[seqfan] Re: Extending A198683
Robert Gerbicz
robert.gerbicz at gmail.com
Wed Nov 23 01:51:01 CET 2011
2011/11/23 Alonso Del Arte <alonso.delarte at gmail.com>
> 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?
> >
> >
> >
> >
> > ______________________________**_________________
> >
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>
>
>
> --
> Alonso del Arte
> Author at SmashWords.com<
> https://www.smashwords.com/profile/view/AlonsoDelarte>
> Musician at ReverbNation.com <http://www.reverbnation.com/alonsodelarte>
>
> _______________________________________________
>
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>
Here it is a quick PARI code:
f(N,prec)=print1("1,");default(realprecision,prec);\
eps=10^(-prec/2);L=A=vector(N);L[1]=1;A[1]=[I];c=random(10^prec)/10^prec;\
for(n=2,N,s=sum(i=1,n-1,L[i]*L[n-i]);v=vector(s);pos=0;\
for(i=1,n-1,for(j=1,L[i],for(k=1,L[n-i],pos++;\
v[pos]=vector(2);v[pos][1]=A[i][j]^A[n-i][k];\
v[pos][2]=real(v[pos][1])+c*imag(v[pos][1]))));v=vecsort(v,2);ct=0;\
for(i=1,s,if(i==1||abs(v[i][2]/v[i-1][2]-1)>eps,ct++));print1(ct",");\
L[n]=ct;A[n]=vector(ct);ct=0;for(i=1,s,\
if(i==1||abs(v[i][2]/v[i-1][2]-1)>eps,ct++;A[n][ct]=v[i][1])))
with f(10,20) you can get the first ten terms, using twenty digits of
precision. For eleven terms I get an overflow in exponent.
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