Divisors that take a prime whole

Numbers that are their own parts began where Euclid did: a number is perfect when its proper divisors add up to it, as 6 = 1 + 2 + 3 and 28 = 1 + 2 + 4 + 7 + 14. The even perfect numbers are exactly Euclid’s $2^{p-1}(2^p - 1)$ with $2^p - 1$ prime, as Euler proved; whether there are infinitely many is unknown, and whether an odd perfect number exists is one of the oldest open problems in mathematics. The shape of a number’s divisors drew why: the divisor sum $\sigma(n)$ is multiplicative, and an odd perfect number would need a very particular balance of prime powers that nobody has found and nobody has ruled out.

Change one word in the definition and the picture changes completely. Call a divisor $d$ of $n$ unitary when $d$ and $n/d$ have no common factor — when $d$ takes each prime of $n$ either completely or not at all. The unitary divisors of $60 = 2^2 \cdot 3 \cdot 5$ are $1, 3, 4, 5, 12, 15, 20$ and $60$: the divisor 2 is left out, because 2 and 30 share the factor 2, and so are 6, 10 and 30. A number is unitary perfect when its proper unitary divisors add up to it. M. V. Subbarao and L. J. Warren defined these numbers in 1966 and found the first four; Charles Wall found the fifth in 1975; nobody has found a sixth.

The unitary divisor sum #

The unitary divisors of $n = p_1^{a_1} \cdots p_k^{a_k}$ are the products of some subset of the whole prime powers $p_i^{a_i}$, so there are exactly $2^k$ of them, one for each subset, and their sum factors as

$$\sigma^*(n) = (1 + p_1^{a_1})(1 + p_2^{a_2}) \cdots (1 + p_k^{a_k}).$$

That is the same shape as the ordinary divisor sum, which is a product of $1 + p + p^2 + \cdots + p^{a}$ over the primes, with all the intermediate powers dropped. A unitary perfect number is one with $\sigma^*(n) = 2n$.

The product formula makes the check for 60 a one-liner: $(1 + 4)(1 + 3)(1 + 5) = 5 \cdot 4 \cdot 6 = 120 = 2 \cdot 60$. For 90 $= 2 \cdot 3^2 \cdot 5$ it is $(1 + 2)(1 + 9)(1 + 5) = 3 \cdot 10 \cdot 6 = 180$. For 6 it is $(1 + 2)(1 + 3) = 12$. No other number below a thousand comes out to exactly twice itself, and the reason is visible in the formula: the factors must supply exactly one more 2 than the number has, and each odd prime’s factor brings at least one. The hero computes $\sigma^*(n)/n$ for every $n$ from the same formula, using a sieve that records each number’s smallest prime factor, so that every $\sigma^*(n)$ costs a handful of multiplications. Over every $n$ below ten million the ratio equals 2 exactly four times.

A Boolean lattice instead of a grid #

The two kinds of divisor arrange themselves differently. The ordinary divisors of $n = p_1^{a_1}\cdots p_k^{a_k}$ form a grid: one coordinate for each prime, running from 0 to its exponent, so $60 = 2^2 \cdot 3 \cdot 5$ has a $3 \times 2 \times 2$ grid of twelve divisors. The unitary divisors form the corners of a cube: each prime power is in or out, so 60 has the $2 \times 2 \times 2$ cube of eight. Divisibility among unitary divisors is the inclusion of subsets, and the unitary divisors of any number with $k$ primes form the same Boolean lattice as the subsets of a $k$-element set, whatever the exponents are.

That uniformity is what makes $\sigma^*$ simpler than $\sigma$. The ordinary divisor sum of $p^a$ is a geometric series, $1 + p + \cdots + p^a$, whose prime factors are hard to predict — much of the theory of ordinary divisor sums is about controlling them. The unitary one is a single sum $1 + p^a$ of two terms. Its prime factors are still hard to predict in general, but its parity is not: for odd $p$ it is always even, and that one fact is the whole proof that odd unitary perfect numbers do not exist.

The cloud and the line #

The ratio $\sigma^*(n)/n$ behaves like the ordinary abundancy $\sigma(n)/n$, only lower: it is 1 for $n = 1$, just over 1 for primes and prime powers, and grows with the number of distinct primes, since each prime $p$ multiplies it by $1 + p^{-a}$. Below 100,000 the largest value is 3.2224, at $30{,}030 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$, the product of the first six primes, each to the first power, which is where every factor $1 + 1/p$ is as large as it can be. The cloud of values is dense near 2, crossing it thousands of times, and almost never landing on it.

That is the same experience as with ordinary perfect numbers — a continuous-looking cloud and an exact equation it almost never satisfies — and for the same reason. $\sigma^*(n) = 2n$ is a statement about prime factorisations, not about sizes: each factor $1 + p^a$ has its own primes, and every one of them must be one of $n$’s primes, to the right power, with the product coming out to exactly $2n$. A ratio of 2.0001 is no nearer to that than a ratio of 3.

