbers was convergent ... i.e. $a_n=l$ from a certain $n_0$2011-10-26
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    "$\forall n \in \mathbb{N}$ the number $a_n$ is represented with $n$ binary digits": You mean "at most $n$ binary digits", right?2011-10-26
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    @TonyK yes, I mean "at most $n$"2011-10-27

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