Strange oscillations in a seemingly simple number sequence

There is a simple sequence of numbers

$a_1=\frac12,\ \ \ a_{n}=\begin{cases}a_{n-1},& if\ n\ even,\\ a_{n-1}-\frac1{n}a_{\frac {n-1}2},& otherwise.\end{cases}$

In principle, nothing special, and very easy to program. Here are the first few elements of this sequence

$a_1=\frac12,\ \ a_2=\frac12,\ \ a_3=\frac13,\ \ a_4=\frac13,\ \ a_5=\frac7{30},\ \ a_6=\frac7{30}, ....$

The sequence (not strictly) tends monotonically to zero. It is easy to estimate the approximate behavior for large n:

$a_n\sim\frac{C}{n},\ \ \ where\ \ \ C=1.62944567664....$

There's a fascinating story behind this constant C. I submitted a question to mathstackexchange.com. Some time later, user Tian Vlašić pointed out that the Wolfram Alpha “knowledge base” determined the possible decimal value as

$C=\frac1{2-\ln4}.$

There were several options offered, and Tiyan chose the right one. But how to prove this? I worked for some time on the generating function for the sequence a_n, and finally found what I thought was a beautiful integral equation for it. I immediately posted another question, this time to mathoverflow.net. A couple of days later, the famous mathematician Fedya Nazarov proved that this function at zero is equal to 1-ln(2), which is what was required for my constant. His method was quite spectacular, in that he went back to the differential equation proposed by another discussant and isolated a discrete subsequence that behaved like ln(2). Inspired by the fact that the problem could be solved, the next day I found my proof, without differential equations, using recursive integration over parts of an integral equation. But let's get back to the sequence. We have found only the first approximation at infinity, but what about the next terms of the asymptotics? Since the sequence tends to zero, and, looking ahead, the most interesting thing happens in the fifth term of the asymptotics, let’s normalize it, immediately by n^2, or rather, define

$\varphi_n=n(n+1)a_n.$

In fact, the sequence phi was originally there, but then I turned it into a_n, which looks very simple. Okay, let's be more specific. The phi sequence is associated with rare events in a branching process in a random environment. This is one of the simplest tasks, but as it turns out, it is already quite difficult. There is one particle, in one step it is divided with probability p into two particles, or with probability 1-p it remains one. The same thing happens with generated particles. In this case, p at each step is selected randomly from the interval (0,1). If X_t is the number of particles at the t-th step, then

$\varphi_n=\lim_{t\to+\infty}\frac{\mathbb{P}(X_t=n)}{\mathbb{P}(X_t=1)}.$

That is, the phi sequence shows how quickly the number of particles grows compared to the case when there was no particle division at any of the steps. This is the initial task, but not the point. One way or another, having calculated a_n, and then phi using the formulas above, we build a graph

comparison of exact values with the first term of the asymptotics

Wonderful, linear growth with a strange multiplier. After tinkering a little with the generating function, I isolated the second term of the asymptotics and constructed it

In fact, this is a sequence with a period of 2. Not surprising, given the shape of the sequence a_n, where adjacent elements are repeated. The two values in the second term of the asymptotics can be counted explicitly; they are expressed through the constant C – the most difficult thing was to find its exact value. Let's go further and construct the third term of the asymptotics

Here the standard double precision is not enough. I program in Delphi, so I used the fast NesLib library, doubling the standard precision. Here is the corrected picture – the fourth term of the asymptotics turns out to be divided by n by a sequence of period 4, which can also be calculated explicitly

Since we have good accuracy, which also works very quickly, let’s go even further

normalized fourth term of the asymptotics

Here is a sequence of period 8 divided by n^2. We see only 6 terms because two coincide with the others. Here are the explicit meanings of this sequence

constants in the fourth term of the asymptotics - period 8 — constants in the fourth term of the asymptotics – period 8

Of course, somewhere further in the asymptotics there will be a sequence of period 16 divided by n^3, and so on. But something strange happened in the fifth term of the asymptotics

normalized fifth term of the asymptotics

It is not at all what we expected. Its order is not 1/n^3 and its period is not constant discrete, but logarithmic and continuous. Note that the analytical expression of the asymptotic parameters through the logarithm in the previous terms is very important, because the remainders are small, and any inaccuracy in the constants will greatly complicate the calculation of the next terms – we would never get to this oscillation. So what are these numbers like 2.54536… and 10.75397…? In fact

degree in the fifth term of the asymptotics

this is one of the complex roots of the equation

$1-2^{\alpha}=-\frac{\alpha}2.$

All roots of this equation, with natural numbers added to them, will contribute to the asymptotics. The peculiarity of complex roots, and there are infinitely many such roots, is their oscillating contribution. Since the imaginary part of complex roots can be arbitrarily large, oscillations will also be with an arbitrarily high frequency. True, the power factor of such oscillations will be small. I will also say that I don’t know how to express the amplitude of oscillations through known constants. Through the logarithm of two I can express only the non-oscillating elements of the asymptotics. But for oscillators this is still an open problem. Yes, frequency and power factor are the roots of the equation above, but what are the amplitude constants themselves? That's how deep the rabbit hole is even for such simple number sequences.