The ratio has no upper limit. Each prime $p$ contributes a factor $1 + p^{-a} \le 1 + 1/p$, and the product of $1 + 1/p$ over the first primes grows without bound, like the logarithm of the logarithm of their product, so numbers built from many small primes have ratios as large as anyone likes. But the growth is glacial: to push the ratio past 4 takes the product of the first eleven primes, 200,560,490,130, a number of twelve digits, and every further unit costs vastly more, since each new prime adds only a factor of $1 + 1/p$. The four unitary perfect numbers below ten million all sit in the busy middle of the cloud, where most even numbers with three or more small primes live.

Their shapes are worth reading side by side. The number 6 is $2 \cdot 3$, with ledger $(1 + 2)(1 + 3) = 3 \cdot 2^2$. The number 60 is $2^2 \cdot 3 \cdot 5$, with $(1 + 4)(1 + 3)(1 + 5) = 5 \cdot 2^2 \cdot (2 \cdot 3)$. The number 90 is $2 \cdot 3^2 \cdot 5$, with $(1 + 2)(1 + 9)(1 + 5) = 3 \cdot (2 \cdot 5) \cdot (2 \cdot 3)$. In each, the odd part of each factor is another of the number’s primes, and the twos add up to one more than the number’s own power of 2. The fourth and fifth do the same thing with more primes and longer chains.

No odd number is unitary perfect #

Here the unitary world parts company with the ordinary one. Suppose $n$ is odd, with $k$ distinct prime factors. Every factor $1 + p^a$ of $\sigma^*(n)$ is then even, because $p^a$ is odd, so $\sigma^*(n)$ is divisible by $2^k$. But $2n$ is divisible by 2 exactly once. So $\sigma^*(n) = 2n$ forces $k = 1$: $n$ is a single prime power $p^a$, and the equation becomes $1 + p^a = 2p^a$, that is $p^a = 1$, which is impossible.

The figure checks the key fact — that $2^{\omega(n)}$ divides $\sigma^*(n)$ for every odd $n$, where $\omega(n)$ counts distinct primes — for every odd number below a million, and finds no exception, as there cannot be. It also records how near an odd number can get. The closest below a million is $442{,}365 = 3 \cdot 5 \cdot 7 \cdot 11 \cdot 383$, where $\sigma^*(n) = 884{,}736 = 2n + 6$: off by six in nearly nine hundred thousand, and off in the wrong way to ever close, since $884{,}736 = 2^{15} \cdot 27$ is divisible by $2^{15}$ and $2n$ only by 2.

The contrast with ordinary perfect numbers is the point. For $\sigma(n)$ the factor $1 + p + \cdots + p^a$ of an odd prime power is odd when $a$ is even, so an odd perfect number escapes the parity argument by having all but one of its prime powers to even exponents — Euler’s form $p^{a} m^2$ — and the question survives every elementary attack. Removing the intermediate powers removes that escape. Divisors that add to three times the number showed the same parity bookkeeping deciding which multiply perfect numbers can be odd; for unitary divisors it decides everything at once.

A closed economy of primes #

What makes an even unitary perfect number work is that the power of 2 absorbs everything. Write $n = 2^m \cdot r$ with $r$ odd and with $k$ distinct odd primes. The factor $1 + 2^m$ is odd, and the $k$ factors $1 + p^a$ from the odd primes are all even, together supplying at least $2^k$; the equation $\sigma^*(n) = 2^{m+1} r$ then needs the odd primes’ factors to deliver exactly $2^{m+1}$ in total, and everything odd in all the factors to be the odd part $r$ again.

For 87,360 the ledger balances exactly. The factor from $2^6$ is $65 = 5 \cdot 13$, supplying the primes 5 and 13. The factors from 3, 5, 7 and 13 are $4 = 2^2$, $6 = 2 \cdot 3$, $8 = 2^3$ and $14 = 2 \cdot 7$, supplying $2^7$ altogether together with the primes 3 and 7. The product is $2^7 \cdot 3 \cdot 5 \cdot 7 \cdot 13 = 174{,}720 = 2 \cdot 87{,}360$. Each prime appears on the right exactly as often as it is needed, and the power $2^6$ was chosen so that $1 + 2^6$ would split into two of the number’s own primes.

Wall’s number is the same economy at a larger scale.

Here $1 + 2^{18} = 262{,}145 = 5 \cdot 13 \cdot 37 \cdot 109$ supplies four odd primes at once, $1 + 5^4 = 626 = 2 \cdot 313$ brings in the prime 313, whose own factor $314 = 2 \cdot 157$ brings in 157, whose factor $158 = 2 \cdot 79$ brings in 79, whose factor $80 = 2^4 \cdot 5$ closes the chain back to 5. Every prime is introduced by some other prime’s factor, and the eleven odd primes’ factors deliver exactly $2^{19}$ between them. The figure multiplies the twelve factors as exact integers and finds $2n$.

