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

[Robert Munafo]

Oooh, I like quickly-growing sequences! Let me make sure I understand what
you're doing.

"tau[n]" appears to be A000005.

You want a monotonically increasing sequence, and the first term is 2.

Given a term X, find the first Y such that tau(Y) is bigger than 2X, and
then the next term is Y-X.

So, tau(5) is 2, giving 5-2=3 as the next term.

tau(9) is 3, giving 9-3=6 as the next term.

tau(12) is 6, but that would give 12-6=6, not increasing ... so we go up to
tau(18)=6, giving 18-6=12 as the next term.

Now we want something with 12 factors. tau(60)=12, giving 60-12=48 as the
next term.

Now we want something with 48 factors. 48=3*2*2*2*2, leading to the
possibility Y=2^2*3*5*7*11=4620, but that is bigger than we need, because we
can take 48=4*3*2, leading to Y=2^3*3^2*5=2520. So we have 2520-48=2472 for
the next term.

Now we want something with 2472 factors. 2472=2^3*3*103, so we're looking
for a Y whose factorization includes p^102 for some prime p. Clearly p
should be 2. For the rest of it we could use q^2*r*s*t or q^3*r^2*s, with
q,r,s,t drawn from the next available primes. The first of these wins out
because 3^2*5*7*11<3^3*5^2*7. So that gives
(2^102*3^2*5*7*11)-2472=17569637319163259504744306426508888
for the next term.

The factors of 17569637319163259504744306426508888 are 2^3 * 3 * 467 *
1567597904993153060737357818211, so clearly we want to go to
2^1567597904993153060737357818210 * 3^466 * N, where N has 24 factors. N
could be 5^2*7*11*13, but 5^3*7 is smaller.

So we get (2^1567597904993153060737357818210 * 3^466 * 5^3 * 7)
- 17569637319163259504744306426508888, which comes out to
about 10^(4.7189399054 * 10^29).

- Robert

On Tue, May 18, 2010 at 10:49, Maximilian Hasler <
maximilian.hasler at gmail.com> wrote:

> I think the next term should be 17569637319163259504744306426508888
>
> M.
>
> On Tue, May 18, 2010 at 3:42 PM, Eric Angelini <Eric.Angelini at kntv.be>
> wrote:
> >
> > Hello SeqFans,
> >
> > 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?
> >
> > 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]
> >
> > Best,
>

-- 
 Robert Munafo  --  mrob.com