[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.


Department of Earth Sciences
University of Hong Kong

More information about the SeqFan mailing list