A ratio nobody else has

Divisors that add to three times the number studies the numbers whose abundancy — the sum of the divisors divided by the number, $I(n) = \sigma(n)/n$ — is a whole number. Most abundancies are not whole numbers. They are fractions: $I(10) = 18/10 = 9/5$, $I(12) = 28/12 = 7/3$, $I(30) = 72/30 = 12/5$. And a fraction invites the question the whole-number case already answered in its own way. The perfect numbers all share the abundancy $2$; which other numbers share an abundancy with someone?

Two numbers with the same abundancy are called friends, and the name, given in the 1970s by Charles Anderson and Dean Hickerson, makes the question sound social. A number with at least one friend is friendly. A number with none is solitary. The perfect numbers are all friends of one another. $30$ and $140$ are friends, both $12/5$. What makes the subject strange is how quickly it runs into problems that nobody can settle: the smallest number of unknown status is $10$.

Friends, drawn #

The cloud of grey dots is every number up to $600$ placed by its abundancy, and the orange segments join numbers with exactly the same value. There are eleven such groups below $600$. The perfect numbers $6$, $28$ and $496$ make one, on the line at two. $12$ and $234$ both have abundancy $7/3$: $\sigma(12) = 28$ and $\sigma(234) = 546$, and $546/234$ reduces to $7/3$. $30$ and $140$ share $12/5$, $40$ and $224$ share $9/4$, $66$ and $308$ share $24/11$.

The pattern in the pairs is visible in the figure: the partners are far apart, and the larger is typically several times the smaller. That is not an accident. Two numbers can only share an abundancy if they are built quite differently — the smaller from few primes with low exponents, the larger from more — so that different combinations of the factors $1 + 1/p + \dots$ happen to multiply to the same fraction. A number’s friend, when it has one, is usually a much larger number, and the smaller the numbers searched, the fewer friendships appear. Below $600$ the grey dots outnumber the orange ones by more than twenty to one.

Clubs below a million #

Running the same comparison over every number up to a million groups them by their abundancy written in lowest terms.

Below a million there are $16{,}828$ shared abundancies. Nearly sixteen thousand are shared by exactly two numbers; six hundred and fifty-three by three; a hundred and ninety-three by four; seventeen by five. The largest club includes $30$ and $140$ — its abundancy is $12/5$ and it has five members, $30$, $140$, $2{,}480$, $6{,}200$ and $40{,}640$. The perfect numbers make a club of four, the most famous club and not the largest.

Clubs like $12/5$ grow for a reason that follows from the product formula. Abundancy is multiplicative: if $n$ and $m$ share no prime, $I(nm) = I(n)\,I(m)$. So multiplying every member of a club by the same number coprime to all of them gives another club. The $12/5$ club is mostly the perfect numbers multiplied by $5$: $30 = 6 \cdot 5$, $140 = 28 \cdot 5$, $2{,}480 = 496 \cdot 5$, $40{,}640 = 8{,}128 \cdot 5$, each with abundancy $2 \cdot \tfrac65$. Its fifth member, $6{,}200 = 2^3 \cdot 5^2 \cdot 31$, is built differently and arrives at $12/5$ by coincidence, as $\tfrac{15}{8} \cdot \tfrac{31}{25} \cdot \tfrac{32}{31}$. Friendship propagates by multiplication and occasionally appears by accident, and the clubs are the record of both.

Why a multiple always has more #

The argument about solitary numbers below rests on one fact that is worth seeing on its own: a proper multiple of $n$ has a strictly larger abundancy than $n$. Every divisor $d$ of $n$ gives a divisor $kd$ of $kn$, and

$$I(kn) = \frac{\sigma(kn)}{kn} \ge \frac{1}{kn}\sum_{d \mid n} kd = \frac{\sigma(n)}{n} = I(n),$$

with strict inequality because $kn$ has at least one more divisor, $1$, not of the form $kd$. It is the same inheritance that makes every multiple of an abundant number abundant, where it sorts the numbers into those above and below two. Here it does something sharper: it says that along any chain of multiples $n, 2n, 4n, \dots$ the abundancy only climbs, so a number can never be a friend of its own multiple. Friends are always numbers neither of which divides the other.

Seen on the lattice of a number’s divisors, the inequality is a statement about boxes: multiplying $n$ by a prime makes its box of divisors longer in one direction, and the abundancy, being a sum of $1/d$ over the box, gains the new terms. It never loses any.

