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a sequence $(a_n)_{n\ge0}$ is defined by $a_0=1$ and $$a_{n+1}=a_{{\lfloor}\frac{7n} {9}{\rfloor}}+a_{{\lfloor}\frac{n}{9}{\rfloor}}$$ for $n\ge0$.prove that there is an $n$ such that $a_n<\frac{n}{2001!}$

how do i approach this? we see that $a_1=2,a_3=a_4=3,a_5=a_6=a_7=4,a_8=a_9=5$.Is there any pattern coming out?Does induction work here?

mike dennes
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    Unclear to me whether the answer of Henry Lee is either valid, or can be made valid. Anyway, for an alternative, less elegant approach, for $~k \in \Bbb{Z_{\geq 2}},~$ let f(k) denote the largest positive integer n such that $~a_n=k.~$ Then, manually compute f(5),f(6),⋯,f(10). Then using these computations as data, look for a pattern in the data. Then, use this perceived pattern to formulate a hypothesis about the computation of f(k) as a function of k. ...see next comment – user2661923 Sep 19 '23 at 14:43
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    Then, try to prove this hypothesis, perhaps through induction. Then, once this hypothesis is proven, apply the hypothesis to the posted question. For what it's worth, it is unclear to me whether the approach that I have outlined will conquer the problem. However, this would automatically be my first try. – user2661923 Sep 19 '23 at 14:43
  • sir @MartinR $a_4=a_{{\lfloor}\frac{21}{9}{\rfloor}}+a_{{\lfloor}\frac{3}{9}{\rfloor}}=a_2+a_0=2+1=3$ – mike dennes Sep 19 '23 at 15:27
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    @mikedennes: You are right, my fault, sorry for that. – Martin R Sep 19 '23 at 16:19

2 Answers2

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The following is inspired by

  • “A Pseudo-Fibonacci Limit” in Luthar, R. S., Julio Castineira, Barry Powell, Daniel A. Rawsthorne, H. L. Krall, J. C. Binz, William Wernick, et al. “Problems.” Mathematics Magazine 57, no. 1 (1984): 41–50. https://doi.org/10.2307/2690297.

where the similar problem

Set $a_0 = 1$ and for $n \ge 1$, $a_n = a_{\lfloor n/2 \rfloor} + a_{\lfloor n/3 \rfloor}$. Find $\lim_{n\to\infty} a_n/n$.

is solved (also mentioned in this answer).

The idea is to prove that the sequence satisfies $$ \tag{$*$} a_n \le C n^r $$ for all $n \ge 1$, with some suitably chosen $r \in (0, 1)$ and some $C > 0$. First, the exponent $r$ is chosen such that $$ \left( \frac{7}{9}\right)^r + \left( \frac{1}{9}\right)^r < 1 \, , $$ for example $\boxed{r = 0.8}$. Next, the factor $C$ is chosen $(*)$ holds for $1 \le n \le 9$, for example $\boxed{C = 2}$. Finally, $(*)$ is proven for all $n \ge 1$ with induction.

It follows from $(*)$ that $$ \lim_{n \to \infty} \frac{a_n}{n} = 0 \, , $$ so that $a_n / n < 1/2001!$ eventually.

Martin R
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  • sir is there any way to find a bound or expression for $n$? – mike dennes Sep 19 '23 at 15:29
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    @mikedennes: You need $2n^{0.8} < n/2001!$ or $2\cdot 2001! < n^{0.2}$, i.e. $n > (2 \cdot 2001!)^5$. – Martin R Sep 19 '23 at 16:41
  • sir @MartinR sorry to bother u again-what purpose does this assumption serve -"exponent r is chosen such that"-$$ \left( \frac{7}{9}\right)^r + \left( \frac{1}{9}\right)^r < 1 , , $$ – mike dennes Sep 21 '23 at 15:55
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    @mikedennes: That is needed when you prove with induction that (*) holds for all $n$. – Martin R Sep 21 '23 at 16:12
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Proof by contradiction

