Multiply perfect numbers

A multiply perfect or pluperfect number $n$ is one the sum of whose divisors, including $n$ and 1, is a multiple of $n$. If the sum is $mn$, $m$ is called the multiplicity of $n$. For brevity, a multiply perfect number of multiplicity $m$ shall be designated by $P_m$. Thus an ordinary perfect number is a $P_2$. Although Robert Recorde³⁹ in 1557 cited 120 as an abundant number, since the sum of its parts is 240, such numbers were first given names and investigated by French writers in the seventeenth century. As a $P_3$ equals one-half of the sum of its aliquot divisors or parts (divisors $<P_3$), it was called a sous-double; a $P_4$ equals one-third of the sum of its aliquot parts and was called a sous-triple; a $P_5$ a sous-quadruple; etc.

Mersenne³⁰⁴ stated that Fermat found the $P_3$ 672 and knew infallible rules and analysis to find an infinitude of such numbers. He³⁰⁵ later gave [Fermat's] method of finding such $P_3$: Begin with the geometric progression $2,4,8,\ldots$. Subtract unity and place the remainders above the former. Add unity and place the sums below.

$$\begin{array}{rrrr}1&3&7&15\cdots\\2&4&8&16\cdots\\3&5&9&17\cdots\end{array}$$

Then if the quotient of the $(n+3)$th number of the top line by the $n$th number of the bottom line is a prime, its triple multiplied by the $(n+2)$th number of the middle line is a $P_3$. Thus if $n=1$, $15/3$ is a prime and $3\cdot5\cdot8=120$ is a $P_3$. For $n=3$, $63/9$ is a prime and $3\cdot7\cdot32=672$ is a $P_3$. [This rule thus states in effect that $3\cdot2^{n+2}p$ is a $P_3$ if $p=(2^{n+3}-1)/(2^n+1)$ is a prime.]

Descartes³¹² gave the following rules for multiply perfect numbers:

I. If $n$ is a $P_3$ not divisible by 3, then $3n$ is a $P_4$.

II. If a $P_3$ is divisible by 3, but by neither 5 nor 9, then $45P_3$ is a $P_4$.

III. If a $P_3$ is divisible by 3, but not by 7, 9 or 13, then $3\cdot7\cdot13P_3$ is a $P_4$.

IV. If $n$ is divisible by $2^9$ but by no one of the numbers $2^{10},31,43,127$, then $31n$ and $16\cdot43\cdot127n$ are proportional to the sums of their aliquot parts.

V. If $n$ is not divisible by 3 and if $3n$ is a $P_{4k}$, then $n$ is a $P_{3k}$.

Multiplicity

A number $n$ is multiply perfect, or pluperfect, if $\sigma(n)=kn$ for an integer $k$, its multiplicity. Here $\sigma$ includes both 1 and $n$. Dickson writes $P_k$ for a number of multiplicity $k$; an ordinary perfect number is a $P_2$.

Recorde³⁹ cited 120 as abundant in 1557 because its proper divisors sum to 240. Dickson dates the naming and investigation of such numbers to seventeenth-century French writers. A $P_3$ is half the sum of its proper divisors and was called a sous-double; a $P_4$ is one-third of that sum, a sous-triple; a $P_5$ is a sous-quadruple; and so on.

Fermat’s rule as reported by Mersenne

Mersenne³⁰⁴ states that Fermat found the $P_3$ 672 and possessed infallible methods for finding infinitely many such numbers. He later gives the construction:³⁰⁵ start with $2,4,8,\ldots$, place each term minus 1 above it and each term plus 1 below it:

$$\begin{array}{rrrr}1&3&7&15\cdots\\2&4&8&16\cdots\\3&5&9&17\cdots\end{array}$$

If the $(n+3)$rd term of the upper row divided by the $n$th term of the lower row is a prime $p$, multiply $3p$ by the $(n+2)$nd term of the middle row to obtain a $P_3$. For $n=1$, the quotient is $15/3=5$ and the result is $120$; for $n=3$, the quotient is $63/9=7$ and the result is $672$. Dickson expresses the rule as

$$p=\frac{2^{n+3}-1}{2^n+1}\text{ prime}\quad\Longrightarrow\quad N=3\cdot2^{n+2}p\text{ is a }P_3.$$

Descartes’ five rules

Descartes³¹² gives the following rules. Write $v_p(n)$ for the exponent of the prime $p$ in $n$.

