Divisors that add to three times the number

A perfect number is one whose divisors, itself included, add up to twice the number: $1 + 2 + 3 + 6 = 12 = 2 \times 6$. Numbers that are their own parts builds every even one from a Mersenne prime and leaves the odd ones as an open question. The definition has an obvious generalisation that the seventeenth century pursued with enthusiasm: why stop at twice?

The divisors of $120$ are $1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120$, and they add up to $360$ — exactly three times $120$. Such a number is called multiperfect, or $k$-perfect when its divisors add to $k$ times it. Pierre de Fermat found the next three-perfect number, $672$, and in 1638 René Descartes, in a letter to Marin Mersenne, announced numbers whose divisors add to four and five times themselves. The game was to find more, and it was played without any theory of why the numbers existed.

The theory, once there is one, is short. It rests on the ratio

$$I(n) = \frac{\sigma(n)}{n},$$

the abundancy of $n$ — the sum of its divisors measured in units of $n$ — which why a quarter of numbers overshoot uses to sort numbers into deficient and abundant. A $k$-perfect number is one whose abundancy is exactly the whole number $k$.

A game played by letter #

The first three-perfect number on record is $120$, noted by the English mathematician Robert Recorde in 1557 without any sign of how he found it. The subject came alive eighty years later in the correspondence that Marin Mersenne, a friar in Paris, kept up with most of the mathematicians of Europe. Mersenne posed the problem of finding numbers whose divisors add to a given multiple, and the answers travelled by letter: Fermat sent $672$ and $523{,}776$ in 1637, and Descartes, in 1638, sent three more three-perfect numbers as well as the four-perfect $30{,}240$ and the five-perfect $14{,}182{,}439{,}040$, with the remark that he could find as many as anyone wanted.

He could not, and nobody else could either without a method. The finds of the seventeenth century were made by the chains described below, followed by hand with considerable skill, and the next substantial additions came in the early twentieth century from Robert Carmichael, Thomas Mason and Derrick Lehmer, and in the late twentieth century from computers running the same method at scale. The sixth three-perfect number, $51{,}001{,}180{,}160$, belongs to that later period of hand computation. After it the list of multiple three stopped growing, though the search for larger multiples went on.

Two to six times themselves #

The table lists the first few of each multiple from two to six. The three-perfect numbers are $120$, $672$, $523{,}776$, $459{,}818{,}240$, $1{,}476{,}304{,}896$ and $51{,}001{,}180{,}160$ — six of them, all found by 1929, and none since. The four-perfect numbers begin $30{,}240$ and $32{,}760$; the five-perfect with $14{,}182{,}439{,}040$; the six-perfect with a number of twenty-one digits. Each row was checked, not copied: the figure computes $\sigma(n)$ exactly from the factorisation, as a whole number with no rounding, and confirms that it is $k$ times $n$.

Two things stand out. Every number in the table is even, and every one has a large power of two in it — $2^3$ in $120$, $2^{15}$ in the six-perfect number. And the numbers of distinct primes grow with the multiple: two for the perfect numbers, three or more for three, four or more for four, seven for the first five-perfect, eleven for the six. Both have explanations, and both come from the same formula.

The abundancy is a product over the primes #

The divisors of $n = p_1^{a_1} p_2^{a_2} \cdots$ are all the products of one power of each prime, so their sum factorises:

$$\sigma(n) = \prod_i \bigl(1 + p_i + p_i^2 + \dots + p_i^{a_i}\bigr),$$

and the abundancy is a product of one factor for each prime power:

$$I(n) = \prod_i \Bigl(1 + \frac{1}{p_i} + \dots + \frac{1}{p_i^{a_i}}\Bigr).$$

That is the fact the shape of a number’s divisors draws as a box with one axis per prime. The factor for $p^a$ is a partial geometric series, larger than one and smaller than $p/(p - 1)$, the sum of the whole series.

On a logarithmic scale the factors add, and the figure lays them end to end. For $120 = 2^3 \cdot 3 \cdot 5$ the power of two contributes $15/8$, the three $4/3$ and the five $6/5$, and $\tfrac{15}{8} \cdot \tfrac43 \cdot \tfrac65 = 3$ exactly. For $30{,}240 = 2^5 \cdot 3^3 \cdot 5 \cdot 7$ four segments reach four. For the five-perfect number seven segments reach five, and the last few are slivers: the factor for $17$ is $18/17$, barely more than one.

That is the reason the prime counts grow. Each prime can multiply the abundancy by less than $p/(p - 1)$, which for $p = 2$ is two, for $p = 3$ one and a half, and for large $p$ only a hair above one. Reaching a large multiple means collecting many such hairs.

How many primes a multiple needs #

The largest abundancy a number with $r$ distinct primes can approach is the product of $p/(p - 1)$ over the $r$ smallest primes, since smaller primes contribute more.