Infinitely many of each #

Both kinds are infinite, and each for a one-line reason. There are infinitely many solitary numbers because every prime is one, and there is no last prime. There are infinitely many friendly numbers because multiplicativity manufactures them: take any $m$ that shares no prime with $6$ or $28$ — that is, any $m$ not divisible by $2$, $3$ or $7$ — and $6m$ and $28m$ have abundancies $2\,I(m)$ and $2\,I(m)$, so they are friends. $6 \cdot 5 = 30$ and $28 \cdot 5 = 140$ are the first such pair, and $6 \cdot 11 = 66$ and $28 \cdot 11 = 308$ the next — both pairs appear in the first figure, joined. The same move works with any two friends in place of $6$ and $28$, so every club spawns infinitely many others, one for each admissible multiplier; the friendly numbers are therefore not a scattering of coincidences but a structure that reproduces itself.

What is not known is whether any particular club is infinite. The perfect numbers form one, and whether it has infinitely many members is exactly whether there are infinitely many Mersenne primes, a question open since antiquity. The clubs manufactured by multiplication are copies of smaller clubs, each as large as the club it was copied from and no larger. No club has been proved to have infinitely many members.

A number that provably has no friend #

Some numbers can be shown to be solitary by a one-line argument.

Suppose $I(n) = a/b$ in lowest terms. Since $\sigma(n)/n = a/b$, the denominator $b$ divides $n$. Now suppose $\sigma(n)$ and $n$ have no common factor: then $\sigma(n)/n$ is already in lowest terms, so $b = n$. If $m$ is a friend of $n$, then $\sigma(m)/m = \sigma(n)/n$, and the denominator $n$ must divide $m$ — so $m$ is a multiple of $n$. But a proper multiple of $n$ has a strictly larger abundancy, because its divisors include every divisor of $n$ scaled up, with more besides. So $m = n$, and $n$ has no friend. Every number coprime to the sum of its own divisors is solitary.

Primes pass the test, since $\sigma(p) = p + 1$ shares no factor with $p$; so do prime powers, since $\sigma(p^a) = 1 + p + \dots + p^a$ leaves remainder one on division by $p$. So do many others — $35$, $49$, $50$, $98$.

The two tests between them decide sixty-three of the first hundred numbers. Twelve have a friend below a million: $6, 12, 28, 30, 40, 42, 56, 60, 66, 78, 80, 84$. Fifty-one are coprime to their divisor sums and so are provably solitary. The remaining thirty-seven are red: neither test says anything about them.

Some of those thirty-seven have friends too large for the search. $24$ has abundancy $60/24 = 5/2$, and its friend is $91{,}963{,}648$ — a number that also has abundancy $5/2$ and would appear only in a search to a hundred million. Others are believed solitary on the evidence of long searches and not proved so. The first red number is $10$.

How common the proof of solitude is #

The coprimality test proves solitude for fifty-one of the first hundred numbers. It proves it for a smaller share as the numbers grow: $40.2\%$ of the first thousand, $36.0\%$ of the first ten thousand, $34.0\%$ of the first hundred thousand and $32.9\%$ of the first million. The share falls because $\sigma(n)$, a product of factors $1 + p + \dots + p^a$, tends to pick up small primes, and a large $n$ usually has small primes too, so the two are less and less often coprime. The decline is slow, as the counts show, and it is not the case that the test stops working; it simply decides a shrinking fraction of the cases while the undecided remainder grows.

Friendship, on the other side, is proved one friend at a time, by finding the friend, and the searches that do it are the same kind as the searches that follow aliquot sequences until they close or give up. Between the two methods there is no third: a number is proved friendly by an example, proved solitary by a divisibility argument, and otherwise left undecided — which, for a great many small numbers, is where the subject stands.

A different kind of friendship #

The word “friend” was used first for a different relation, and the contrast is instructive. Amicable numbers are pairs like $220$ and $284$, which the essay on perfect numbers mentions: the proper divisors of each add up to the other. In terms of $\sigma$, $\sigma(220) = \sigma(284) = 220 + 284 = 504$. Amicable numbers share a divisor sum, and it equals their total; friendly numbers share a divisor sum relative to the number. The two conditions have nothing to do with each other: $220$ has abundancy $504/220 = 126/55$ and $284$ has $504/284 = 126/71$, different fractions.

Both relations make the perfect numbers special, from opposite directions. A perfect number is amicable with itself, since its proper divisors add to it; and all perfect numbers are friends of one another, since they share the abundancy two. The aliquot map of the earlier essay sends amicable pairs round a cycle of length two and perfect numbers to themselves; the abundancy sends every perfect number to the same fraction. The questions left open are also parallel — whether there are infinitely many amicable pairs, and whether $10$ has a friend — and both come down to equations in $\sigma$ that no method can solve except by search.

What a friend of 10 would have to be #

$I(10) = \sigma(10)/10 = 18/10 = 9/5$. A friend of $10$ is a number $m \ne 10$ with $5\,\sigma(m) = 9m$. The one-line test does not apply, since $\sigma(10) = 18$ shares the factor $2$ with $10$. But the equation forces a lot.

