Why do some Fibonacci numbers appear in an approximation for $e^{\pi\sqrt{163}}$?

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.)

Solution 1:

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.

Solution 2:

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:

$e^{\pi \sqrt{19+24\cdot n}} \approx (24k)^3 + 31\cdot24 $

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)

$960=24\cdot 40$, $5280=220\cdot 40$, $640320=24\cdot 26680$

So, similarly, the case 1), that is $e^{\pi \sqrt{19}}$ could also be included into the new representation with the equation $5 x^6-96 x^5-10 x^3+1=0$ which root $x \approx 19.2054...$ also satisfies $e^{\pi \sqrt{19}}\approx (5 x)^3 + 6$

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 $e^{\pi \sqrt{b(n)}}$ to more values $ b(n) = \{19, 25, 43, 58, 67, 163, 232, \ldots\} $ then the expression $e^{\pi \sqrt{b(n)}}/m$ (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: $\{2, 6, 5, 3, 32, 33, \ldots\}$

Note also that $e^{\pi \sqrt{19 + 24k}}=$

exp(Pisqrt(19+24(4^(k-2)*Pochhammer[1/2,k-2])/Pochhammer[1,k-2])) for k={1,2,3,4}

Another plain text notation is:

exp(π sqrt(19 + 24×(4^(k - 2) (1/2)(k - 2))/(1)(k - 2))) where k = {1, 2, 3, 4}

And in Wolfram Language Code notation:

ReplaceAll[Exp[Pi Sqrt[19 + 24 ((4^(k - 2) Pochhammer[1/2, k - 2])/Pochhammer[1, k - 2])]], {k -> {1, 2, 3, 4}}]

Also note 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}