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Superperfect Numbers form

\(n\) is called a superperfect number when \(\sigma(\sigma(n))=2n\).I saw every superperfect numbers is in the form \(2^k\) when \(2^{k+1}-1\) is a Mersenne prime.I can prove that numbers in this form will be superperfect but cannot prove the converse that all superperfect numbers are in this form. can anybody help me to prove? If this is not true then also tell me I am not sure about it.

Note by Kalpok Guha
1 year, 10 months ago

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Can you please provide a proof that if \(2^{k+1} -1 \) is a Mersenne prime then \(2^{k}\) is a perfect number because that might help us when we try to prove the converse. Curtis Clement · 1 year, 10 months ago

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@Curtis Clement Here let \(n=2^k\)

\(\sigma(\sigma(2^k)\)

=\(\sigma(\frac {2^{k+1}-1}{2-1})\)

=\(\sigma(2^{k+1}-1)\)

As \(2^{k+1}-1\) is a prime \(\sigma(2^{k+1}-1)=(2^{k+1}-1)+1=2^{k+1}=2.2^k=2n\)

Proved that it is always a superperfect number. Kalpok Guha · 1 year, 10 months ago

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@Curtis Clement Certainly I will post it .just wait Kalpok Guha · 1 year, 10 months ago

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This might be helpful...check out perfect Numbers, as explained by Numberphile: (http://m.youtube.com/watch? v=q8n15q1v4Xo) B.S.Bharath Sai Guhan · 1 year, 10 months ago

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@B.S.Bharath Sai Guhan Thank you very much it became really useful to me .Though here they only proved the form of perfect number but listening to the class I got idea to the form of superperfect number.Identified the series of super perfect number Kalpok Guha · 1 year, 10 months ago

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I read it on Elementary Number Theory by David Burton,then opened wiki and saw but there is no proof there neither direct one nor converse Kalpok Guha · 1 year, 10 months ago

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You mean you saw it on wiki? 'Cause its there... Danny Kills · 1 year, 10 months ago

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