[seqfan] Re: A335071 question.

[michel.marcus at free.fr]

I had an idea like this: 

Some relevant articles: 


https://cs.uwaterloo.ca/journals/JIS/VOL10/Holdener/holdener7.pdf 


https://www.researchgate.net/publication/265661827_Results_concerning_uniqueness_for_sxxsp_n_q_m_p_n_q_m_and_related_topics/link/5b258279458515270fd40f88/download 


https://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/the-abundancy-ratio-a-measure-of-perfection 




Using 1st article: 

Property 2.2 If ab(n) = k/m with gcd(k, m) = 1, then m|n. 
This follows directly from setting σ(n)/n = k/m. 
Clearly, m|(nk) and since k and m are relatively prime, it must be that m|n. 


Now if ab(n) = ab(n+1) = k/m with gcd(k, m) = 1, then m|n and also m|(n+1). 
But n and n+1 are coprime. So it is not possible. 


But there is a flaw there if they are multiperfect, that is with m=1, then m|n and m|(n+1) . 
So ...