# Descartes number

In number theory, a Descartes number is an odd number which would have been an odd perfect number, if one of its composite factors were prime. They are named after René Descartes who observed that the number D = 32⋅72⋅112⋅132⋅22021 = (3⋅1001)2⋅(22⋅1001 − 1) = 198585576189 would be an odd perfect number if only 22021 were a prime number, since the sum-of-divisors function for D would satisfy, if 22021 were prime,

{\begin{aligned}\sigma (D)&=(3^{2}+3+1)\cdot (7^{2}+7+1)\cdot (11^{2}+11+1)\cdot (13^{2}+13+1)\cdot (22021+1)=(13)\cdot (3\cdot 19)\cdot (7\cdot 19)\cdot (3\cdot 61)\cdot (22\cdot 1001)\\&=3^{2}\cdot 7\cdot 13\cdot 19^{2}\cdot 61\cdot (22\cdot 7\cdot 11\cdot 13)=2\cdot (3^{2}\cdot 7^{2}\cdot 11^{2}\cdot 13^{2})\cdot (19^{2}\cdot 61)=2\cdot (3^{2}\cdot 7^{2}\cdot 11^{2}\cdot 13^{2})\cdot 22021=2D,\end{aligned}} where we ignore the fact that 22021 is composite (22021 = 192⋅61).

A Descartes number is defined as an odd number n = mp where m and p are coprime and 2n = σ(m)⋅(p + 1) , whence p is taken as a 'spoof' prime. The example given is the only one currently known.

If m is an odd almost perfect number, that is, σ(m) = 2m − 1 and 2m − 1 is taken as a 'spoof' prime, then n = m⋅(2m − 1) is a Descartes number, since σ(n) = σ(m⋅(2m − 1)) = σ(m)⋅2m = (2m − 1)⋅2m = 2n . If 2m − 1 were prime, n would be an odd perfect number.

## Properties

Banks et al. showed in 2008 that if n is a cube-free Descartes number not divisible by $3$ , then n has over a million distinct prime divisors.