Let $u_0\ge 2$ and $\alpha\ge 1$ integers.

I'm trying to study the sequence $(u_n)_{n\ge 0}$ defined by : $$\forall n\ge 0,\quad u_{n+1}=d(u_n)^{\alpha}, $$ where $d$ is the number-of-divisors function.

If $\alpha=1$, it's not hard to prove that $$\lim\limits_{n\rightarrow +\infty}u_n=2,$$ And if $\alpha=2$ we also have : $$\lim\limits_{n\rightarrow +\infty} u_n=9. $$ However, if $\alpha\ge 3$, the sequence doesn't always converge.

For example, when $\alpha=3$, if $u_0=2$, then for $n\ge 1$, $u_{2n}=2^6$ and $u_{2n+1}=7^3$.

Do we have more results about the behaviour of the sequence $(u_n)_{n\ge 0}$ when $\alpha\ge 3$ ?

$\textit{Edit 2 -}$ After many simulations, I have the following conjecture :

$\boxed{\textbf{Conjecture -} \text{ For all $\alpha\ge 1$, the sequence $(u_n)_{n\ge 0}$ is eventually periodic.}}$

When $\alpha=4$, the sequence always seems to converge, but the limit depends on $u_0$ (for $u_0=2$ the limit is $625$ and for $u_0=6$ the limit is $6561$).

For example, when $\alpha=5$, if $u_0=2$, then the sequence converges to $3^52^5$, but if $u_0=2^53^511^{10}$, then the sequence is $2$-periodic (with $u_1=2^{10}3^{10}11^{5}$).

We can easily prove that :

$\boxed{\text{For all $\alpha\ge 1$, there are infinitelty many numbers such that $(u_n)_{n\ge 0}$ converges.}} $

Indeed, if $u_0=p_1...p_l$ is a squarefree integer, then : $$u_1=2^{\alpha l} $$ and $$u_2=(\alpha l+1)^{\alpha}. $$ By Dirichlet's theorem on arithmetic progressions, there are infinitely many integers $l\ge 1$ such that $\alpha l+1$ is a prime number. If we choose a such integer $l\ge 1$, then for all $n\ge 2$ : $$u_n=(\alpha l+1)^{\alpha}, $$ and the sequence $(u_n)_{n\ge 0}$ converges. $\blacksquare$

  • $\begingroup$ Does it being equivalent to modifying the number of divisors in a specific weird way count. Even if predictable ? $\endgroup$ – Roddy MacPhee Aug 3 at 2:18
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    $\begingroup$ Since $\tau(n)$ grows slower than any power of $n$, the sequence will eventually converge to a limit point or a limit cycle for any fixed $\alpha$. Infinite growth is impossible, that much we can be sure of. $\endgroup$ – Ivan Neretin Aug 3 at 13:38

I'll show that the conjecture that the sequence will end up periodic (or converge which just means of period $1$).

Let's first express the sequence in terms of $v_n=d(u_n)$ so that $v_{n+1}=d(v_n^\alpha)$.

I'll start off by proving that, for any integer $\alpha>0$, there is at most a finite number of positive integers $v$ for which $d(v^\alpha)\ge v$, and then use this to prove that the sequence must eventually be periodic.

Prime factorise $v=p_1^{m_1}\cdots p_k^{m_k}$. Then, $d(v^\alpha) = (\alpha m_1+1)\cdots(\alpha m_k+1)$. So, I wish to prove that for $v$ sufficiently large, $d(v^\alpha)/v<1$. We can write this fraction as $$ \frac{d(v^\alpha)}{v}=\prod_{i=1}^k \frac{\alpha m_i+1}{p_i^{m_i}}. $$ For any given prime number $p$, the largest value of $(\alpha m+1)/p^m$ is $(\alpha+1)/p$ if $p\le\alpha$ (corresponding to $m=1$), and $1$ for $p>\alpha$ (corresponding to $m=0$). This gives us an upper bound $d(v^\alpha)/v\le A$ where $A=\prod_{\text{prime } p\le\alpha} (\alpha+1)/p$. Since picking $m_i$ which maximise $d(v^\alpha)/v$ makes it at most $A$, if there is any term with $(\alpha m_i+1)/p_i^{m_i}<1/A$, that will make $d(v^\alpha)/v<1$. However, for any prime $p$, the restriction that $(\alpha m+1)/p^m\ge 1/A$ places an upper bound $m\le M(p)$, as well as an upper bound $p\le 2A$ om primes for which $m\ge 1$.

