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By playing around, I noticed that $F_n$, the $n$th Fibonacci number, equals $$\frac{\cosh\left(n\operatorname{arcsinh}\left(\frac12\right)\right)}{\cosh\left(\operatorname{arcsinh}\left(\frac12\right)\right)}\text{ for odd $n$,}\quad\frac{\sinh\left(n\operatorname{arcsinh}\left(\frac12\right)\right)}{\cosh\left(\operatorname{arcsinh}\left(\frac12\right)\right)}\text{ for even $n$.}$$ This can be proved by showing that it is equivalent to Binet's formula.

I imagine this formula is almost certainly already known. Assuming so, does it have a name? What is it? (I'm turning to Math Stack Exchange because it's not easy to Google formulas.

For a quick proof sketch, using Binet's formula, one can show that $$\begin{align*}\cosh\left(n\operatorname{arcsinh}\left(\frac{1}{2}\right)\right)=\begin{cases}\frac{\sqrt5}2F_n,&\text{$n$ odd}\\\frac12L_n,&\text{$n$ even}\end{cases}\\ \sinh\left(n\operatorname{arcsinh}\left(\frac{1}{2}\right)\right)=\begin{cases}\frac12L_n,&\text{$n$ odd}\\\frac{\sqrt5}2L_n,&\text{$n$ even}\end{cases}\end{align*}$$ where $L_n:=F_{n-1}+F_{n+1}$ is the $n$th Lucas number. Since $\cosh\left(\operatorname{arcsinh}\left(\frac12\right)\right)=\frac{\sqrt5}2$, the above formula follows readily.

As a bonus, the Lucas numbers satisfy a similar formula as well. They equal $$2\sinh\left(n\operatorname{arcsinh}\left(\frac{1}{2}\right)\right)\text{ for odd $n$,}\qquad2\cosh\left(n\operatorname{arcsinh}\left(\frac{1}{2}\right)\right)\text{ for even $n$.}$$

Akiva Weinberger
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    I am not aware of a specific name. See, for example: Piero Filipponi and Herta T. Freitag, "Conversion of Fibonacci identities into hyperbolic identities valid for an arbitrary argument." In Proceedings of The Fourth International Conference on Fibonacci Numbers and Their Applications, Aug. 1990 (Applications of Fibonacci Numbers: Volume 4, Springer 1991), pp. 91-98. – njuffa Sep 26 '24 at 08:41
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    An early paper that considers the relationship between Fibonacci numbers and hyperbolic functions: Édouard Lucas, "Sur la théorie des fonctions numériques simplement périodiques.", Nouvelle Correspondance Mathématique, Vol. 3, 1877, pp. 369-376. See section "II Relations avec les fonctions circulaires ou hyperboliques" – njuffa Sep 26 '24 at 08:42

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I don't believe it has a name and I would describe it as just a restatement of Binet's formula. You could say conceptually the point is that the Fibonacci matrix $F = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$, whose powers have entries given by Fibonacci numbers, has determinant $-1$ and real eigenvalues. So its square has determinant $1$ and positive real eigenvalues, which therefore have the form $e^r, e^{-r}$ for some $r$; in other words, $F^2$ is conjugate to a hyperbolic rotation.

Qiaochu Yuan
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    This description does not explain why the formula depends on the parity of $n$. – Akiva Weinberger Sep 26 '24 at 01:27
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    @AkivaWeinberger: It does. A linear recurrence admits an explicit solution which is readily derived from the Jordan form of its matrix and the initial data. Passing from odd to even does not change the matrix, but it does change the initial data. – tomasz Sep 26 '24 at 12:20
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    Oh, I didn't read carefully. $F$ is not conjugate to a hyperbolic rotation, but $F^2$ is. I see now. – Akiva Weinberger Sep 26 '24 at 13:18