Determine the least natural number $k$ such that $a(k)>1$ Let $a(n)$ be a sequence with $a(0)=1/2$ and $a(n+1)=a(n)+(a(n)^2)/2013$, $n$ natural number.
Determine the least natural number $k$ such that $a(k)>1$.
This problem is from Poland proposed to Romanian Masters of Mathematics.
Can you give me some hints? I don't want a complete solution.
Thank you!
 A: 
Motto: To iterate a  quadratic function is basically impossible, to iterate a homographic function is trivial.

Consider any sequence $(x_n)$ defined by $x_{n+1}=x_n+\frac1cx_n^2$ for some $0\lt x_0\lt 1\lt c$. Then, for every $0\lt x\lt 1$, 
$$
\frac{c+1}{c+1-x}\lt1+\frac{x}c\lt\frac{c}{c-x},
$$ 
hence, as long as $x_n\lt 1$, 
$$
\frac{c+1}{c+1-x_n}x_n\lt x_{n+1}\lt\frac{c}{c-x_n}x_n,
$$
that is, 
$$
\frac1{x_n}-\frac1c\lt \frac1{x_{n+1}}\lt\frac1{x_n}-\frac1{c+1}.
$$
Thus, 
$$
\frac1{x_0}-\frac{n}c\lt\frac1{x_n}\lt\frac1{x_0}-\frac{n}{c+1},
$$ 
for every $n$ such that 
$$
\frac1{x_0}-\frac{n-1}c\geqslant1,
$$
that is, such that $n\leqslant(c/x_0)-c+1$. Assume that $c$ is an integer and that $x_0\leqslant1/2$. Then $n=c$ and $n=c+1$ are admissible. For $n=c$, the lower bound of $1/x_n$ is $1/x_0-1$. For $n=c+1$ the upper bound is $1/x_0-1$, hence 
$$
\frac1{x_{c+1}}\lt\frac1{x_0}-1\lt\frac1{x_c}.
$$

For every integer $c\geqslant2$ and $x_0\leqslant1/2$, the least $n$ such that $x_n\gt x_0/(1-x_0)$ is $n=c+1$. In particular, if $x_0=1/2$, the least $n$ such that $x_n\gt1$ is $n=c+1$.

A: I was working out the details when Did posted basically the same solution that I had in mind, so I won't repeat it here. But I thought I could say a thing or two about how one can come to a solution like this. At least that is how I came to it.
Our sequence is given by relation $a_{n+1} - a_n = a_n^2 / c$. Notice that $a_n^2/c$ is very small in our case, so we have a kind of a discrete system that moves step by step, making very small steps.
We could try to approximate it by looking at its continuous version, given by the differential equation
$$
    f'(t) = f^2(t) / c.
$$
and the boundary condition $f(0) = a(0)$.
The equation can be rewritten as $df/f^2 = dt/c$, and the solution then is:
$$
    \frac{1}{a(0)} - \frac{1}{f(t)} = \frac{t}{c}.
$$
So it is logical to expect that $\frac{1}{a(0)} - \frac{1}{a(n)}$ is somewhere around $\frac{n}{c}$. Then one can try and prove the inequalities as in Did's answer.
