[seqfan] Re: a(n) is tau[a(n)+a(n+1)]
Robert Munafo
mrob27 at gmail.com
Tue May 18 18:34:57 CEST 2010
Oh, darn. I got the last bit wrong, it should have been 5^2*7*11*13
versus 5^3*7^2*11, and the first is smaller. So the next term after
17569637319163259504744306426508888
should be
(2^1567597904993153060737357818210 * 3^466 * 5^2 * 7 * 11 * 13) -
17569637319163259504744306426508888
However this still comes out to 10^(4.7189399054 * 10^29). If I use 40
digits of precision for this, and I
get 4.676024*10^471893990542954871479908400227.
On Tue, May 18, 2010 at 12:21, Robert Munafo <mrob27 at gmail.com> wrote:
> 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
>>
>> On Tue, May 18, 2010 at 3:42 PM, Eric Angelini <Eric.Angelini at kntv.be>
>> wrote:
>> > 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?
>>
>
--
Robert Munafo -- mrob.com
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