[seqfan] Re: a(n) is tau[a(n)+a(n+1)]

Douglas McNeil mcneil at hku.hk
Tue May 18 17:25:45 CEST 2010


> Yeah, but they're big.  You can use the fact that tau(n) =
> (e_1+1)*(e_2+1)..(e_k+1) where e_i are the prime factors of n to
> construct candidates.

Oh, good grief.  Correcting the obvious mistake in the above is left
as an exercise for the reader.


Doug

-- 
Department of Earth Sciences
University of Hong Kong




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