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The following sequence appeared in IMC 2012 (a math competition): $$a_1 = \frac{1}{2}, \qquad a_{n+1} = \frac{n a_n^2}{1+(n+1)a_n}$$

I am trying to find an explicit formula for the sequence. It seems to be nicer to look at $b_n = a_n^{-1}$: $$b_1 = 2, \qquad b_{n+1} = \frac{b_n(b_n +n +1)}{n}$$ Can some closed formula be derived?

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$$b_n=a_n^{-1}\Longrightarrow b_1=a_1^{-1}=\frac{1}{2}...?$$ – DonAntonio Aug 4 '12 at 1:03
The multiplicative structure of the $a_n$ is extremely interesting. If $a_n=p_n/q_n$, then the numerators $p_n$ are all powers of $2$ - in fact $p_{n+1} = p_n + 2^n + \nu_2(n)$, where $\nu_2(n)$ is the exponent of $2$ in the prime power factorization of $n$. The $q_n$ also have a multiplicative-like structure: $q_{n+1}/q_n$ is very close to an integer, in that it is a rational number whose denominator is either $1$ or else a very small prime (at most $n$, in the first $25$ terms of the sequence). – Greg Martin Aug 4 '12 at 2:26
@DonAntonio: I fixed it, $a_1 = \frac{1}{2}$. – Ofir Aug 4 '12 at 8:54
On purely empirical grounds, I would argue that no closed formula can be found. The point is, this sequence was stared at rather hard by a lot of contestants and team leaders. If anybody had found a closed formula for $a_n$, then this would probably make its way into the solution, and would surely be quickly spread after the contest via small talk. Since neither of these was the case, I suspect that closed formula either does not exist, or is very complex/hard to find. – Jakub Konieczny Apr 28 '13 at 8:59


$$ b(n) \sim n c^{2^n} $$ where c = 1.36534926036757464312824443040683531215776134381623126072...

The asymptotic ratio

I created also a new sequence in OEIS, see

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