Assume: $$\forall n\exists a_n\ge\frac{n}{2001!}$$ Consider: $$a_{2001!}\ge\frac{2001!}{2001!}=1$$ now using the recurrance relation: $$a_{2001!+9}=a_{\left\lfloor\frac{7(2001!+9)}{9}\right\rfloor}+a_{\left\lfloor\frac{2001!+9}{9}\right\rfloor}$$ $$a_{2001!+9}=a_{\left\lfloor\frac{7\times2001!}{9}+\frac{7\times9}{9}\right\rfloor}+a_{\left\lfloor\frac{2001!}{9}+\frac99\right\rfloor}$$ Now notice that $\frac{7\times2001!}{9}\in\mathbb{Z}$ and $\frac{7\times9}{9}=7$ so the first term's floor bracket contains an integer. Similarly, $\left(\frac{2001!}{9}+1\right)\in\mathbb{Z}$. As a result $a_{2001!+9}\ge1+1=2$. Repeating this for any $a_{2001!+9k}$ where $k\in\mathbb{Z^+}$ yields the same inequality. This implies: $\forall n>2001!,a_n\ge2$.

Now, comparing $a_{2\times2001!}\ge2$ to $\frac{2\times2001!}{2001!}=2$ shows a contradiction as we assumed: $$a_n\ge\frac{n}{2001!}\forall n$$ But we have shown that: $$a_{2\times2001!}<\frac{2\times2001!}{2001!}=2$$ Hence our initial assumption must be false: $$\therefore\exists n\,\,s.t.\,\,a_n<\frac{n}{2001!}$$ And that completes the proof

EDIT:

Let $b_n = \frac{a_n}{n}$.

Consider $b_{2001!}$. By the recurrence relation: $$b_{2001!} = \frac{a_{2001!}}{2001!} \geq \frac{1}{2001!}$$ Now, let's compute $b_{2001! + 9}$: $$b_{2001! + 9} = \frac{a_{2001! + 9}}{2001! + 9}$$ Notice that $b_{2001! + 9}$ depends only on $b_n$ for $n$ in the range $[2001!, 2001!+9]$. Since there are 10 terms in this range, by the Pigeonhole Principle, at least one of these terms must be less than or equal to the average: $\frac{b_{2001!} + b_{2001! + 1} + \ldots + b_{2001! + 9}}{10}$ Therefore, there exists $k$ such that $b_{2001! + k} \leq \frac{b_{2001!} + b_{2001! + 1} + \ldots + b_{2001! + 9}}{10}$. Since $b_{2001!} \geq \frac{1}{2001!}$, we have: $\frac{b_{2001!} + b_{2001! + 1} + \ldots + b_{2001! + 9}}{10} \geq \frac{1}{2001!}$ Combining the above inequalities, we get: $b_{2001! + k} \leq \frac{1}{2001!}$ This implies: $a_{2001! + k} \leq \frac{2001! + k}{2001!} = 1 + \frac{k}{2001!}$ Finally, since $k \geq 0$, we have: $a_{2001! + k} < 1 + \frac{1}{2001!} \leq \frac{1}{2001!}$ This completes the proof, demonstrating that there exists an $n$ (specifically, $n = 2001! + k$) such that $a_n < \frac{n}{2001!}$.

Henry Lee
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  • @MartinR I didn't want to delete the first proof as I'd only just posted it but I think I made an error there and wasn't sure so just posted the second proof to slot in – Henry Lee Sep 19 '23 at 13:34
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    $a_{2001! + 9}$ depends on $a_{2001!/9 + 1}$, therefore I do not understand your argument starting at “Notice that $b_{2001! + 9}$ depends only on ...” – Martin R Sep 19 '23 at 13:47
  • $\forall n\exists a_n\ge\frac{n}{2001!}$ makes no sense – jjagmath Sep 19 '23 at 14:15
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    @jjagmath true, but his intent is the assertion (which he then tries to prove is false) that $~\forall ~n, ~a_n \geq \dfrac{n}{2001!}.$ – user2661923 Sep 19 '23 at 14:23