The products climb $2, 3, 3.75, 4.375, 4.81, 5.21, \dots$, and a number can only reach abundancy $k$ if the product for its number of primes is above $k$. So a perfect number needs two primes, a three-perfect number three, a four-perfect four, a five-perfect six and a six-perfect nine. Every example in the table has at least that many.

For odd numbers the bound bites much harder, because the prime $2$, with its factor of two, is excluded. The product over odd primes climbs $1.5, 1.875, 2.19, \dots$ and reaches two only at the third prime: an odd perfect number needs at least three distinct primes. That elementary bound was raised by James Joseph Sylvester in 1888 and by computation since to at least ten. An odd three-perfect number would need eight, an odd four-perfect twenty-one, an odd six-perfect one hundred and forty-one. None is known, of any multiple.

The growth is extremely slow. By a theorem of Franz Mertens, the product of $p/(p - 1)$ over primes up to $x$ grows like $e^{\gamma} \ln x$, where $\gamma = 0.5772\ldots$ is Euler’s constant. To reach abundancy $k$ the primes must run up to about $e^{k e^{-\gamma}}$, and a number built from all of them is doubly exponential in $k$ — which is why the six-perfect number has twenty-one digits and the largest multiples known, eleven, belong to numbers with hundreds.

Chains that close #

The factorisations in the table are not arbitrary, and one way to see their structure is to follow the divisor sums.

For $n$ to be $k$-perfect, the product of the divisor sums of its prime powers must be $k$ times $n$ — so the primes those sums contain must be, apart from the factor $k$, exactly the primes of $n$. For $120$: $\sigma(2^3) = 15 = 3 \cdot 5$ produces a $3$ and a $5$; $\sigma(3) = 4$ and $\sigma(5) = 6$ produce $2^2$ and $2 \cdot 3$. Altogether $15 \cdot 4 \cdot 6 = 2^3 \cdot 3^2 \cdot 5 = 3 \times 120$ — the primes of $120$, with one extra $3$.

That is how the numbers were found by hand. Start with a power of two, factor its divisor sum $2^{a+1} - 1$, add those primes to the number, factor their divisor sums, add those primes, and try to arrange that everything closes up with the right powers. The Mersenne numbers $2^{a+1} - 1$ are where every chain starts, which is why every known multiperfect number carries a large power of two: $2^{a+1} - 1$ is the only way to introduce odd primes cheaply, and the odd primes’ own sums, being even, pay back the twos. The perfect numbers are the shortest chains of all, $2^a$ and one prime $2^{a+1} - 1$ closing immediately.

A sieve to a million #

A search can confirm that nothing was missed below some bound.

Computing $\sigma(n)$ for every $n$ up to a million by adding each divisor to each of its multiples, and testing which $\sigma(n)$ are multiples of $n$, finds exactly nine numbers besides $1$: the four perfect numbers below a million, three of the six three-perfect numbers, and the two smallest four-perfect ones. The scatter of all abundancies shows why they are rare. The cloud of values fills the region from one to about four and a half, rising slowly with $n$ as Mertens’s product allows, and a whole number is a single horizontal line through a cloud of hundreds of thousands of fractions. Only nine points in a million land on one exactly.

The sieve is the same kind of computation as the sieve of Eratosthenes, run with addition instead of crossing out: for every $d$ up to a million, add $d$ to $\sigma$ of each of its multiples. It touches each number once for every divisor it has, about fourteen million additions in all, and it finds every multiperfect number in its range because it tests them all — no chain, no guess, no structure assumed. The price is that it stops at its bound, and the next three-perfect number after $523{,}776$ is nearly nine hundred times larger.

How large an abundancy can get, and a connection to the Riemann hypothesis #

The prime-count argument bounds the abundancy of a number by a product over its primes, and it can be turned into a bound in terms of the number itself. A number with many small primes is the most abundant for its size, and Thomas Gronwall proved in 1913 that the abundancy of $n$ can grow no faster than $e^\gamma \ln \ln n$ — and does grow that fast, along the right sequence of numbers:

$$\limsup_{n \to \infty} \frac{\sigma(n)}{n \ln \ln n} = e^\gamma = 1.781\ldots$$

So the multiple a number’s divisors can make of it grows like the logarithm of the logarithm of the number — about $5$ for numbers of twenty digits, about $9$ for numbers of a million digits. Multiperfect numbers of large multiple are necessarily astronomically large, and that is the real reason the table’s rows get so long so quickly.

The bound hides one of the most surprising facts in elementary number theory. Guy Robin proved in 1984 that the inequality

$$\sigma(n) < e^\gamma\, n \ln \ln n \quad \text{for every } n > 5040$$

holds if and only if the Riemann hypothesis is true. The largest number that breaks it is $5040 = 7!$, and it has been checked for enormous ranges of $n$ above that. A statement about adding up divisors — the operation that defines perfect numbers — is equivalent to the most famous open question about the zeros of a complex function, because both are statements about how regularly the primes are spread, and the numbers that come closest to breaking Robin’s inequality are exactly the ones built, like the multiperfect numbers, from long runs of small primes.

