[seqfan] Re: a(n) is tau[a(n)+a(n+1)]
Douglas McNeil
mcneil at hku.hk
Tue May 18 17:22:58 CEST 2010
> If we decide that a(n) is tau[a(n)+a(n+1)], then
> a monotonically increasing sequence S of such terms
> could start like this:
>
> S = 2,3,6,12,48,2472,...
>
> Are more terms of S computable?
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.
We have 2472 = tau[2472 + next], so we just need to find the next
smallest number with 2472 divisors and subtract 2472. (I assume the
'smallest greater than seen so far' is understood.) I find the next
term is 2^102 * 3^2 * 5 * 7 * 11 - 2472.
sage: a
[2, 3, 6, 12, 48, 2472, 17569637319163259504744306426508888]
sage: pairs = zip(a[:-1], a[1:])
sage: all(x == number_of_divisors(x+y) for x,y in pairs)
True
To compute the next term I'd need to rewrite to work with the prime
decomposition instead.
> If we drop the "monotonically increasing" constraint
> but add "all terms of T must be different -- always
> take the smallest available integer not leading to
> a contradiction", would T be different of S? [of course
> 0 and 1 are not allowed in T]
Seems to be the same, modulo implementor error.
Doug
PS: M. Hasler's email just arrived, but that's okay, because he seems
to agree. :^)
--
Department of Earth Sciences
University of Hong Kong
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