See http://www.kalva.demon.co.uk/imo/isoln/isoln783.html This is IMO 1978 problem 3. Surely discussed on ML.

hmm...i posted a topic today not seeing that it has been posted already, i`m interested to see a solution of finding $f(n)$

I can't read the kalva solution very well. Hoping to see a LATEX solution up here soon.

hmmm...i will post a solution a bit diferent from the one in kalva so we go on: since the $n$-th number missing is $f(f(n))+1$ and $f(f(n))$ is a member of the sequence, it results that there are exactly $n-1$ "gaps" less than $f(f(n))$, which leads us to: $f(f(n))=f(n)+n-1$ $(\star)$ now with a simple induction we can prove that $f(F_{n}+1) = F_{n+1}+f(1)$ , where $F_{k}$ is the Fibonacci sequence. our next step now is to prove that $f(F_{n}+x) =F_{n+1}+f(x)$, for all $x$ with $1\leq x\leq F_{n-1}$....and agaiiiiin induction(on $n$) : for $n=0$ and $n=1$ it`s trivial and now we supose that it`s true for $n-1$ and to prove that holds for $n$ , in other words $P(n-1) \Rightarrow P(n)$ $(i)$ If $x=f(y)$ for some $y$, then by the inductive assumption and $(\star)$ we have: $f(F_{n}+x)=f(F_{n}+f(y))=f(f(F_{n-1}+y))=F_{n}+f(y)+F_{n-1}+y-1=F_{n+1}+f(x)$ $(ii)$ If $x=f(f(y))+1$ is a gap, then $f(F_{n}+x-1)+1=F_{n+1}+f(x-1)+1$ is a gap also: $F_{n+1}+f(x)+1=F_{n+1}+f(f(f(y)))+1=f(F_{n}+f(f(y)))+1=f(f(F_{n-1}+f(y)))+1$ It follows that : $f(F_{n}+x)=F_{n+1}+f(x-1)+2=F{n+1}+f(x)$ now since we know that each positive integer $x$ is expressible as: $x=F_{k_{1}}+F_{k_{2}}+...+F_{k_{r}}$ , where $0<k_{r}\neq 2$, $k_{i}\geq k_{i+1}+2$ we obtain that $f(x) = F_{k_{1}+1}+...F_{k_{r}+1}$ and now that we have the general form, pe particulary calculate for $f(240)$: we can write $240=233+5+2$, therefore $f(240)= 377+8+3=388$ Remark: it can be shown now that $f(x)=[ax]$ , where $a=\frac{1+\sqrt{5}}{2}$

pohoatza wrote: since the $n$-th number missing is $f(f(n))+1$ and $f(f(n))$ is a member of the sequence, it results that there are exactly $n-1$ "gaps" less than $f(f(n))$, which leads us to: $f(f(n))=f(n)+n-1$ $(\star)$ I don't understand this part. Could anyone help? (Shouldn't it be explained by induction?)

enescu wrote: See http://www.kalva.demon.co.uk/imo/isoln/isoln783.html This is IMO 1978 problem 3. Surely discussed on ML. Why don't these links work anymore?

ZETA_in_olympiad wrote: enescu wrote: See http://www.kalva.demon.co.uk/imo/isoln/isoln783.html This is IMO 1978 problem 3. Surely discussed on ML. Why don't these links work anymore? Really old links tend to die because the hosting server of the website is no longer maintained. Fortunately, the link can still be accessed on the Wayback Machine if you still want to view it: https://web.archive.org/web/20060925135040/http://www.kalva.demon.co.uk/imo/isoln/isoln783.html

P1kachu wrote: Really old links tend to die because the hosting server of the website is no longer maintained. Fortunately, the link can still be accessed on the Wayback Machine if you still want to view it: https://web.archive.org/web/20060925135040/http://www.kalva.demon.co.uk/imo/isoln/isoln783.html Actually as far as I see, then moved the site to here, thus it didn't technically die yet.

pohoatza wrote: now since we know that each positive integer $x$ is expressible as: $x=F_{k_{1}}+F_{k_{2}}+...+F_{k_{r}}$ , where $0<k_{r}\neq 2$, $k_{i}\geq k_{i+1}+2$ we obtain that $f(x) = F_{k_{1}+1}+...F_{k_{r}+1}$ I think it's not correct. We need to redefine the conditions for the representation of $F_i$. Let me give an example. $x=8=F_6=F_1+F_3+F_5$. However, $$ f(8)=12=F_2+F_4+F_6 \neq F_7=13 $$