Generic recursive function infers incorrect return type

### 🔎 Search Terms

generic, recursive, return, infer

### 🕗 Version & Regression Information

- This changed between versions 5.4.5 and 5.5.4
- This bug continues to exist in nightly

### ⏯ Playground Link

https://www.typescriptlang.org/play/?ts=6.0.3#code/MYewdgzgLgBAZjAvDAPAFQDQD4AUBDALhkxgAciwBXAWwCMBTAJwEoksYBvAKBl5gEsEOAISlWjelEqMwMPAG4efCVJnwcpDGRgBaGAEZmXAL5cuoSLAAeSdQCI4IEHa0BWIxYggANvQB03iAA5jhWzEA

### 💻 Code

```ts
const f = (a: T, p: number) => {
if (!p) return a;
return f(p, p - 1) // f
}

const x = f("foo", 5) // inferred type "foo"
console.log(x) // 1
```

### 🙁 Actual behavior

The compiler infers `f`'s return type to be `T`. Then `x` is inferred to have type `"foo"` (or `string`), even though the actual return value is `1` (type `number`).

### 🙂 Expected behavior

In the example `f` is a recursive and generic function. Importantly, when recursing `f` calls itself on a _non-generic_ type.

If `p` is falsy then `f` returns `a`, which has the generic type `T`. This is the "base case". However if `p` is truthy then `f` calls itself with type parameter `number`, so in all the recursive cases the return type is actually `number`.

So the actual return type of `f` should be `T | number`.

### Additional information about the issue

In earlier versions of the compiler (I tested 5.4.5) recursive functions without explicit return type annotations are flagged as an error, so we sidestep this bug by requiring the return type to be manually declared:

> 'f' implicitly has return type 'any' because it does not have a return type annotation and is referenced directly or indirectly in one of its return expressions. (7023)

In later versions however the compiler seems to think it can infer the return type, leading to this incorrect inferred result. Interestingly even though the compiler infers `f` to return type `T`, if we add the explicit return type annotation `T` then the compiler actually becomes unhappy (geez, make up your mind):

```ts
const f = (a: T, p: number): T => {
if (!p) return a;
return f(p, p - 1) // Type 'number' is not assignable to type 'T'. 'T' could be instantiated with an arbitrary type which could be unrelated to 'number'. (2322)
}
```

Commencez par la reproduction liée dans TypeScript Playground et comparez le comportement de l'inférence entre les versions du compilateur signalées. Suivez la manière dont les types de retour génériques récursifs sont inférés lorsque l'appel récursif utilise un type concret, puis ajoutez un test de régression pour cet exemple. C'est terminé lorsque la fonction est inférée comme retournant T | number et que le résultat inféré incorrect n'est plus accepté.

Rédigé par le modèle d'indexation à partir du texte de l'issue.