Sum the Divisors and Go Again

Begin with a positive whole number and list the smaller positive whole numbers that divide it without a remainder. These are its proper divisors; the number itself is excluded. For 12, the list is 1, 2, 3, 4 and 6, whose sum is 16.

Use that sum as the next starting value. The proper divisors of 16 are 1, 2, 4 and 8, giving 15. Repeating produces 12 → 16 → 15 → 9 → 4 → 3 → 1 → 0. The sum for 1 is zero because it has no smaller positive divisors, and the laboratory ends the chain there rather than applying the divisor rule to zero.

Not every chain ends this way. The proper divisors of 28 add to 28, so it stays fixed. Those of 220 add to 284, and those of 284 add to 220, making a two-number cycle. Other starts, including 276, reach this laboratory's computation limit without resolving. That limit tells us where the calculation stopped, not what the chain will eventually do.

Two rough observations explain most of what you see, and neither of them is enough to settle anything.

Primes collapse instantly. A prime has exactly one proper divisor, namely 1, so any chain that lands on a prime is over in two more steps: prime, then 1, then 0. Since a majority of the numbers a chain passes through are not especially rich in divisors, most chains find their way onto something small fairly quickly and end.

Numbers with many divisors grow. The divisor sum of $n$ can easily exceed $n$. Take 12: its divisors add to 16, which is a third larger. Numbers like this are called abundant, and they are not rare, because having a lot of small factors is not rare. A chain that lands on an abundant number goes up, and if it keeps landing on abundant numbers it keeps going up.

Put those together and the picture is a tug of war with no referee. There is no quantity here that must decrease, which is the thing every other iteration on this site has and this one does not. Kaprekar's routine cannot leave four digits. The digit-power rule cannot outrun itself. Ducci's rule can only shrink. Each of those is trapped in a finite set, so it must end in a fixed point or a cycle, and the only remaining question is which one.

The aliquot rule has no such trap. A chain is free to wander upwards forever as far as anything elementary can tell, and that single missing guarantee is why the sentence "every chain ends or repeats" is a conjecture from 1888 rather than a two-line argument.

The sentence worth carrying away is this one.

A chain that passed the ceiling has been shown to climb. It has not been shown to escape.

Those are different claims and only the first one is true here. When the laboratory stops at the ceiling it prints how many steps it took and how high the chain got, and that is the entire content of the result. There is no step limit being read as an answer and no picture being read as a trend.

Three things on this page are being established in three different ways, and it is worth knowing which is which.

• The chain from 12 ends, and the chain from 138 ends after climbing to a hundred and eighty billion. Both are proved, because both are finite computations carried out to completion in exact arithmetic. Nothing is being trusted.
• Every start below 1,000 was run to a ceiling of a million million. Nine hundred and fifty-eight reach zero, nineteen settle on a perfect number, three enter the loop containing 220, and nineteen exceed the ceiling. That is a complete check of a finite list under stated bounds, and it says nothing whatever about what those nineteen do next.
• Whether every chain ends or repeats is open. It is cited, it is not proved here, and nothing on this page is evidence either way.

The temptation this page is built to resist is the second bullet quietly becoming the first. A hundred and seventeen steps of unbroken climbing looks like an answer. It is worth knowing that chains exist which climb for as long as you like and then stop climbing, so no amount of watching a chain go up can settle where it goes.

For a positive integer $n$, let $\sigma (n)$ be the sum of all divisors of $n$ including $n$ itself, and define the aliquot map

the sum of the proper divisors, with $s(1)=0$. The aliquot sequence of $n$ is $n,s(n),s(s(n)),\dots$, stopping if it reaches $0$.

1. If $n=p_1^{a_1}\cdots p_k^{a_k}$ then $\sigma (n)={\prod }_i\left(1+p_i+\cdots +p_i^{a_i}\right)$, and $\sigma$ is multiplicative on coprime arguments. Proved.
2. $s(p)=1$ for every prime $p$. Proved.
3. $n$ is a fixed point of $s$ exactly when $n$ is perfect. The only such $n<10^4$ are $6,28,496,8128$. Exhaustively checked over exactly that range.
4. The orbits of $220$, $12496$ and $14316$ are cycles of length $2$, $5$ and $28$. Proved, by exhibiting each cycle.
5. $s^{178}(138)=0$, and ${\max }_k{s}^k(138)=179,931,895,322$, attained at $k=117$. Proved, by carrying out all 178 steps exactly.
6. $s^{57}(276)>10^{12}$. Proved, by carrying out those 57 steps exactly. No statement is made about $s^k(276)$ for $k>57$.
7. Every aliquot sequence is bounded. Open (Catalan 1888, Dickson 1913), and the smallest start for which it is unknown is $276$.

