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Hi I'm stuck at yet another question.

$a_{0}=5$. Given $a_{n+1}a_{n} = a_{n}^{2} + 1$ for all $n \ge 0$, determine $\left \lfloor{a_{1000}}\right \rfloor$.

So I got $a_{n+1}=a_{n} + \frac{1}{a_{n}}$ and then:

$a_{1000}=a_{0}+ \frac{1}{a_{0}} + \frac{1}{a_{1}} + \frac{1}{a_{2}} + ... + \frac{1}{a_{999}}$

and I even tried writing the relation as $a_{n}^{2} - a_{n+1}a_{n} + 1 = 0$ but I'm still not getting anywhere. Can someone just tell me how to deal with this recurrence relation, to start off? Thanks.

Vladislav Kharlamov
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shgdh fgxbcv
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  • I'm not sure what the correct approach is here. However, after writing to program to brute force it, I have found that $a_{1000} = 45.0023$, and this is the first element of the (increasing) sequence to come out over $45$. – Ben Grossmann May 31 '18 at 13:26
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    Well yes the answer is 45 – shgdh fgxbcv May 31 '18 at 13:27
  • It seemed from your question like you didn't have the answer – Ben Grossmann May 31 '18 at 13:28
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    Oh I do but just don't know how the method – shgdh fgxbcv May 31 '18 at 13:29
  • @Omnomnomnom I think the number you mentioned is $a_{999}$, but the answer is nevertheless $45$. – Peter May 31 '18 at 13:29
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    @peter you're right, thanks – Ben Grossmann May 31 '18 at 13:30
  • Look at $a_n^2$ – Empy2 May 31 '18 at 13:34
  • Let $a_0=x$ to make it simpler to write. Let's look at the expansions of $a_1, a_2, a_3, a_4$ in terms of $x$:

    $$a_1 = \dfrac{x^2+1}{x}$$

    $$a_2 = \dfrac{x^4+3x^2+1}{x(x^2+1)}$$

    $$a_3 = \dfrac{x^8+7x^6+13x^4+7x^2+1}{x(x^2+1)(x^4+3x^2+1)}$$

    $$a_4 = \dfrac{x^{16}+15x^{14}+83x^{12}+220x^{10}+303x^8+220x^6+83x^4+15x^2+1}{x(x^2+1)(x^4+3x^2+1)(x^8+7x^6+13x^4+7x^2+1)}$$

    Does anyone recognize those coefficients from some pattern that might be easier to figure out?

     – SlipEternal May 31 '18 at 13:49
  • @Michael haha sorry I'm still not very sure. Would u mind dropping another hint. Thanks – shgdh fgxbcv May 31 '18 at 14:09

1 Answers1

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Squaring the given relation $a_{n+1} = a_{n} + \frac{1}{a_n}$, we get $$ a_{n+1}^2 = a_n^2 + \frac{1}{a_n^2} + 2 \Rightarrow a_{n+1}^2 > a_n^2 +2 $$ and iterating we get (for $n = 999$ we get the right side) $$ a_{n+1}^2 > a_0^2 + 2n + 2 \Rightarrow a_{1000} > 2025 = 45^2 $$

Also, using the terms ($\frac{1}{a_i^2} $ terms) we neglected for relation above, and the fact that $a_{n+1}^2 > 2n + 27 > 2n + 2$ or just $ a_n^2 > 2n$, $$ a_{n+1}^2 = 2025 + \sum_{i=1}^{n} \frac{1}{a_i}^2 < 2025 + \frac{1}{2} \sum_{i=1}^{n} \frac{1}{i}$$ $$ \Rightarrow a_{1000}^2 < 2025 + \frac{1}{2} \sum_{i=1}^{999} \frac{1}{i} = 2025 + H_{999} < 2025 + \ln(1000) < 46^2 $$ where $H_n$ is nth harmonic number. Try to prove by yourself that $H_n < \ln(n+1)$. (Hint Riemann sums).

It follows that $\lfloor a_{1000} \rfloor = 45$.

mateeeeeee
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