I. If $\sigma(n)=3n$ and $3\nmid n$, then $\sigma(3n)=4(3n)$.

II. If $\sigma(n)=3n$, $v_3(n)=1$ and $5\nmid n$, then $45n$ has multiplicity 4.

III. If $\sigma(n)=3n$, $v_3(n)=1$, and neither 7 nor 13 divides $n$, then $3\cdot7\cdot13n$ has multiplicity 4.

IV. If $v_2(n)=9$, and none of $31,43,127$ divides $n$, the two numbers $31n$ and $16\cdot43\cdot127n$ have the same ratio to their respective proper-divisor sums.

V. If $3\nmid n$ and $3n$ has multiplicity $4k$, then $n$ has multiplicity $3k$.

Integer values of the abundancy ratio

For $n>1$ and an integer $k\ge2$, call $n$ $k$-perfect when $A(n)=k$, or equivalently $\sigma(n)=kn$. Then $s(n)=(k-1)n$. This explains the historical sous-double terminology for multiplicity 3: the number is half its proper-divisor sum. The harmless edge case $n=1$, $A(1)=1$, is excluded in this unit.

The examples $120=2^3\cdot3\cdot5$ and $672=2^5\cdot3\cdot7$ satisfy

$$\sigma(120)=15\cdot4\cdot6=360,\qquad\sigma(672)=63\cdot4\cdot8=2016.$$

Thus both are 3-perfect. The same local factors that classify deficiency now have to multiply to a prescribed integer.

Why Fermat’s displayed rule has only two outputs

The construction reported on pp. 33–34 asks for a prime

$$p=\frac{2^{n+3}-1}{2^n+1},\qquad n\ge1,$$

and then produces $3\cdot2^{n+2}p$. Rewrite the quotient as

$$p=8-\frac9{2^n+1}.$$

For $p$ even to be an integer, $2^n+1$ must divide 9. Since $n\ge1$, the only possibilities are 3 and 9, giving $n=1,3$ and $p=5,7$. These give exactly 120 and 672, verified above. Descartes already states the two-output limitation on p. 34; this division exposes its reason. The report of an infinitude on p. 33 cannot be justified by this particular formula.

Transformations as ratios of local factors

Write $L_p(a)=1+p^{-1}+\cdots+p^{-a}$, as in Divisor sums and abundance. If a transformation changes only a few prime exponents, all the unchanged factors cancel in $A(N')/A(N)$. This proves Descartes' five rules and explains their exclusions.

Rule I. If $3\nmid N$, adjoining 3 multiplies $A$ by $L_3(1)=4/3$. Hence $A(N)=3$ implies $A(3N)=4$.

Rule II. If $v_3(N)=1$ and $5\nmid N$, multiplying by 45 changes the exponent of 3 from 1 to 3 and adjoins 5. Therefore

$$\frac{A(45N)}{A(N)}=\frac{L_3(3)}{L_3(1)}L_5(1)=\frac{40/27}{4/3}\frac65=\frac43.$$

For $N=672$, this gives $45N=30240$, of multiplicity 4. The rule cannot be applied to 120 because 5 already divides it.

Rule III. If $v_3(N)=1$ and neither 7 nor 13 divides $N$, multiplication by $3\cdot7\cdot13$ gives

$$\frac{A(273N)}{A(N)}=\frac{13/9}{4/3}\frac87\frac{14}{13}=\frac43.$$

With $N=120$ the result is $32760$, also 4-perfect. Reapplying the same rule is not licensed: the new number has acquired the excluded factors.

Rule IV. Suppose $v_2(N)=9$ and $31,43,127$ do not divide $N$. Adjoining 31 multiplies $A$ by $32/31$. On the other hand,

$$\frac{A(16\cdot43\cdot127N)}{A(N)}=\frac{(2^{14}-1)/2^{13}}{(2^{10}-1)/2^9}\frac{44}{43}\frac{128}{127}=\frac{32}{31},$$

because $2^{14}-1=3\cdot43\cdot127$ and $2^{10}-1=3\cdot11\cdot31$. Both transformed numbers have the same $A$, hence the same $s(n)/n=A(n)-1$. This is the source's proportionality assertion; it does not require that $N$ be multiply perfect.

Rule V. If $3\nmid N$, then $A(3N)=\frac43A(N)$. Thus $A(3N)=4k$ forces $A(N)=3k$.

These are conditional transformations of known numbers. The conditions are part of the mechanism, not incidental restrictions. For a failure example, $A(120)=3$ but $A(3\cdot120)=13/4$, not 4: 3 already divides 120. An unlimited supply of formally written rules does not by itself provide an unlimited supply of admissible inputs.