Statement 7 is the exploration. Statements 5 and 6 are what this page can establish on its own, and they are consistent with either answer to 7 without being evidence for either.

The divisor-sum formula. Every divisor of $n=p_1^{a_1}\cdots p_k^{a_k}$ is $p_1^{b_1}\cdots p_k^{b_k}$ with $0\le b_i\le a_i$, and each such choice occurs exactly once. Expanding the product ${\prod }_i(1+p_i+\cdots +p_i^{a_i})$ produces exactly one term for each such choice, so the product is $\sigma (n)$. Multiplicativity on coprime arguments follows, since coprime numbers share no prime and the two products are over disjoint sets of primes. ∎

Primes. A prime $p$ has divisors $1$ and $p$ only, so $\sigma (p)=p+1$ and $s(p)=1$. The next step is $s(1)=0$, so any chain reaching a prime is over within two further steps. ∎

The cycles. Run the rule. $s(220)=284$ and $s(284)=220$, so the pair repeats. From $12496$ the chain is $14288,15472,14536,14264,12496$, closing after five steps. From $14316$ it closes after twenty-eight. In each case a value has occurred twice, and since $s$ depends on nothing but the number it is given, everything after the repeat repeats too. Exhibiting the cycle is the whole proof, and this is the one place on the page where a finite computation settles an infinite statement outright. ∎

138 and 276. Both are single finite computations in exact integer arithmetic, and the laboratory carries them out in full. For 138 the chain reaches $0$ after 178 steps, so the statement "138 terminates" is established. For 276 the chain exceeds $10^{12}$ after 57 steps, so the statement "$s^{57}(276)>10^{12}$" is established. Note carefully that the second computation proves a statement about step 57 and stops there. ∎

What is not proved here. That 276 is unbounded, that it is bounded, or anything about the nineteen unresolved starts below 1,000 beyond the height they reached. It is worth being explicit about why the evidence does not accumulate: arbitrarily long strictly increasing aliquot chains are known to exist, so a chain that has climbed for a hundred steps and a chain that will climb forever are not distinguishable by watching either of them climb. This is not a limitation of the ceiling chosen here. Raising the ceiling moves the number at which the computation stops and changes nothing about what stopping means.

What this laboratory computes. Chains from any start up to a million, followed for at most 400 steps, with a height ceiling you choose from a million, a thousand million or a million million. Divisor sums are computed by trial-division factorisation in exact integer arithmetic, so every number printed is exact. Trial division is the slow way to factorise, which is why the ceiling exists at all: it bounds how large a number the page will ever be asked to take apart. The endings view enumerates every start below 10,000, 20,000 or 100,000 in a background thread with a working progress bar and a cancel that really stops it.

The window is not the space. Every other state-space figure on this site enumerates something genuinely closed. This one cannot, because a chain is free to leave any window drawn around it, and about a fifth of the starts below 10,000 do exactly that. Those starts are drawn apart from the endings rather than sorted into one, because sorting them would be reporting where the range was cut as though it were a property of the rule. Widen the window and the counts all change, which is the honest way to notice that they were never counts about numbers in general.

Where the mathematics is still moving. The Catalan-Dickson conjecture is open. Five independent unresolved starts below 1000, 276, 552, 564, 660 and 966, are called the Lehmer five, and 276 has been pursued to chains of well over two thousand terms and numbers of more than two hundred digits without resolving. Guy and Selfridge suggested the opposite conjecture, that many such chains are in fact unbounded. This page takes no side, computes nothing that would let it, and stops at the ceiling.

Read this next to the Collatz orbit atlas. The two rules look alike from a distance: both climb, both fall, both have a universal statement nobody has proved. The difference is where the uncertainty sits. Every Collatz orbit anyone has tested has returned to 1, so the conjecture there is a statement about numbers nobody has reached. Here the unresolved cases are small enough to write on your hand, and 276 has been resisting since before anyone had a computer to resist with.