How a fifth is found #

The ledger also says how to search for unitary perfect numbers without testing every number. Choose the power of 2 first, say $2^m$. The factor $1 + 2^m$ is odd and its prime factors must all be primes of $n$, so factor it: for $m = 6$ it is $65 = 5 \cdot 13$, for $m = 18$ it is $262{,}145 = 5 \cdot 13 \cdot 37 \cdot 109$. Each of those primes brings its own factor $1 + p^a$, for some exponent $a$ still to be chosen, and every prime of that factor must again be a prime of $n$. Following these demands from prime to prime builds a tree of choices, and the search succeeds when some branch closes — when every prime introduced has had its factor paid for and the powers of 2 add up to exactly $m + 1$.

For $m = 6$ the tree closes almost at once: 5 and 13 bring 6 and 14, which bring 3 and 7, which bring 4 and 8, and the powers of 2 from those four factors are $1 + 1 + 2 + 3 = 7 = m + 1$. For $m = 18$ it takes eleven odd primes and a choice of exponent 4 for the prime 5, whose factor 626 begins the long chain through 313, 157 and 79 back to 5. Wall’s search in 1975 explored these trees for many values of $m$ and found exactly one new closure, and no sixth has been reported since. The number of branches grows rapidly with $m$, because $1 + 2^m$ tends to have more prime factors as $m$ grows, and each brings a choice of exponent.

Cousins with other rules #

Changing which divisors count changes which numbers are perfect, and some variations have been settled completely. A divisor $d$ of $n$ is bi-unitary when the largest common unitary divisor of $d$ and $n/d$ is 1, a weaker condition that admits more divisors; Wall proved in 1972 that the only bi-unitary perfect numbers are 6, 60 and 90 — the first three unitary perfect numbers, and no others. At the other extreme, the ordinary perfect numbers are the case where every divisor counts, and there the even ones are known exactly while the odd ones are not known to exist.

The same family of questions runs through the rest of this subject’s divisor problems. Practical numbers ask that the divisors can make up every smaller amount; weird numbers are abundant without any subset of divisors summing to the number; friendly numbers share their abundancy with another. Each is a statement about how a number’s divisors add, and each is decided by the same multiplicative bookkeeping that the ledgers above make visible.

Why the search stops at four #

The sieve is exhaustive below ten million and finds exactly 6, 60, 90 and 87,360. Beyond that, the structure above says what a search must look for: an even number whose power of 2 is chosen so that $1 + 2^m$ factors into odd primes that can, in turn, be fed back into the chain. Searches organised that way, rather than number by number, are how Wall found the fifth, and they have pushed the possible sixth beyond very large bounds; Subbarao conjectured that there are only finitely many unitary perfect numbers, and nobody has proved or disproved it.

The finiteness is plausible for the same reason the ordinary odd perfect numbers are implausible: the ledger must balance exactly, and as the numbers grow, the factors $1 + p^a$ bring in new primes faster than the closing conditions can be met. But “plausible” is all it is. The methods that rule out small cases — bounds on how many odd primes a sixth would need, and on its power of 2 — grow with the effort spent and do not extend to all sizes.

A finite search and a quoted theorem #

The search is finite. Below ten million there are exactly four unitary perfect numbers; that the fifth is Wall’s, and that nothing lies between, is the result of searches organised by the structure of the ledger, not of the sieve drawn here.

The cloud is thinned. The hero draws every $n$ up to 2,000 and one in seven beyond, so that the picture shows the shape of the cloud rather than a solid band; every $n$ was computed, and the four hits are found from the full computation.

Unitary perfection is not perfection. Of the five unitary perfect numbers only 6 is perfect in the ordinary sense; 60 and 90 are abundant, their proper divisors adding to 108 and 144. The two notions coincide only when every prime of $n$ appears to the first power, and then $n$ is perfect in both senses only if it is 6.

Near misses prove nothing. The odd number 442,365 misses by six, and even numbers come closer still in relative terms, but a near miss in size is not a near miss in structure: the ledger either balances exactly or not at all, and the size of the imbalance says nothing about how far a number is from a solution in the sense that matters, which is how many primes would have to change.

Still open: a sixth #

Are there only finitely many unitary perfect numbers? Subbarao conjectured so, and the conjecture is open. It sits beside the questions the aliquot sequences and the abundant numbers raise, all of them about how the primes of $\sigma(n)$ relate to the primes of $n$, and all of them hard for the same reason: the divisor sum is multiplicative, its factors are additive objects $1 + p^a$, and nothing in number theory controls the prime factors of numbers of the form $1 + p^a$ well enough to decide whether a closed economy of primes can exist at every size. The unitary version has the advantage that oddness is settled, which leaves the even case as a pure question about chains of factors — and it is still open.