[seqfan] Re: Extending A198683

[Alonso Del Arte]

Thanks for looking into it, Hans.

Of course we should never assume things to be so just because Mathematica
says they are, as much as we might like the program.

Now I wish that I had thought to look in the Messages window for the
computations of a(11) and a(12); would those messages be saved with in the
notebook file somehow?

However, I will take your word that I^(I^I)^(I^I^(I^I)^I^I^I^I)^I triggers
an underflow error, and your logic after that strikes me as a correct
deduction from that fact.

Perhaps Eric Weisstein might shed some light on this matter?

Al

On Tue, Nov 22, 2011 at 10:09 AM, Hans Havermann <gladhobo at teksavvy.com>wrote:

> Alonso Del Arte:
>
>  With Vladimir's Mathematica program, I've been able to extend A198683 with
>> two more terms, a(11) and a(12). After almost 16 hours of calculation, I
>> have a(13) = 7543 but I don't trust the result because several overflow
>> and
>> exceeding of machine precision error messages occurred in the course of
>> that computation...
>>
>
> I wanted to investigate the overflow errors, which eventually led me back
> to Mathematica's calculation of a(11). There are, I think, 2020 candidates
> for the count, but only 1152 after the 'SameTest -> Equal' is applied in
> 'Union' with no noted Overflow errors.
>
> However, one of the 2020 candidates is I^I^I^(-I)^I^(I^I)^I^I and
> N[I^I^I^(-I)^I^(I^I)^I^I]] yields Overflow[] but this is not a problem
> because it is the only candidate with an Overflow approximation, so
> a(11)=1152 should be correct, regardless.
>
> For Mathematica's calculation of a(12) there are, I think, 5139 candidates
> for the count, but only 2926 after the 'SameTest -> Equal' is applied in
> 'Union' with no noted Overflow errors.
>
> However, there are six problematic candidates:
>
> 1. I^I^(-I)^I^(-I)^I^I^I
> 2. I^I^I^I^(-I)^I^(I^I)^I^I
> 3. I^((-I)^I)^I^(-I)^I^I^I
> 4. I^(I^I)^(I^I^(I^I)^I^I^I^I)^I
> 5. (I^I)^I^I^(-I)^I^(I^I)^I^I
> 6. (I^I^I^(-I)^I^(I^I)^I^I)^I
>
> Approximating #4 yields Underflow[] but this is the only candidate with an
> Underflow approximation and this differentiates #4 from the others,
> suggesting it is distinct.
>
> Approximating #1, #3, and #5 yields Overflow[]. Approximating #2 and #6
> yields Indeterminate, with an Overflow noted in the 'messages' window. Now,
> I realize the application of 'SameTest -> Equal' suggests that these five
> candidates are in fact distinct from each other (they all end up in the
> 2926 final tally) but *how* does Mathematica actually know this, and should
> we just assume it to be so?
>
>
>
>
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-- 
Alonso del Arte
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Musician at ReverbNation.com <http://www.reverbnation.com/alonsodelarte>