$ m=1\mod 2^{2007}$ work.

Rust wrote: $ m = 1\mod 2^{2007}$ work. Post your solution, please

because $ a_{2}$ must not be an integer so $ a_{1}=\frac{m}{2}$ must'nt be an integer too because if it is an integer then $ \left\lceil a_{1}\right\rceil$ is an integer too and... this tells us that $ m\equiv1(\text{ mod }2)$ (or in other words $ m$ must be odd and for a number $ k_{1}\in\mathbb{Z}$, $ m = 2k_{1}+1$) then we have $ a_{2}=\frac{2k_{1}+1}{2}\times(k_{1}+1)$ now $ a_{3}$ must'nt be integer and similarly we must have $ (2k_{1}+1)(k_{1}+1) = 2k^{2}_{1}+3k_{1}+1\equiv1(\text{ mod }2)\Longrightarrow k_{1}\equiv0(\text{ mod }2)$ and this means there exists $ k_{2}\in\mathbb{Z}$ such that $ k_{1}= 2k_{2}$ and then as we know $ \left\lceil\frac{8k^{2}_{2}+6k_{2}+1}{2}\right\rceil = 4k^{2}_{2}+3k_{2}+1$ we have: $ a_{3}=\frac{(4k_{2}+1)(2k_{2}+1)(4k^{2}_{2}+3k_{2}+1)}{2}$ now $ a_{4}$ must'nt be an integer and similarly we must have $ (4k_{2}+1)(2k_{2}+1)(4k^{2}_{2}+3k_{2}+1)\equiv1(\text{ mod }2)\Longrightarrow k_{2}\equiv0(\text{ mod }2)$ and so there exists $ k_{3}\in\mathbb{Z}$ such that $ k_{2}= 2k_{3}$ and so on... and atlast we can prove that if for $ n\in\mathbb{N}$, $ a_{n}$ is the first integer appearing in the sequence this means instead of $ a_{1},a_{2},\cdots,a_{n-2}$, the $ a_{n-1}$ must be noninteger too then there exists $ k_{n-2}\in\mathbb{Z}$ such that $ m = 2^{n-1}k_{n-2}+1$ but because $ a_{n}$ must be integer then $ k_{n-2}\equiv1(\text{ mod }2)$ and there exists $ k_{n-1}\in\mathbb{Z}$ such that $ k_{n-2}= 2k_{n-1}+1$ and $ m = 2^{n-1}\left(2k_{n-1}+1\right)+1$. So the answer is all numbers in the form $ \boxed{m = 2^{2006}\left(2s+1\right)+1}$ where $ s\in\mathbb{Z}$.Hope theres no miscalculations!

I get a slightly different solution, but maybe there is a typo Consider the map $ \Psi:\frac{1}{2}\mathbb{N}\rightarrow\frac{1}{2}\mathbb{N}\,,\,\Psi(x)=x\,\lceil x\rceil$ and define for $ k\geq 1$ the sets $ T_{k,+}=2^{k}\mathbb{N}+\frac{1}{2}-2^{k-1}$ and for $ k\geq 0$ similar sets $ T_{k,-}=2^{k}\mathbb{N}+\frac{1}{2}-2^{k}$. We observe the following obvious properties: (1) We have the disjoint decomposition $ \frac{1}{2}\mathbb{N}=\mathbb{N}\cup T_{0,-}$ (2) We have for all $ k\geq 0$ the disjoint decomposition $ T_{k,-}=T_{k+1,+}\cup T_{k+1,-}$ (hint: substitute $ x=2y$ resp $ x=2y-1$ in $ 2^{k}x+\frac{1}{2}-2^{k}$) Application of (1),(2) yields the generalization (2') $ \forall k\geq0:\frac{1}{2}\mathbb{N}=\mathbb{N}\cup T_{k,-}\cup\bigcup_{v=1}^{k}T_{v,+}$ which allows to immediately verify: (3) $ \Psi^{-1}(\mathbb{N})=\mathbb{N}\cup T_{1,+}$ (4) $ \Psi^{-1}(T_{k,+})=T_{k+1,+}$ (hint: simple induction and (2')) Application of the previous properties requires $ a_{2006}\in T_{1,+}$ and pulling recursively back with (4) we get ultimately $ a_{1}\in T_{2600,+}$ which is equivalent to $ m\equiv 1-2^{2006}\bmod{2^{2007}}$

