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Pentation roots self but please you do...
#1
Smile 









By the results have 5 cases decimais (minimum).

Please you think to calculate on up... I have known its by program Pari/GP (it's very fast). Tongue x)
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#2
(10/18/2010, 12:09 AM)nuninho1980 Wrote:








By the results have 5 cases decimais (minimum).

Please you think to calculate on up... I have known its by program Pari/GP (it's very fast). Tongue x)
Pentation is hard to understand.... Here's my results. I used "b" for the base.
, for each "n", calculate b
, calculate b
, calculate b
, calculate b
and so on, limit as
n= 2 1.63221539635499
n= 3 1.73480823757765
n= 4 1.73013167405422
n= 5 1.71198477313212
n= 6 1.69588829898111
n=70 1.63599652477221

I calculated these values by simple binary search, but I used "\p 28", which is accurate to ~14 digits, but very fast, 4 seconds for init(B);loop. Only problem is its very easy to get an overflow, so the initial starting based needs to readjusted; for n=70, I used a more complicated algorithm.
Code:
\r kneser.gp
\p 28;
{ curbase=1.6;
  curstep=0.1;
  while ((curstep>1E-14),
    init(curbase);loop;
    y = sexp(sexp(B));
    if (y>3, curbase=curbase-curstep, curbase=curbase+curstep);
    curstep=curstep/2;
  ); }

As n goes to infinity, I would expect the value for b to go to Nuinho's constant, the base for which the upper fixed point of sexp is parabolic,
b=1.635324496715276399345344618306171
- Sheldon
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#3
It's excellent!! Congratulations! Big Grin But you are 2 weeks later. I already could calculate pentaroots Tongue but I tried to "explore" each digit by "lottery" up to 5 digits from days 15 to 17, october because I used only "kneser.gp". Smile

Nuinho's constant - bad but yes Nuninho Smile

thank! Wink
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