Some numbers are defined so precisely that we know almost everything about them, except whether a single example exists. This article explores quasiperfect numbers, almost perfect numbers, Descartes numbers, and superperfect numbers, all built from one simple idea: adding up the divisors of a number.
Perfect, Abundant, and the Sum of Divisors
The divisors of 6 are 1, 2, 3, and 6, and they sum to 12, which is exactly 2 × 6. A number whose divisors sum to twice itself is called perfect. The divisors of 12 are 1, 2, 3, 4, 6, and 12, and they sum to 28, which exceeds 24. Such numbers are called abundant, and the excess, 28 − 24 = 4, is the abundance. The OEIS defines abundance as the sum of divisors minus 2n [1].
Writing σ(n) for the sum of the divisors of n, a perfect number satisfies the equation below. Unless stated otherwise, every definition that follows is a variation on this one.
σ(n) = 2n
Quasiperfect Numbers
A quasiperfect number is one with abundance exactly 1, meaning σ(n) = 2n + 1 [2]. No example is known. Cattaneo showed in 1951 that any such number must be an odd square [3]. Hagis and Cohen later showed it would need at least seven distinct prime factors and would exceed 10³⁵ [2]. MathWorld confirms that none have been found [4]. These constraints connect naturally to unsolved prime number problems.
σ(n) = 2n + 1
Almost Perfect Numbers
If instead σ(n) = 2n − 1, the number is called almost perfect, also known as least deficient. The name is confusing, since "quasi" already means "almost." Every power of two works: the divisors of 2ᵏ are 1, 2, 4, …, 2ᵏ, which sum to 2ᵏ⁺¹ − 1 = 2n − 1. For example, σ(8) = 1 + 2 + 4 + 8 = 15 = 2 × 8 − 1 [1].
The open question is the converse: are there almost perfect numbers that are not powers of two? The only known examples are powers of two, and it has not been shown that no others exist. An odd almost perfect number m other than 1 would need at least six prime factors, and m(2m − 1) would then be a Descartes number [5].
σ(n) = 2n − 1
Descartes Numbers and Spoof Primes
To compute σ quickly, factor n into primes and multiply together, for each prime power, the sum of its powers up to the exponent. For 72 = 2³ × 3², the sum of divisors is (1 + 2 + 4 + 8)(1 + 3 + 9) = 15 × 13 = 195.
In 1638 Descartes noted the number D = 198,585,576,189 = 3² × 7² × 11² × 13² × 22021. If 22021 were prime, the formula would give σ(D) = 2D, making D an odd perfect number. But 22021 = 19² × 61 is not prime, so D is not perfect. Treating 22021 as a prime anyway makes it a spoof prime [6].
A Descartes number is an odd number n = m × p, where m and p are coprime and 2n = σ(m)(p + 1) [7]. D is the only known Descartes number [8], and whether others exist is an open problem. It is another example of a simple-sounding unsolved mathematical problem.
2n = σ(m) × (p + 1)
Superperfect Numbers
Applying σ twice gives a new notion. A number n is superperfect if σ(σ(n)) = 2n [9]. For n = 4, σ(4) = 7 and σ(7) = 8 = 2 × 4. Other examples are 2, 16, and 64, all powers of two.
σ(σ(n)) = 2n
For n = 2ᵏ we have σ(n) = 2n − 1 = 2ᵏ⁺¹ − 1. If that is a Mersenne prime, then σ of it is 2ᵏ⁺¹ = 2n, so n is superperfect. Not every power of two works: 32 gives σ(32) = 63 = 3² × 7, and σ(63) = 104, not 64. In fact, even superperfect numbers are exactly the powers 2ᵏ for which 2ᵏ⁺¹ − 1 is a Mersenne prime [9]. Mersenne primes also appear in other deceptively simple open problems.
Odd superperfect numbers are the final open problem. If one exists, it must be a square [10]. A computer search found none below 7 × 10²⁴ [11].
Sources
- A033880 – OEIS
- /. Austral. Math. Soc. (Series A) 33 (1982), 275-286 SOME RESULTS CONCERNING
- Quasiperfect numbers with the same exponent
- Quasiperfect Number — from Wolfram MathWorld
- Almost perfect number
- Descartes number – OeisWiki
- Descartes number
- There are no Cube-free Descartes Numbers with Exactly Seven Distinct Prime Factors
- Superperfect Number — from Wolfram MathWorld
- Generalized perfect numbers
- Superperfect number
Leave a Reply