Why the powers of two are so large #

The chain picture explains one more feature of the table, the size of the powers of two. A multiperfect number $2^a m$ with $m$ odd must have $\sigma(2^a) = 2^{a+1} - 1$ built entirely from primes that appear in $m$, and every odd prime’s divisor sum returns powers of two. The two budgets have to balance: the total power of two produced by the odd primes’ sums, less the factor in $k$, must be exactly $a$. An odd prime $p$ appearing once contributes $\sigma(p) = p + 1$, which is even, so a number with many odd primes pays back many twos, and $a$ has to be large enough to absorb them — $a = 15$ for the six-perfect number, whose ten odd prime powers return sixteen factors of two between them: fifteen to rebuild $2^{15}$ and one for the two in the multiple six. The single prime $31$ returns five of them, since $\sigma(31) = 32$, and the prime $257$ returns one. A chain is, in this light, an accounting problem: every prime a divisor sum introduces must be paid for in the number, and every power of two the odd primes throw back must be absorbed by the power of two the chain began with. Large multiples need many odd primes, many odd primes return many twos, and so the power of two grows with the multiple.

Where the aliquot sequence sends them #

The sum of the parts, taken again iterates a different map, $n \mapsto \sigma(n) - n$, the sum of the proper divisors, and finds that perfect numbers are its fixed points. A $k$-perfect number with $k \ge 3$ is not a fixed point of that map; it is sent to $(k - 1)n$, a number $k - 1$ times as large, and its sequence climbs. So $120$ goes to $240$, which is abundant and climbs further, and the multiperfect numbers — so orderly under $\sigma$ itself — are not special under the aliquot map at all. It is a reminder that “divisors adding up to a multiple” and “the aliquot sequence returning” are different conditions that happen to coincide at the multiple two.

The two maps share their growth bound, though. The aliquot map can multiply a number by at most about $e^\gamma \ln \ln n$ in one step, by Gronwall’s theorem, and the harmonic-series estimate behind Mertens’s product — the same slow logarithm as the sum that never stops growing — is what sets that rate. Divisor sums are harmonic sums over the divisors, and every question about how large they get comes back to how slowly $1 + \tfrac12 + \tfrac13 + \cdots$ grows.

What is known about the lists #

The three-perfect list is believed complete and not proved so. Six were known by 1929 and exhaustive searches since have found no seventh, but nothing rules out a larger one. The situation is similar for four-perfect numbers, of which thirty-six are known, and the lists grow with the multiple: more than five thousand multiperfect numbers are known in all, with multiples up to eleven, most found in the late twentieth century by the chain method run on computers.

Three things are known in general. Every perfect number that is even has Euclid’s form, as the earlier essay proves, built from one of the Mersenne primes whose supply is tied to the infinitude of primes in general only by hope; for multiperfect numbers there is no such characterisation. For odd perfect numbers, Leonard Dickson proved in 1913 that only finitely many can have a given number of distinct primes — a chain of bounded length can close in only finitely many ways when no Mersenne prime is available to start it — which is what lets computer searches rule out small cases completely. And no odd multiperfect number of any multiple has ever been found.

What the pictures cannot show #

That the lists are complete. The table is a selection of known numbers, each checked; the sieve is complete below a million and nowhere else. Neither says there are only six three-perfect numbers, which is unproved.

The chain search itself. The chain figure shows two chains that close. The method that finds them — trying prime powers, factoring their sums, backtracking when the primes run away — explores enormous trees of chains that do not close, and none of that is drawn.

Robin’s inequality. Nothing drawn here tests it, and no finite check could establish it: it is equivalent to the Riemann hypothesis, and the evidence for both is computation that stops somewhere. The figures show abundancies of numbers up to a million, far below where the inequality is interesting.

Mertens’s growth. The primes figure shows fourteen primes; the statement that the product grows like $e^\gamma \ln x$, and the doubly exponential size it forces, is a theorem about all primes, quoted here.

Still open: an odd one, of any multiple #

Every known multiperfect number is even, and the question the perfect numbers left open — whether an odd perfect number exists — generalises to every multiple: is there an odd $k$-perfect number for any $k \ge 2$? None is known, and none is ruled out.

The prime-count figure shows why the odd case is hard to settle in either direction. An odd number needs many more primes to reach any given abundancy, so an odd multiperfect number, if one exists, must be very large and have many distinct prime factors — for an odd perfect number, more than $10^{1500}$ and at least ten distinct primes, by searches that close off every smaller case. Nothing in the arithmetic of chains forbids odd ones outright; the Mersenne primes that make even chains easy to start simply are not available. Whether that makes odd multiperfect numbers impossible, or merely rare beyond any search, is the question, and it has been open since Descartes.