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Groupes de discussion : sci.math.research
Date : Sat, 12 Apr 2008 22:28:14 -0700 (PDT)
Date/heure locale : Dim 13 avr 2008 01:28
Objet : More on e^(pi*sqrt(163))
Hello all, It is quite well-known that:
e^(pi*sqrt(19)) ~ 96^3 + 744 e^(pi*sqrt(43)) ~ 960^3 + 744 e^(pi*sqrt(67)) ~ 5280^3 + 744 e^(pi*sqrt(163)) ~ 640320^3 + 744
using the four highest Heegner numbers. But it is not so well-known that the expression e^(pi*sqrt(d)) can be given *another* internal structure:
e^(pi*sqrt(19)) ~ 12^3(3^2-1)^3 + 744 e^(pi*sqrt(43)) ~ 12^3(9^2-1)^3 + 744 e^(pi*sqrt(67)) ~ 12^3(21^2-1)^3 + 744 e^(pi*sqrt(163)) ~ 12^3(231^2-1)^3 + 744
The reason for the squares are due to certain Eisenstein series -- but that's another story. :-)
Beautifully consistent, aren't they?
I'm working on a new webpage about this and, er, other Ramanujan- related stuff. But I'm having a devil of a time finishing it due to my day job. I'll post the link here when it's done.
Yours,
Titus
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Groupes de discussion : sci.math.research
De :
Alexander R.Povolotsky <apovo... @gmail.com>
Date : Sat, 27 Jun 2009 15:08:21 +0100 (BST)
Date/heure locale : Sam 27 juin 2009 10:08
Objet : Re: More on e^(pi*sqrt(163))
I suggest to consider following for above, namely: the last four of Class Number 1 expressions in http://www.geocities.com/titus_piezas/Ramanujan_a.htm could be generalized as: exp(Pi*sqrt(19+24*n) =~ (24*k)^3 + 31*24
PARI confirmation is below ..
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 <<======== n= 1 k= 39.99999999999664632214064072 <<======== n= 2 k= 219.9999999999999993336409313 <<======== n= 3 k= 908.2994607084626509324663895 n= 4 k= 3139.719720204852366879238790 n= 5 k= 9587.574481226312121129336932 n= 6 k= 26680.00000000000000000000000 <<======== n= 7 k= 69020.39408641981880200520050 n= 8 k= 168277.4270764306998213353795 n= 9 k= 390498.9836593266367110562264 n= 10 k= 868910.8509221483459190206684
Cheers, Alexander R. Povolotsky
---------- Forwarded message ----------
From: tpie
... @gmail.com
Date: Apr 13 2008, 1:28 am
Subject: More on e^(pi*sqrt(163))
To: sci.math.research
Hello all,
It is quite well-known that:
e^(pi*sqrt(19)) ~ 96^3 + 744 e^(pi*sqrt(43)) ~ 960^3 + 744 e^(pi*sqrt(67)) ~ 5280^3 + 744 e^(pi*sqrt(163)) ~ 640320^3 + 744
using the four highest Heegner numbers. But it is not so well-known that the expression e^(pi*sqrt(d)) can be given *another* internal structure:
e^(pi*sqrt(19)) ~ 12^3(3^2-1)^3 + 744 e^(pi*sqrt(43)) ~ 12^3(9^2-1)^3 + 744 e^(pi*sqrt(67)) ~ 12^3(21^2-1)^3 + 744 e^(pi*sqrt(163)) ~ 12^3(231^2-1)^3 + 744
The reason for the squares are due to certain Eisenstein series -- but that's another story. :-)
Beautifully consistent, aren't they?
I'm working on a new webpage about this and, er, other Ramanujan- related stuff. But I'm having a devil of a time finishing it due to my day job. I'll post the link here when it's done.
Yours,
Titus
Vous ne disposez pas de l'autorisation nécessaire pour publier un message.
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))
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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