[seqfan] Re: Extending A198683

[Hans Havermann]

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?