There is a simple sequence of rational numbers. It starts from $a_1=1$, and then $$ a_{n}=\begin{cases} a_{n-1} &\text{for even }n \\ a_{n-1}-\frac1n a_{\frac{n-1}2} &\text{for odd }n \end{cases}. $$ It seems that it converges to $0$ as $$ a_n\sim \frac{A}n,\ \ \ A=3.258891.... $$ Has anyone seen a sequence like this before? Any fast algorithm for the computation of $A$? Any expression for $A$ through known constants? Thanks for any links.
There are a few first elements $$ a_1=1,\ a_2=1,\ a_3=\frac23,\ a_4=\frac23,\ a_5=\frac7{15},\ a_6=\frac7{15}, .... $$
Update 27/08/2023. Let me say a few words about the source of the problem. Suppose that at each time step, a particle generates one new particle with probability $p$ or stays alone with probability $1-p$, where $p\in\mathbb{U}(0,1)$ is the uniform random variable on the unit interval. Let $X_t$ be the number of particles at time $t\in\mathbb{N}$. At the beginning $X_1=1$ - one particle. Denote
$$
\varphi_n:=\lim_{t\to+\infty}\frac{\mathbb{P}(X_t=n)}{\mathbb{P}(X_t=1)},
$$
$$
\Phi(z)=\varphi_1z+\varphi_2z^2+\varphi_3z^3+...,
$$
$$
F(z)=\int_0^z\Phi(\zeta)d\zeta.
$$
It can be shown that $F$ satisfies the functional equation with the initial conditions
$$
\frac{F(z)-F(z^2)}{z-z^2}=\frac12F'(z),\ \ F(0)=F'(0)=0,\ \ F''(0)=1,
$$
which leads to the generating equation for $\varphi_n$ similar to that for $a_n$. It seems that the main part of the asymptotics for $\varphi_n$ is the following, see the Figure
At the moment, I still can not prove that the constant before $n$ is $1/2(1-\ln2)$. This problem is equivalent to the problem from the question above. However, if it is true then other terms in the asymptotics can be expressed through this first constant. If I have time then I will try to write all the details and put them in, e.g., arxiv - because it can be lengthy to put it here. I will add other problems related to branching processes in a random environment. Of course, the link to math.stackexchange.com will be added as well - everybody can see who (Tian Vlašić) suggested $1/(1-\ln2)$.
Update 26/09/2023
The power series coefficients $\varphi_n$ of $\Phi(z)$, defined above, satisfy
$$
\varphi_{n}=\begin{cases}
\frac{n+1}{n-1}\varphi_{n-1}
&\text{for even }n
\\
\frac{n+1}{n-1}\varphi_{n-1}-\frac{4}{n-1} \varphi_{\frac{n-1}2}
&\text{for odd }n
\end{cases},
$$
which is similar to the functional equation for $a_n$. It seems that the first few asymptotic terms for $\varphi_n$ contain the linear growth and power law decaying coefficients multiplied by short-phase periodic factors of period $2$, $4$, $8$, etc., see the next Fig.
$8$" />
However, the next asymptotic term contains a long-phase oscillation of exponential period, see the Fig.
Here,
$$
\rho_{8n}=\frac{11\ln2-9}{4\ln2-2},\ \ \ \rho_{8n+1}=\frac{19-31\ln2}{4\ln2-2},\ \ \ \rho_{8n+2}=\frac{27-45\ln2}{4\ln2-2},\ \ \ \rho_{8n+3}=\frac{\ln2-5}{4\ln2-2},
$$
$$
\rho_{8n+4}=\frac{11\ln2-9}{4\ln2-2},\ \ \ \rho_{8n+5}=\frac{33\ln2-13}{4\ln2-2},\ \ \ \rho_{8n+6}=\frac{19\ln2-5}{4\ln2-2},\ \ \ \rho_{8n+7}=\frac{\ln2-5}{4\ln2-2},
$$
and
$$
\alpha=2.545364930374021...\pm10.75397517526887...\mathbf{i}
$$
is the root of
$$
1-2^{\alpha}=-\frac12\alpha.
$$
As it is mentioned above, I still cannot prove that the primary constant is $1/(2-\ln4)$. As a bonus, there is, perhaps, an equivalent problem: is it true that
$$
\lim_{n\to+\infty}\int_0^1\int_{z_n^2}^1...\int_{z_3^2}^1\int_{z_2^2}^1\frac{2z_1}{1+z_1}\prod_{k=2}^n\frac{2z_k}{(1+z_k)(1-z_k^2)}dz_1dz_2...dz_n=1-\ln2?
$$
Some details are available in https://arxiv.org/abs/2309.13765
The link to this question and the explicit mention of the help made by Tian Vlašić with using of WolframAlpha is added to the arxiv version.
There is also a maybe relevant paper, where $1/1-\ln2$ appears in some context https://www.jstor.org/stable/3689501 but I did not read it carefully yet.