Let me try, put $ b_{n}= 2a_{n}$ then $ b_{n}\in\mathbb Z$ for all $ n\geq1$. $ a_{2007}$ is the first integer appearing in the sequence iff $ 2\mid b_{2007}$ and $ 2\nmid b_{2006}$ iff $ b_{2006}\equiv1\pmod 4$. Now we need some auxiliary steps: Lemma 1. For each $ s\in\mathbb Z$ then $ s\big[\frac{s}{2}\big]\equiv 1\pmod 4$ iff $ s\equiv-1\pmod 8$. Lemma 2. For each $ s\in\mathbb Z$ then $ s\big[\frac{s}{2}\big]\equiv-1\pmod 8$ iff $ s\equiv-5\pmod{16}$. Lemma 3. For each $ s\in\mathbb Z$ then $ s\big[\frac{s}{2}\big]\equiv-5\pmod{16}$ iff $ s\equiv3\pmod{32}$. Lemma 4. For each $ s\in\mathbb Z$ then $ s\big[\frac{s}{2}\big]\equiv 3\pmod{2^{n}}$ iff $ s\equiv3\pmod{2^{n+1}}$ for all $ n\geq3$. So we have $ b_{2006}\equiv1\pmod 4$ iff $ b_{2005}\equiv-1\pmod 8$ iff $ b_{2004}\equiv-5\pmod{16}$ iff $ b_{2003}\equiv3\pmod{2^{5}}$ iff $ b_{2002}\equiv3\pmod{2^{6}}$ ... iff $ b_{1}\equiv3\pmod{2^{2007}}$ iff $ m\equiv 3\pmod{2^{2007}}$. I solved this problem when I was very sleepy!

hjbrasch wrote: I get a slightly different solution, but maybe there is a typo Application of the previous properties requires $ a_{2006}\in T_{1,+}$ and pulling recursively back with (4) we get ultimately $ a_{1}\in T_{2600,+}$ which is equivalent to $ m\equiv 1-2^{2006}\bmod{2^{2007}}$ well your answer is similar to what i've got! $ m\equiv 1-2^{2006}\bmod{2^{2007}}$ means there exists $ s\in\mathbb{Z}$ such that $ m = 2^{2007}s+1-2^{2006}= 2^{2006}(2s-1)+1$ i think the answer of both of us is correct

Oh I though the problem is for the integer part $ [x]$ . Sorry!

Note that if $ a_{i}$ is not an integer it is of the form $ \frac{k}{2}$ where $ k$ is an integer. Now suppose that $ a_{j}=\frac{2^{n}(2s_{j}+1)+1}{2}$ with $ n\ge 1$. We have that: $ a_{j+1}=\frac{2^{n}(2s_{j}+1)+1}{2}\left(\frac{2^{n}(2s_{j}+1)+2}{2}\right)$ $ \implies 2a_{j+1}= 2^{2n-1}(2s_{j}+1)^{2}+2^{n}(2s_{j}+1)+2^{n-1}(2s_{j}+1)+1$ So we see that: i) $ 2a_{j+1}\neq 1 (\bmod 2^{n})$ ii) $ 2a_{j+1}\equiv 1 (\bmod 2^{n})$ Or equivalently we have that there is an integer $ s_{j+1}$ such that $ a_{j+1}=\frac{2^{n-1}(2s_{j+1}+1)+1}{2}$. We can right $ m = 2^{k}(2s+1)+1$ for some integers $ s$ and $ k\ge 0$. The last result tells us that the first integer of the secuence is $ a_{k+1}$. So in order to have $ a_{2007}$ as the first integer of the secuence, we need to have $ k = 2006$. So we have that all the possible values of $ m$ are: $ m = 2^{2006}(2s+1)+1$ for any integer $ s$.

I think the most natural way in which to look at the problem is base 2. It's easy to see that m is odd in particular 4k+1 and that it cannot be only 1. So we have something like m= .....1000000...0001 with some ammount of consecutive digits equal to zero. if we call k this ammount of zeros we can see that as a_1 has k zeros between the coma and the first one, and every time you iterate the function that ammount of zeros decreases exactly one time. And when we have no more zeros de ceiling will be even so the next term will be an integer. So the ammount of zeros is 2005 which yields m=2^2006+1 mod 2^2007 sory about my english and no latex, julian

it's identical with 2010 China Second Round,Test 2,problem 2.