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It is rather well-known that,

$e^{\pi\sqrt{43}} \approx 960^3 + 743.999\ldots$

$e^{\pi\sqrt{67}} \approx 5280^3 + 743.99999\ldots$

$e^{\pi\sqrt{163}} \approx 640320^3 + 743.999999999999\ldots$

Not so well-known is,

$e^{\pi\sqrt{43}} \approx (5x_1)^3 + 6.000000010\ldots$

$e^{\pi\sqrt{67}} \approx (5x_2)^3 + 6.000000000061\ldots$

$e^{\pi\sqrt{163}} \approx (5x_3)^3 + 6.000000000000000034\ldots$

where the $x_i$ is the appropriate root of the sextics,

$5x^6-960x^5-10x^3+1 = 0$

$5x^6-5280x^5-10x^3+1 = 0$

$5x^6-640320x^5-10x^3+1 = 0$

One can see the j-invariants (or at least their cube roots) appearing again. These sextics are solvable in radicals, factoring over $Q(\sqrt{5})$. However, a more interesting field is $Q(\phi)$, with the golden ratio $\phi = (1+\sqrt{5})/2$. Hence, these sextics have the relevant cubic factor,

$5x^3 - 5(53+86\phi)x^2 + 5(\color{blue}{8}+\color{blue}{13}\phi)x - (\color{red}{18}+\color{red}{29}\phi) = 0$

$5x^3 - 20(73+118\phi)x^2 - 20(\color{blue}{21}+\color{blue}{34}\phi)x - (\color{red}{47}+\color{red}{76}\phi) = 0$

$5x^3 - 20(8849+14318\phi)x^2 + 20(\color{blue}{377}+\color{blue}{610}\phi)x - (\color{red}{843}+\color{red}{1364}\phi) = 0$

respectively. Compare the x term with the Fibonacci numbers,

$F_n = 0, 1, 1, 2, 3, 5, \color{blue}{8, 13, 21, 34}, 55, 89, 144, 233, \color{blue}{377, 610},\dots$

and the constant term with the Lucas numbers,

$L_n = 2, 1, 3, 4, 7, 11, \color{red}{18, 29, 47, 76}, 123, 199, 322, 521, \color{red}{843, 1364},\dots$

Why, oh, why?

P.S. These can be easily verified in Mathematica using the Resultant[] function,

Resultant[$5x^3 - 5(53+86\phi)x^2 + 5(8+13\phi)x - (18+29\phi)$, $\phi^2-\phi-1$, $\phi$]

which eliminates $\phi$ and restores the original sextic. (Similarly for the other two.)

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18  
Would it be slightly less mysterious if I remarked that $F_{n-1} + F_n \phi = \phi^n$ for $n \ge 1$ (and presumably a similar identity holds for the Lucas numbers)? –  Qiaochu Yuan May 22 '11 at 19:58
10  
I don't get the bit about ${\bf Q}(\phi)$ being a more interesting field than ${\bf Q}(\sqrt5)$ --- it's the same field, innit? –  Gerry Myerson May 23 '11 at 0:07
2  
1 To Yuan: Oh, I should have seen that! Thus, $5x^3 - 5(53+86\phi)x^2 + 5\phi^7x - \sqrt{5}\phi^7 = 0$, $5x^3 - 20(73+118\phi)x^2 - 20\phi^9x - \sqrt{5}\phi^9 = 0$, $5x^3 - 20(8849+14318φ)x^2 + 20\phi^{15}x - \sqrt{5}\phi^{15} = 0$. 2. To Gerry: My apologies. You are right. My intent was that if one factors it over $Q(\phi)$ vs $Q(\sqrt{5})$, the polynomial $F_{n-1}+F_n\phi$ is more recognizable than some polynomial $a+b\sqrt{5}$. –  Tito Piezas III May 23 '11 at 2:49
    
@QiaochuYuan Check this out. –  Pedro Tamaroff Apr 5 '12 at 15:31
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2 Answers

As described in my comments to Tito's posting in

https://groups.google.com/group/sci.math.research/browse_thread/thread/3d24137c9a860893?hl=en&pli=1

approximations could be re-written as:

exp(Pi*sqrt(19+24*n)) =~ (24*k)^3 + 31*24

which gives 4 (four) "almost integer" solutions:

1) n=0, k= 4 ;

2) n=1, k= 40 ;

3) n=2, k= 220 ;

4) n=6, k = 26680 ; -this of course is the case for Ramanujan constant vs its integer counterpart approximation

Note that 960, 5280, 640320

mentioned in the original question, posted here by Tito

are related to above cases 2), 3), 4)

960=24*40, 5280=220*40, 640320=24*26680

So, similarly, the case 1), that is

exp(Pi*sqrt(19))

could also be included into the new representation

with the equation

5*x^6-96*x^5-10*x^3+1=0

which root

x ~= 19.2054...

also satisfies

exp(Pi*sqrt(19)) ~= (5*x)^3 + 6

So possibly the root solution of

5*x^6-96*x^5-10*x^3+1=0

could be also presented over "golden ratio" ?

Also note that if to expand

exp(Pi*sqrt(b(n)))

to more values b(n) = {19, 25, 43, 58, 67, 163, 232, ...}

then the expression

(exp(Pi*sqrt(b(n))))/m

(where m is either integer 1 or 8 )

yields values being very close to whole integer value :

Note, that the first differences of b(n) are all dividable by 3, giving after the division:

{2, 6, 5, 3, 32, 33, ...}

Note also that

exp(Pi*sqrt(19+24*k))=

=exp(Pi*sqrt(19+24*(4^(k-2)*Pochhammer[1/2,k-2])/Pochhammer[1,k-2]))

for k={1,2,3,4}

and that coefficients at x^5 for the sextics given above in the question by Tito

could be derived from

solve n=floor(exp(Pi/3*sqrt(19+24*(4^(k-2)*Pochhammer[1/2,k-2])/Pochhammer[1,k-2]))),{k=1,2,3,4}

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Solve $5x^6-Ax^5-10x^3+1$ for A, and you get $A=5x-\frac{10}{x^2}+\frac{1}{x^5}$

Solve $A^3+744 = (5x)^3+6$ for A, and you get $A=(125x^3-738)^{\frac13}$

These are almost equal as $x\rightarrow\infty$. In fact, if the difference is $D(x)$, then taking the taylor series of $D(\frac1y)$ around $y=0$ gives $D(x)=-\frac{4}{25x^2}+\frac{63641}{3125y^5}+\dots$

This solves half your mystery.

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Did you mean $63641/3125x^5$ ? –  grok_it Sep 30 '11 at 10:59
    
The roots "x" actually are the EXACT values of certain Dedekind eta quotients. What I found rather surprising was how the polynomials that defined the "x's" neatly contained high powers of phi (as pointed out by Yuan) in its terms. –  Tito Piezas III Dec 5 '11 at 8:32
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