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{align}
\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\cdot19)\cdot(7\cdot19)\cdot(3\cdot61)\cdot(22\cdot1001) \\
&= 3^2\cdot7\cdot13\cdot19^2\cdot61\cdot(22\cdot7\cdot11\cdot13) = 2 \cdot (3^2\cdot7^2\cdot11^2\cdot13^2) \cdot (19^2\cdot61) = 2 \cdot (3^2\cdot7^2\cdot11^2\cdot13^2) \cdot 22021 = 2D,
\end{align}

where we turn a blind eye to the fact that 192⋅61 = 22021 reveals that 22021 is composite!

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,[1] 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!

Notes

  1. Currently, the only known almost perfect numbers are the nonnegative powers of 2, whence the only known odd almost perfect number is 20 = 1.

References


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