So, to summarise, in order to get $d(v^\alpha)/v\ge1$, we must have $v=\prod p_i^{m_i}$ where $p_i$ are prime numbers $\le A/2$ and $m_i\le M(p_i)$. There are only a finite number of such numbers. Thus, we can conclude that for all but a finite number of $v$, we have $d(v^\alpha)<v$. Since there is only a finite number of $v$ with $d(v^\alpha)\ge v$, these must have an upper bound $W\ge d(v^\alpha)\ge v$. We can therefore conclude that the sequence $v_n$ will eventually drop and stay below $W$ which means it must eventually become periodic.

Have corrected the explanation of the upper bound $W$. Hope the rest should be correct.


For $\alpha = 49,$ we can find $u_0 = \left(2^53^45^{10}11^2\right)^\alpha$ yields a $20$-periodic sequence.

Specifically, $u_n^{1/\alpha}$ for $0\leq n\leq 20$ is $$\begin{array}{c|c} n & u_n^{1/\alpha} \\ \hline 0 & 2^5\times 3^4\times 5^{10}\times 11^2 \\ 1 & 2\times 3^3\times 11\times 41\times 197\times 491 \\ 2 & 2^7\times 5^{10}\times 37 \\ 3 & 2^4\times 5^2\times 43\times 491 \\ 4 & 2^2\times 3^2\times 5^4\times 11\times 197 \\ 5 & 2^2 \times 3^4 \times 5^4 \times 11^2 \times 197 \\ 6 & 2 \times 3^4 \times 5^2 \times 11^2 \times 197^2 \\ 7 & 2 \times 3^6 \times 5^2 \times 11^3 \times 197 \\ 8 & 2^4 \times 3^2 \times 5^5 \times 11 \times 37 \times 59 \\ 9 & 2^4 \times 3^3 \times 5^6 \times 11 \times 41 \times 197 \\ 10 & 2^5 \times 5^7 \times 37 \times 59 \times 197 \\ 11 & 2^7 \times 3 \times 5^6 \times 41 \times 43 \\ 12 & 2^6 \times 5^7 \times 43 \times 59 \\ 13 & 2^5 \times 5^5 \times 43 \times 59 \\ 14 & 2^4 \times 3^2 \times 5^4 \times 41^2 \\ 15 & 3^4 \times 11^2 \times 197^2 \\ 16 & 3^4 \times 11^2 \times 197 \\ 17 & 2 \times 3^2 \times 5^2 \times 11 \times 197 \\ 18 & 2^3 \times 3^4 \times 5^6 \times 11^2 \\ 19 & 2^2 \times 3^2 \times 5 \times 11 \times 37 \times 59 \times 197 \\ 20 & 2^5\times 3^4\times 5^{10}\times 11^2 \end{array}$$

It still might be interesting to consider the relationship between $\alpha$ and the possible periods, but I don't think there's an upper bound on the period that is independent of $\alpha$ itself.

  • $\begingroup$ Thanks ! How did you find such values of $u_0$ and $\alpha$ ? Do you have a particular method or was it by chance ? $\endgroup$ – Mishikumo2019 Aug 10 at 9:17
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    $\begingroup$ For $\alpha$, I just chose one such that $\alpha+1, 2\alpha+1, 3\alpha+1$ are "fairly" composite, since the more composite they are, the greater chance to have larger exponents. I wasn't aware $u_n^{1/\alpha}$ was going to have prime powers as high as $10$ until I started, so there perhaps is some degree of luck there. I wouldn't know without trying other values. $\endgroup$ – Brian Moehring Aug 10 at 13:13
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    $\begingroup$ As for $u_0$? I actually chose a less exotic-looking initial value (i.e. one of $2, 4, 6$---can't remember which) and then just ran the sequence until the sizes of exponents in the prime factorization seemed to stabilize. Only then did I start writing down the sequence. Since this means I might have run through the entire period once already, the earliest might not have even been my eventual choice of $u_0,$ but it really matter which element you choose inside the period itself. $\endgroup$ – Brian Moehring Aug 10 at 13:19

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