Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|cite|improve this question
$$b_n=a_n^{-1}\Longrightarrow b_1=a_1^{-1}=\frac{1}{2}...?$$ – 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

1 Answer 1


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

The asymptotic ratio

I created also a new sequence in OEIS, see

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.