\(n\) 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.

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TopNewestCan 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 · 2 years ago

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

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

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

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 · 2 years ago

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– Kalpok Guha · 2 years ago

Certainly I will post it .just waitLog in to reply

This might be helpful...check out perfect Numbers, as explained by Numberphile: (http://m.youtube.com/watch? v=q8n15q1v4Xo) – B.S.Bharath Sai Guhan · 2 years ago

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– Kalpok Guha · 2 years ago

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 numberLog in to reply

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 · 2 years ago

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You mean you saw it on wiki? 'Cause its there... – Danny Kills · 2 years ago

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