The bounded descent conjecture for the recursive divisor-sum sequence

Unbounded descent times for the recursive divisor-sum map

In On a Recursive Integer Sequence Implying the Nonexistence of Odd Perfect Numbers, Ritesh Dwivedi and Rohit Yadav define

$$\mathfrak R(n)= \begin{cases} \sigma(n),&n\text{ odd},\\ n/2,&n\text{ even}, \end{cases} \qquad \sigma(n)=\sum_{d\mid n}d. \tag{1}$$

Their Conjecture 3.6 asserts the existence of a constant $c\geq1$ such that every positive integer $n$ satisfies

$$\mathfrak R^j(n)\leq n \qquad\text{for some }1\leq j\leq c. \tag{2}$$

This uniform bounded-descent assertion is false. In fact, an exact formula for the first descent time holds for every odd squarefree integer and produces arbitrarily long counterexamples.

For $n>1$, define

$$\tau(n)=\min\{j\geq1:\mathfrak R^j(n)\leq n\}, \tag{3}$$

whenever the indicated set is nonempty.

Theorem. If $n>1$ is odd and squarefree, then

$$\boxed{\displaystyle \tau(n)=1+ \left\lceil\log_2\frac{\sigma(n)}{n}\right\rceil.} \tag{4}$$

In particular, the values of $\tau(n)$ on odd squarefree integers are unbounded.

Write

$$n=p_1p_2\cdots p_r, \qquad p_1,\ldots,p_r\text{ distinct odd primes}. \tag{5}$$

Multiplicativity gives

$$\sigma(n)=\prod_{i=1}^r(p_i+1), \qquad Q:=\frac{\sigma(n)}{n} =\prod_{i=1}^r\left(1+\frac1{p_i}\right). \tag{6}$$

Each factor in the second product lies strictly between $1$ and $2$, and each factor in the first product is even. Consequently,

$$1<Q<2^r, \qquad \nu_2(\sigma(n)) =\sum_{i=1}^r\nu_2(p_i+1)\geq r. \tag{7}$$

Set

$$t=\lceil\log_2Q\rceil. \tag{8}$$

Then $1\leq t\leq r$, and (7) ensures that the first application of $\mathfrak R$ is followed by at least $t$ valid halving steps. Therefore

$$\mathfrak R^j(n)=\frac{\sigma(n)}{2^{j-1}} \qquad(1\leq j\leq t+1). \tag{9}$$

For $1\leq j\leq t$ we have $2^{j-1}<Q$, whereas $2^t\geq Q$. Hence

$$\mathfrak R^j(n)>n\quad(1\leq j\leq t), \qquad \mathfrak R^{t+1}(n)\leq n. \tag{10}$$

This proves (4), including the case in which $Q$ is an exact power of two.

To show that these descent times are unbounded, take the odd primorial

$$N_x=\prod_{3\leq p\leq x}p, \qquad Q_x=\frac{\sigma(N_x)}{N_x} =\prod_{3\leq p\leq x}\left(1+\frac1p\right). \tag{11}$$

The divergence needed here follows directly from the finite Euler product, without any prime-distribution asymptotic. Indeed,

$$Q_x =\prod_{3\leq p\leq x}\left(1-\frac1{p^2}\right) \prod_{3\leq p\leq x}\left(1-\frac1p\right)^{-1}. \tag{12}$$

The first product is bounded below by the elementary telescoping product

$$\prod_{3\leq p\leq x}\left(1-\frac1{p^2}\right) \geq\prod_{m=2}^{\infty}\left(1-\frac1{m^2}\right) =\frac12. \tag{13}$$

Expanding the second finite Euler product includes every positive odd integer at most $x$. Therefore

$$Q_x\geq \frac12\sum_{\substack{1\leq m\leq x\m\text{ odd}}}\frac1m \geq\frac14\sum_{m=1}^{\lfloor x\rfloor}\frac1m \longrightarrow\infty. \tag{14}$$

Now let $c\geq1$ be arbitrary. Choose $x$ large enough that

$$Q_x>2^{c-1}. \tag{15}$$

Formula (4) gives

$$\tau(N_x)=1+\lceil\log_2Q_x\rceil>c. \tag{16}$$

Equivalently, this single positive integer violates every proposed descent step simultaneously:

$$\mathfrak R^j(N_x)>N_x \qquad(1\leq j\leq c). \tag{17}$$

Since the proposed constant $c$ was arbitrary, no universal bounded-descent constant exists, and Conjecture 3.6 is disproved.

The distinct Conjecture 3.5 asks whether every orbit eventually reaches $1$. Unbounded first-descent times do not imply the existence of a nonconvergent orbit, so this counterexample makes no claim concerning that separate conjecture or the existence of odd perfect numbers.