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If {{math| ''m'' }} is an odd ],<ref>Currently, the only known almost perfect numbers are the nonnegative powers of 2, whence the only known odd almost perfect number is {{math| 2<sup>0</sup> {{=}} 1. }}</ref> that is, {{math| σ(''m'') {{=}} 2''m'' − 1 }} and {{math| 2''m'' − 1 }} is taken as a 'spoof' prime, then {{math| ''n'' {{=}} ''m''⋅(2''m'' − 1) }} is a Descartes number, since {{math| σ(''n'') {{=}} σ(''m''⋅(2''m'' − 1)) {{=}} σ(''m'')⋅2''m'' {{=}} (2''m'' − 1)⋅2''m'' {{=}} 2''n'' }}. If {{math| 2''m'' − 1 }} were prime, {{math| ''n'' }} would be an odd perfect number. | If {{math| ''m'' }} is an odd ],<ref>Currently, the only known almost perfect numbers are the nonnegative powers of 2, whence the only known odd almost perfect number is {{math| 2<sup>0</sup> {{=}} 1. }}</ref> that is, {{math| σ(''m'') {{=}} 2''m'' − 1 }} and {{math| 2''m'' − 1 }} is taken as a 'spoof' prime, then {{math| ''n'' {{=}} ''m''⋅(2''m'' − 1) }} is a Descartes number, since {{math| σ(''n'') {{=}} σ(''m''⋅(2''m'' − 1)) {{=}} σ(''m'')⋅2''m'' {{=}} (2''m'' − 1)⋅2''m'' {{=}} 2''n'' }}. If {{math| 2''m'' − 1 }} were prime, {{math| ''n'' }} would be an odd perfect number. | ||
==Properties== | |||
Banks et al. showed in 2008 that if {{math| ''n'' }} is a cube-free Descartes number not divisible by <math>3</math>, then {{math| ''n'' }} has over a million distinct prime divisors. | |||
==See also== | ==See also== |
Revision as of 08:39, 11 May 2019
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 = 3⋅7⋅11⋅13⋅22021 = (3⋅1001)⋅(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,
where we ignore the fact that 19⋅61 = 22021 reveals that 22021 is composite.
A Descartes number is defined as an odd number n = m⋅p 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 , then n has over a million distinct prime divisors.
See also
- Erdős–Nicolas number, another type of almost-perfect number
Notes
- Currently, the only known almost perfect numbers are the nonnegative powers of 2, whence the only known odd almost perfect number is 2 = 1.
References
- Banks, William D.; Güloğlu, Ahmet M.; Nevans, C. Wesley; Saidak, Filip (2008). "Descartes numbers". In De Koninck, Jean-Marie; Granville, Andrew; Luca, Florian (eds.). Anatomy of integers. Based on the CRM workshop, Montreal, Canada, March 13--17, 2006. CRM Proceedings and Lecture Notes. Vol. 46. Providence, RI: American Mathematical Society. pp. 167–173. ISBN 978-0-8218-4406-9. Zbl 1186.11004.
- Klee, Victor; Wagon, Stan (1991). Old and new unsolved problems in plane geometry and number theory. The Dolciani Mathematical Expositions. Vol. 11. Washington, DC: Mathematical Association of America. ISBN 0-88385-315-9. Zbl 0784.51002.
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