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

It is instead well - recognized 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 - recognized 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 ideal 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 the very least their dice origins) showing up once more. These sextics are understandable in radicals, factoring over $Q(\sqrt{5})$. Nonetheless, an extra intriguing area is $Q(\phi)$, with the gold proportion $\phi = (1+\sqrt{5})/2$. Therefore, these sextics have the pertinent cubic variable,

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

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

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

specifically. Contrast 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 also 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 conveniently validated in Mathematica making use of 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 removes $\phi$ and also recovers the initial sextic. (Similarly for the various other 2.)

38
2022-06-07 14:32:31
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Answers: 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)

$960=24\cdot 40$, $5280=132\cdot 40$, $640320=24\cdot 26680$

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 $x \approx 19.2054\dots$ 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 very close to integers.

Note that the first differences of $b(n)$ are all divisible by 3, giving after the division: $\{2, 6, 5, 3, 32, 33, \ldots\}$.

Note also that $$\exp(\pi \sqrt{19 + 24k}) = \exp\Big(\pi\sqrt{19+24(4^{k-2} [1/2]_{(k-2)})/[1]_{(k-2)}}\Big)$$ for $k={1,2,3,4}$. $a_{(b)}$ is the Pochhammer symbol aka falling factorial.

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/3sqrt(19+24(4^(k-2)*Pochhammer[1/2,k-2])/Pochhammer[1,k-2]))),k=1,2,3,4

1
2022-07-19 21:14:30
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Solve $5x^6-Ax^5-10x^3+1$ for A, and also you get $A=5x-\frac{10}{x^2}+\frac{1}{x^5}$

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

These are virtually equivalent as $x\rightarrow\infty$. Actually, if the distinction is $D(x)$, after that taking the taylor collection of $D(\frac1y)$ around $y=0$ offers $D(x)=-\frac{4}{25x^2}+\frac{63641}{3125y^5}+\dots$

This addresses half your enigma.

4
2022-06-10 17:46:28
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