APMO

Note that $$a_{i+1}+1=2\left(a_i+1\right)\text{ or }{a}_{i+1}+1=\frac{a_i+a_i+2}{a_i+2}=\frac{2\left(a_i+1\right)}{a_i+2}.$$ Hence $$\frac{1}{a_{i+1}+1}=\frac{1}{2}\cdot \frac{1}{a_i+1}\text{ or }\frac{1}{a_{i+1}+1}=\frac{a_i+2}{2\left(a_i+1\right)}=\frac{1}{2}\cdot \frac{1}{a_i+1}+\frac{1}{2}.$$ Therefore, $$\begin{array}{c}\frac{1}{a_k+1}=\frac{1}{2^k}\cdot \frac{1}{a_0+1}+\sum _{i=1}^k\frac{{\epsilon }_i}{2^{k-i+1}}\end{array}\text{\hspace{1em}(1)}$$ where ${\epsilon }_i=0$ or $1$. Multiplying both sides by $2^k\left(a_k+1\right)$ and putting $a_k=2014$, we get $$2^k=\frac{2015}{a_0+1}+2015\cdot \left(\sum _{i=1}^k{\epsilon }_i\cdot 2^{i-1}\right)$$ where ${\epsilon }_i=0$ or $1$. Since $\gcd (2,2015)=1$, we have $a_0+1=2015$ and $a_0=2014$. Therefore, $$2^k-1=2015\cdot \left(\sum _{i=1}^k{\epsilon }_i\cdot 2^{i-1}\right)$$ where ${\epsilon }_i=0$ or $1$. We now need to find the smallest $k$ such that $2015\mid 2^k-1$. Since $2015=5\cdot 13\cdot 31$, from the Fermat little theorem we obtain $5\mid 2^4-1$, $13\mid 2^{12}-1$ and $31\mid 2^{30}-1$. We also have $\mathrm{lcm}[4,12,30]=60$, hence $5\mid 2^{60}-1$, $13\mid 2^{60}-1$ and $31\mid 2^{60}-1$, which gives $2015\mid 2^{60}-1$. But $5\nmid 2^{30}-1$ and so $k=60$ is the smallest positive integer such that $2015\mid 2^k-1$. To conclude, the smallest positive integer $k$ such that $a_k=2014$ is when $k=60$.

Alternative solution 1: Clearly all members of the sequence are positive rational numbers. For each positive integer $i$, we have $a_i=\frac{a_{i+1}-1}{2}$ or $a_i=\frac{2a_{i+1}}{1-a_{i+1}}$. Since $a_i>0$ we deduce that $$a_i=\{\begin{array}{ll}\frac{a_{i+1}-1}{2}& \text{ if }{a}_{i+1}>1\\ \frac{2a_{i+1}}{1-a_{i+1}}& \text{ if }{a}_{i+1}<1\end{array}$$ Thus $a_i$ is uniquely determined from $a_{i+1}$. Hence starting from $a_k=2014$, we simply run the sequence backwards until we reach a positive integer. We compute as follows. $$\begin{array}{rl}& \frac{2014}{1},\frac{2013}{2},\frac{2011}{4},\frac{2007}{8},\frac{1999}{16},\frac{1983}{32},\frac{1951}{64},\frac{1887}{128},\frac{1759}{256},\frac{1503}{512},\frac{991}{1024},\frac{1982}{33},\frac{1949}{66},\frac{1883}{132},\frac{1751}{264},\frac{1487}{528},\frac{959}{1056},\frac{1918}{97},\frac{1821}{194},\frac{1627}{388},\\ & \frac{1239}{776},\frac{463}{1552},\frac{926}{1089},\frac{1852}{163},\frac{1689}{326},\frac{1363}{652},\frac{711}{1304},\frac{1422}{593},\frac{829}{1186},\frac{1658}{357},\frac{1301}{714},\frac{587}{1428},\frac{1174}{841},\frac{333}{1682},\frac{666}{1349},\frac{1332}{683},\frac{649}{1366},\frac{1298}{717},\frac{581}{1434},\frac{1162}{853},\\ & \frac{309}{1706},\frac{618}{1397},\frac{1236}{779},\frac{457}{1558},\frac{914}{1101},\frac{1828}{187},\frac{1641}{374},\frac{1267}{748},\frac{519}{1496},\frac{1038}{977},\frac{61}{1954},\frac{122}{1893},\frac{244}{1771},\frac{488}{1527},\frac{976}{1039},\frac{1952}{63},\frac{1889}{126},\frac{1763}{252},\frac{1511}{504},\frac{1007}{1008},\frac{2014}{1}.\end{array}$$ There are 61 terms in the above list. Thus $k=60$.

Alternative solution 2: Start with $a_k=\frac{m_0}{n_0}$ where $m_0=2014$ and $n_0=1$ as in alternative solution 1. By inverting the sequence as in alternative solution 1, we have $a_{k-i}=\frac{m_i}{n_i}$ for $i\ge 0$ where $$\left(m_{i+1},n_{i+1}\right)=\{\begin{array}{ll}\left(m_i-n_i,2n_i\right)& \text{ if }{m}_i>n_i\\ \left(2m_i,n_i-m_i\right)& \text{ if }{m}_i<n_i\end{array}$$ Easy inductions show that $m_i+n_i=2015$, $1\le m_i,n_i\le 2014$ and $\gcd \left(m_i,n_i\right)=1$ for $i\ge 0$. Since $a_0\in {\mathbb{N}}^{+}$ and $\gcd \left(m_k,n_k\right)=1$, we require $n_k=1$. An easy induction shows that $\left(m_i,n_i\right)\equiv \left(-2^i,2^i\right)(\mathrm{mod}2015)$ for $i=0,1,\dots ,k$. Thus $2^k\equiv 1(\mathrm{mod}2015)$. As in the official solution, the smallest such $k$ is $k=60$. This yields $n_k\equiv 1(\mathrm{mod}2015)$. But since $1\le n_k,m_k\le 2014$, it follows that $a_0$ is an integer.