Divisor sums and abundance

T. M. Putnam¹⁸² proved that not all of the $r$ distinct prime factors of a perfect number exceed $1+r/\log_e2$ and hence do not all equal or exceed $1+3r/2$.

L. E. Dickson¹⁸³ gave an immediate proof that every even perfect number is of Euclid's type. Let $2^nq$ be perfect, where $q$ is odd and $n>0$. Then $(2^{n+1}-1)s=2^{n+1}q$, where $s$ is the sum of all the divisors of $q$. Thus $s=q+d$, where $d=q/(2^{n+1}-1)$. Hence $d$ is an integral divisor of $q$, so that $q$ and $d$ are the only divisors of $q$. Hence $d=1$ and $q$ is a prime.

L. E. Dickson¹⁹³ called a non-deficient number primitive if it is not a multiple of a smaller non-deficient number, and proved that there is only a finite number of primitive non-deficient numbers having a given number of distinct odd prime factors and a given number of factors 2. As a corollary, there is not an infinitude of odd perfect numbers with any given number of distinct prime factors. There is no odd abundant number with fewer than three distinct prime factors; the primitive ones with three are

$$3^3 5\cdot7,\quad 3^2 5^2 7,\quad 3^2 5\cdot7^2,\quad 3^3 5^2 11,\quad 3^5 5^2 13,\quad 3^4 5^3 13,\quad 3^4 5^2 13^2,\quad 3^3 5^3 13^2.$$

There is given a list of the numerous primitive odd abundant numbers with four distinct prime factors and lists of even non-deficient numbers of certain types. In particular, all primitive non-deficient numbers $<15000$ are determined (23 odd and 78 even). In view of these lists, there is no odd perfect number with four or fewer distinct prime factors (cf. Sylvester^148–153).

A bound on the least prime factor

Putnam¹⁸² proves that if a perfect number has $r$ distinct prime factors, at least one is at most $1+r/\log 2$. Consequently they cannot all be at least $1+3r/2$.

Dickson’s converse

Dickson¹⁸³ gives a short proof of the classification of even perfect numbers. Suppose $2^a m$ is perfect, where $m$ is odd and $a>0$. Let $\sigma(m)$ be the sum of all positive divisors of $m$. Then

$$(2^{a+1}-1)\sigma(m)=2^{a+1}m.$$

Thus $\sigma(m)=m+d$, where $d=m/(2^{a+1}-1)$. The integer $d$ divides $m$. Since the sum of all divisors is already $m+d$, these are its only divisors. Therefore $d=1$ and $m$ is prime.

Primitive non-deficient numbers

Dickson¹⁹³ calls a number non-deficient when $\sigma(n)\ge2n$, and primitive when none of its proper divisors is non-deficient. He proves that fixing the number of distinct odd prime factors and the exponent of 2 leaves only finitely many primitive non-deficient numbers. The corollary is that there are only finitely many odd perfect numbers with a fixed number of distinct prime factors.

He reports that an odd abundant number has at least three distinct prime factors, and gives the following primitive ones with exactly three:

$$3^3\cdot5\cdot7,\quad3^2\cdot5^2\cdot7,\quad3^2\cdot5\cdot7^2,\quad3^3\cdot5^2\cdot11,$$ $$3^5\cdot5^2\cdot13,\quad3^4\cdot5^3\cdot13,\quad3^4\cdot5^2\cdot13^2,\quad3^3\cdot5^3\cdot13^2.$$

His paper lists many primitive odd abundant numbers with four distinct prime factors, and even non-deficient numbers of specified types. It determines all primitive non-deficient numbers below 15000: 23 odd and 78 even. Dickson concludes from these lists that no odd perfect number has four or fewer distinct prime factors, and compares Sylvester.^148–153

Divisor sums relative to size

For $n\ge1$, define the abundancy ratio

$$A(n)=\frac{\sigma(n)}n=\sum_{d\mid n}\frac1d.$$

The second equality follows because $d\mapsto n/d$ permutes the divisors. From the divisor product,

$$A(n)=\prod_{p^a\parallel n} L_p(a),\qquad L_p(a)=1+\frac1p+\cdots+\frac1{p^a}.$$

The notation $p^a\parallel n$ means that $a$ is the exact exponent of $p$. For every finite $a$,

$$L_p(a)<\frac p{p-1}.$$

Proposition. An odd integer with at most two distinct prime factors is deficient.

Proof. For one prime factor the bound is at most $3/2<2$. For two, the largest possible upper bound uses the smallest odd primes, 3 and 5:

$$A(n)<\frac32\frac54=\frac{15}{8}<2.$$

Thus arbitrary large exponents cannot overcome a shortage of prime factors. The argument explains the kind of restriction in Dickson's p. 31 report; it does not prove that report's stronger four-prime exclusion.

Putnam's bound on p. 30 follows from the same product with logarithms. If $n$ is perfect with distinct primes $p_1,\ldots,p_r$, then

$$\log2=\log A(n)<\sum_{i=1}^r\log\left(1+\frac1{p_i-1}\right)<\sum_{i=1}^r\frac1{p_i-1}.$$

Here $\log(1+x)<x$ for $x>0$ follows by integrating $1/t<1$ from 1 to $1+x$. If every $p_i>1+r/\log2$, the right side would be less than $\log2$, a contradiction. Thus some $p_i\le1+r/\log2$. Also $\log2>2/3$: the function $\log x-2(x-1)/(x+1)$ is zero at 1 and has derivative $(x-1)^2/[x(x+1)^2]>0$ for $x>1$. This gives Putnam's weaker bound $p_i<1+3r/2$. The logarithmic proof is optional for subsequent units; the finite-product bound is all they need.

Positivity and divisor containment

Lemma. If $m\mid n$ and $m<n$, then $A(m)<A(n)$.

Proof. Every divisor of $m$ is a divisor of $n$, and $n$ itself is an additional divisor. In the reciprocal sums defining $A$, all summands are positive.

Consequently every proper multiple of a perfect number is abundant and every proper divisor of a perfect number is deficient. The qualifier proper matters: the number itself is neither a deficient divisor nor an abundant multiple of itself.

Dickson's short converse uses positivity differently. If the sum of all positive divisors is already the sum of two distinct known divisors, there is no room for a third. Divisor containment proves monotonicity; exhaustion of a divisor sum proves the converse. Neither needs estimates about large primes.

Primitive non-deficient numbers

A number is non-deficient when $A(n)\ge2$. It is primitive non-deficient if all of its proper divisors are deficient. Every non-deficient number has a primitive non-deficient divisor: choose the smallest non-deficient divisor from its finite divisor set. Thus these primitive numbers generate the non-deficient numbers by divisibility.

One can check primitiveness efficiently once the factorization is known. It suffices to check $A(n/p)<2$ for every prime $p\mid n$. Every proper divisor divides at least one such $n/p$, and monotonicity handles all smaller divisors.

For example, $945=3^3\cdot5\cdot7$ has $\sigma(945)=40\cdot6\cdot8=1920>1890$. Removing one factor of 3, 5 or 7 gives $315,189,135$, each deficient. So 945 is primitive abundant.

Dickson reports a further theorem: with the number of distinct odd prime factors and the exponent of 2 fixed, there are only finitely many primitive non-deficient numbers. This theorem is an external input here; the monotonicity argument is not its proof. Every perfect number is primitive, so his stated corollary for odd perfect numbers follows from it. The eight three-prime examples and historical bounds remain in Modernized. Their exact arithmetic can be checked without turning those checks into an exhaustive classification proof.