Since $5\,\sigma(m) = 9m$ and $5$ does not divide $9$, $5$ divides $m$. If $m$ were even it would be a multiple of $10$, and every proper multiple of $10$ has abundancy above $9/5$; so $m$ is odd. Then $\sigma(m) = 9m/5$ is odd. The divisor sum of $m$ is odd exactly when $m$ is a square or twice a square — it is odd when $m$ has an odd number of odd divisors, and the odd divisors pair off as $d$ and $m'/d$ for the odd part $m'$ of $m$ unless $m'$ is a square — and $m$ is odd, so $m$ is an odd square, divisible by $25$.

The rest is computation. Searches that run through the possible prime factorisations of such an $m$, closing off each small case the way the searches for odd perfect numbers do, have shown that a friend of $10$ would have to exceed $10^{30}$. The figure’s own direct search checks every multiple of $5$ up to a million and finds only $10$ itself. Nothing rules a friend out, and nothing suggests where one would be.

The problem has the same shape as the odd perfect number problem, one step removed. An odd perfect number is an odd number with abundancy exactly $2$; a friend of $10$ is an odd square with abundancy exactly $9/5$. Both are equations of the form $\sigma(m)/m =$ a given fraction, with the easy even solutions excluded, and both resist every method except searching and bounding.

Fractions that are nobody’s abundancy #

The reverse question also has content: which fractions are the abundancy of anything at all?

The argument that proved solitude also proves impossibility. If $I(n) = a/b$ in lowest terms, then $b$ divides $n$, so $I(n) \ge I(b)$, because a number’s abundancy is at least that of each of its divisors. So any fraction $a/b$ smaller than $I(b)$ is not the abundancy of anything. $5/4$ is the simplest example: its denominator is $4$, so $n$ would be a multiple of $4$, and every multiple of $4$ has abundancy at least $I(4) = 7/4$, far above $5/4$. Every fraction below $3/2$ with an even denominator is excluded the same way, since every even number has abundancy at least $I(2) = 3/2$.

Such fractions have been called abundancy outlaws. Below two with denominators up to twelve the figure finds eighteen fractions that are some number’s abundancy, fifteen ruled out by the divisor argument, and fourteen that neither a search to a million nor the argument settles. The abundancies of all numbers are known to be dense in the numbers above one — every interval, however small, contains one, a theorem of Richard Laatsch from 1986 — so the outlaws are not a region but a scatter of individual fractions, and deciding which fractions they are is, like deciding which numbers are solitary, largely open.

What the pictures cannot show #

Friendship beyond a million. Every “friendly” verdict here comes from a search up to a million, and every red cell is a number whose friend, if it has one, lies beyond it. The figure calls $24$ undecided although its friend $91{,}963{,}648$ is known, because the search does not reach it; the red colour means “not decided by these tests”, not “unknown to mathematics”.

The squares condition. The chain for $10$ uses the fact that $\sigma(m)$ is odd exactly for squares and twice squares. The figure checks that fact for every $m$ up to two thousand and proves it in a sentence; it is true for all $m$ by the pairing argument, not by the check.

Whether the undecided numbers are solitary. The red cells are a statement about two tests, not about the numbers. For most of them the evidence of large searches points towards solitude, and no amount of it is a proof; for a few, like $24$, the evidence was misleading until a search went far enough.

Density. Laatsch’s theorem that abundancies are dense is quoted. The figures show abundancies of numbers up to a million, which fill the range they cover thickly but say nothing about arbitrarily small intervals.

Still open: whether 10 has a friend #

The status of $10$ is the cleanest open question in the subject, and it has been open since the terms were coined. It is not known whether there is an $m$ with $\sigma(m)/m = 9/5$ other than $10$. The same is true of $14$, $15$, $20$, $22$, $26$ and many others in the red cells — numbers that are believed, on the evidence of searches, to be solitary, with no proof available for any of them.

The underlying difficulty is the one the odd perfect numbers share. To prove a number solitary one must show an equation $\sigma(m) = (a/b)\, m$ has no solution except the obvious one; the coprimality argument does that when the equation’s denominator is the whole number, and for every other number it leaves an equation that looks just like the one defining odd perfect numbers, with a different right-hand side. The multiperfect numbers of the previous essay are the same kind of problem with a whole number on the right, and they show what a solution tends to look like when one exists: a long chain of primes whose divisor sums supply one another. A friend of $10$ would be such a chain for the fraction $9/5$, built entirely from odd primes, with no Mersenne prime to start it — which is the reason the searches have found nothing, and also the reason nobody can prove there is nothing to find. In a subject where adding up divisors is the only operation, $10$ is the first number about which nobody can say whether it is alone.