I stumbled upon the following identity, which I have not tried to prove, but seems true:
the function $$f(t):={}\_2F\_1(1/2,2t;1-t;4)$$
is periodic of period 1, and more precisely
$$f(t)=\dfrac{1+2e^{-2\pi it}}{3}\;.$$
This begs several questions:


(1). Is this true? (2). Are there other (infinitely many?) formulas of this type (of course outside of trivial manipulations of this) ? (3) More generally, call a function $f(t)$ scale-periodic if there exists $A>0$ such that $A^{-t}f(t)$ is $1$-periodic (so periodic if $A=1$). I have found a number of such $f(t)$ of the form ${}\_2F\_1(a(t),b(t);c(t);z)$: is there some way to find them? (this last question was inspired by a paper of Beukers and Forsgard arXiv:2004.08117).



Periodic Gauss hypergeometric function
Gauss' contiguous relations provide a basis for finding a linear relationship between three functions of the form ${}\_2F\_1(a+k, b+l, c+m, z)$, henceforth $\mathbf{F}\left(\begin{matrix}a+k, b+l \\ c+m\end{matrix}\right)$. Various papers have been published on calculating these relationships; I'm using the notation of Vidūnas's paper [*Contiguous relations of hypergeometric series*](https://arxiv.org/abs/math/0109222):


$$\textbf{F}\left(\begin{matrix}a+k, b+l \\ c+m\end{matrix}\right) = \textbf{P}(k,l,m) \textbf{F}\left(\begin{matrix}a, b \\ c\end{matrix}\right) + \textbf{Q}(k,l,m) \textbf{F}\left(\begin{matrix}a+1, b \\ c\end{matrix}\right)$$


Vidūnas recommends finding the coefficients by first chaining up $k$ with his formula (9), but a naïve approach runs into bootstrap problems (as I discovered by trying it). Once $\textrm{Q}(2,0,0)$ is found, I think the approach works, but here we have such small shifts that we can do directly.


Rephrasing his (2), (3), (4) in terms of $\textbf{Q}$ we obtain


$$\mathbf{Q}(k+1,l,m) = \tfrac{(2a+2k-c-m+(b+l-a-k)z)}{(a+k)(1-z)}\mathbf{Q}(k,l,m) + \tfrac{(c+m-a-k)}{(a+k)(1-z)}\mathbf{Q}(k-1,l,m)$$
$$\mathbf{Q}(k,l,m-1) = \tfrac{a+k}{c+m-1} \mathbf{Q}(k+1,l,m) + \tfrac{c+m-a-k-1}{c+m-1} \mathbf{Q}(k,l,m)$$
$$\mathbf{Q}(k,l+1,m) = \tfrac{a+k}{b+l} \mathbf{Q}(k+1,l,m) + \tfrac{b+l-a-k}{b+l}\mathbf{Q}(k,l,m)$$


and by repeated application of these I calculate that


$$\mathbf{Q}(0,2,-1) = a \tfrac{(a+b-c+1)(a+b-c+2) + (c-a-1)(2a+b-c+1)(1-z) + (b-a+1)(c-a-1)(1-z)^2}{(c-1)b(b+1)(1-z)^2} \\
$$


$$\mathbf{P}(0,2,-1) = (c-a-1) \tfrac{a(a+b-c+2) + a(-2a+b+c-1)(1-z) + (b-a)(b-a+1)(1-z)^2}{(c-1)b(b+1)(1-z)^2}$$


Substituting your parameters $a=\tfrac12$, $b=2t$, $c=1-t$, $z=4$ I get respectively $0$ and $1$, which establishes the periodicity of your $f$.




---


To search systematically for scale-periodic hypergeometric functions, I propose the following approach:


1. Pick $k$, $l$, $m \in \mathbb{Z}$ (taking into account symmetries for efficiency).
2. Compute $\mathbf{Q}(k,l,m)$ and $\mathbf{P}(k,l,m)$.
3. Substitute $a = a\_0 + kt$, $b = b\_0 + lt$, $c = c\_0 + mt$ into $\mathbf{P}(k,l,m) = A$ and $\mathbf{Q}(k,l,m) = 0$. Equate coefficients of $t$ in each to get a collection of polynomial constraints over $a\_0,b\_0,c\_0,z,A$. Use Gröbner bases or other techniques to solve these constraints.


To take the obvious example of $k=0$, $l=2$, $m=-1$, and (for simplicity) applying the substitution $x = 1-z$, $\textbf{Q}(0,2,-1) = 0$ gives


$$
a\_0 (9-3x-2x^2) = 0 \\
6a\_0+6b\_0-6c\_0+9+(-5a\_0-b\_0+4c\_0-4)x+(-a\_0-b\_0+2c\_0-3)x^2 = 0 \\
(a\_0+b\_0-c\_0+1)(a\_0+b\_0-c\_0+2) +
(-a\_0+c\_0-1)(2a\_0+b\_0-c\_0+1)x +
(-a\_0+b\_0+1)(-a\_0+c\_0-1)x^2 = 0
$$


Ignoring the trivial $a\_0 = 0$, we see that $x \in \{-3, \tfrac32\}$.


### Case $x=-3$


$a\_0 = \tfrac12$ drops out of the next constraint, leaving the final constraint to reduce to $(b\_0+2c\_0-3)(b\_0+2c\_0-2) = 0$. If $b\_0+2c\_0-2 = 0$ then we're going to recover your $f$, so consider instead $b\_0+2c\_0-3 = 0$. This turns out to have no solutions: substituting into $\mathbf{P}$ and expanding we get a contradiction in the coefficients of $t^0$.


### Case $x = \tfrac32$


Comparing coefficients for $t^3$ in $\mathbf{P}$ gives $A = 1$; subsequently comparing coefficients for $t^2$ in $\mathbf{P}$ gives $a\_0 = 0$, so there is no non-trivial solution.


I therefore withdraw my blithe assertion in an earlier comment that question (2) is easily answered in the affirmative; it remains plausible, but with larger shifts we're going to have more coefficients of $t$ to impose constraints on a fixed number of variables and so there might not be very many sets of shifts which admit solutions.


