More on e^(pi*sqrt(163)) - sci.math.research

Message de la discussion More on e^(pi*sqrt(163))

Alexander Povolotsky

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Autres options 17 sep 2009, 15:00

Groupes de discussion : sci.math.research

De : Alexander Povolotsky <apovo...@gmail.com>

Date : Thu, 17 Sep 2009 19:00:02 +0000 (UTC)

Date/heure locale : Jeu 17 sep 2009 15:00

Objet : Re: More on e^(pi*sqrt(163))

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As a follow-up to my previous posting re the "near integer" values of
k (obtained for n=0,1,2,6 ), it is interesting (IMHO) that we could
subtract (from those k)
3.<near one fractional part>
and observe that the subtraction result is dividable by 36

exp(Pi*sqrt(19+24*n) =~ (24*k)^3 + 31*24
gp > for(n=0,10,print1("n= ",n," k= ",
((exp(Pi*sqrt(19+24*n))/24-31)/24/24)^(1/3),"\n"))
n= 0 k= 3.999999664954872711861691865 <<= - 3.9... 0
/ 36 = 0
n= 1 k= 39.99999999999664632214064072 <<= - 3.9... 36 /
36 = 1
n= 2 k= 219.9999999999999993336409313 <<= - 3.9... 216 /
36 = 6
n= 6 k= 26680.00000000000000000000000 <<= - 4 26676 / 36 = 741

Using PARI/GP the results of above described division by 36 could be obtained as
gp >b(n)=((exp(Pi*sqrt(19+24*n))/24-31)/24/24)^(1/3)
gp > for (n=0,3,print1((ceil(b((abs(n-1))!*n))-4)/36,"\n"))
0
1
6
741

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