I'm trying to solve the general formula for nth term: $a_n = a_{n-1} + 2^{-a_{n-1}}, a_0 = 0$. And want to if there exists $f(n)$, s.t. $lim_{n\rightarrow \infty}a_n = f(n)$.

First, I've tried to see what do first terms look like:

$a_0 = 0$
$a_1 = 1$
$a_2 = 1.50000000000000$
$a_3 = 1.85355339059327$
$a_4 = 2.13026337562636$
$a_5 = 2.35867953516332$

But I can't find any clue.

Then I tried to directly write down the general formula

$a_1 = 2^{-a_0}$
$a_2 = 2^{-a_0} + (2)^{2^{-a_0}}$
$a_3 = 2^{-a_0} + (2)^{2^{-a_0}} + (2)^{(2)^{2^{-a_0}}}$

In this way, I can definitely write down the general formula in a very silly way. However, I can't see what $f(n)$ would $a_n$ goes to as $n\rightarrow \infty$.

Apparently, it's not arithmetic or geometric sequence. And I don't think the method of characteristic functions would help.

Any hint or detailed steps would be appreciated!

  • $\begingroup$ It doesn't change much, but it should be $a_3 = 2^{-a_0} + (2)^{-2^{-a_0}} + (2)^{-(2^{-a_0} + (2)^{-2^{-a_0}})}$ $\endgroup$ – Ross Millikan Apr 30 '19 at 23:17

(I think I recently answered a similar question. Oh well, here I go again.)

If $a_n = a_{n-1} + c^{-a_{n-1}}, a_0 = 0 $ where $c > 1$, my conjecture is that $a_n \sim \dfrac{\ln(u)}{u}+\dfrac{\ln(n)}{u} $ where $u = \ln(c) $.

I will first show that the $a_n$ are unbounded.

If the $a_n$ are bounded, say $a_n < M$, then $a_n = a_{n-1} + 2^{-a_{n-1}} \gt a_{n-1} + 2^{-M} $ or $a_n - a_{n-1} \gt 2^{-M} $. Summing this, $a_m-a_0 =\sum_{n=1}^m(a_n - a_{n-1}) \gt m2^{-M} $ which is a contradiction for large enough $m$.

If $a_n \sim f(n)$, then $f'(n) \sim c^{-f(n)} = u^{-uf(n)} $ where $u = \ln c$.

Then $f''(n) =(-uf'(n))e^{-uf(n)} =(-uf'(n))f'(n) =-uf'^2(n) $.

Letting $g = f'$, then $g' = -ug^2$ so $\dfrac{-g'}{g^2} =u $ or $(1/g)' = u $ or $1/g = dn+r$ or $g = \dfrac1{dn+r}$.

Therefore $f'(n) = \dfrac1{un+r}$ so

$\begin{array}\\ f(n) &= \int \dfrac{dn}{un+r}\\ &=\frac1{u}\ln(un+r)\\ &=\frac1{u}(\ln(u)+\ln(n+r/u))\\ &=\frac{\ln(u)}{u}+\frac1{u}\ln(n+r/u)\\ \end{array} $

For simplicity's sake I will assume that $r = 0$. Then

$\begin{array}\\ f(n) &=\frac{\ln(u)}{u}+\frac1{u}\ln(n)\\ \text{so}\\ f(n+1) &=\frac{\ln(u)}{u}+\frac1{u}\ln(n+1)\\ &=\frac{\ln(u)}{u}+\frac1{u}(\ln(n)+\ln(1+1/n))\\ &\sim\frac{\ln(u)}{u}+\frac1{u}(\ln(n)+\frac1{n}+O(\frac1{n^2}))\\ &=f(n)+\frac1{u}(\frac1{n}+O(\frac1{n^2}))\\ &=f(n)+\frac1{un}+O(\frac1{n^2})\\ \end{array} $

So we want

$\begin{array}\\ \frac1{un} &\sim c^{-(\ln(u)/u+\ln(n)/u)}\\ &\sim e^{\ln(c)(-\ln(u)/u-\ln(n)/u)}\\ &\sim e^{u(-\ln(u)/u-\ln(n)/u)}\\ &\sim e^{-\ln(u)-\ln(n))}\\ &\sim e^{-\ln(un)}\\ &=\frac1{un}\\ \end{array} $

So it works.

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  • $\begingroup$ It works very well ! $\endgroup$ – Claude Leibovici May 1 '19 at 6:10
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    $\begingroup$ I think there is a typo on 6th line. Should be $f'(n) \sim c^{-f(n)} = e^{-uf(n)}$ $\endgroup$ – Donny May 1 '19 at 23:58
  • $\begingroup$ Thanks! This perfectly solved my problem! $\endgroup$ – Donny May 2 '19 at 0:10
  • $\begingroup$ Fortunately, that typo didn't affect anything following. $\endgroup$ – marty cohen May 2 '19 at 11:10

It will not converge. If it converged to a limit $L$, it would solve $$L=L+2^{-L}$$ which has no solution.

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  • $\begingroup$ Thanks! I see why it's not convergent. However, I still want to see how big is $a_n$, i.e. to find it's upper bound or lower bound if they exist. It seems $log_2(n)$ is an upper bound. Do you have any advice on finding a good upper/lower bound? $\endgroup$ – Donny May 1 '19 at 0:04
  • $\begingroup$ It is monotonically increasing, so there can't be an upper bound. If there were it would converge. It goes off to infinity $\endgroup$ – Ross Millikan May 1 '19 at 0:11


Note that the difference between $a_n$ and $a_{n-1}$ is given in terms of $a_{n-1}$...


Shift the indices by $1$, and you have a difference equation...

$$\Delta a(n)=\frac{1}{2^{a(n)}}$$

...with initial condition $a(0)=0$.

I know nothing about solving difference equations, but from what you have written and the result I got when I put this into Wolfram Alpha, it has no closed-form solution. (If it does have a closed form solution, it will probably be in terms of the natural logarithm, but again, this is not my area of expertise)

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