# Why do we divide the regularization term by the number of examples in regularized logistic regression?

So this is the formula for the regularized logistic regression cost function: $$x^{(i)}$$ - the $$i$$'th training example

$$\theta_j$$ - the parameter of the $$j$$'th feature

$$m$$ - the number of training examples

$$n$$ - the number of features

$$y_{(i)}$$ - the actual outcome of the $$i$$'th training example, such that $$y \in \{0, 1\}$$

$$\lambda$$ - the regularization term

$$h_\theta$$ - the hypothesis function that produces a prediction in the interval $$(0, 1)$$

$$h_\theta(x^{(i)})$$ - predicted value of the $$i$$'th training example, such that:

$$h_\theta(x^{(i)}) = \sigma(\theta^{T}x^{(i)})$$, where $$\sigma(z) = \frac{1}{1+e^{-z}}$$ (the sigmoid/logistic function)

My question is about that last term: From my understanding, in order to do regularization, we need to find the sum of all the squares of the parameters $$\theta$$. That is clear enough. Also we multiply this sum by the regularization term $$\lambda$$, which we can change if we think the data is overfit/underfit. Good. Then, for convenience we divide this term by $$2$$ such that when we take the derivative we will get rid of that $$2$$ that would come from the exponent of $$\theta$$. All clear so far. BUT, why in the world do we also divide by $$m$$ (the number of training examples)? We divide the left term by $$m$$ in order to find the average error, and that makes sense since we have $$m$$ examples, therefore $$m$$ errors and after we find the sum of these $$m$$ errors, we would need to divide by $$m$$ to get the average error. But in this right term that I am confused about, we find the sum of the squares of the features, and the number of features is $$n$$. If we want to find the average of all the $$\theta_j^2$$ wouldn't we need to divide by $$n$$ instead of $$m$$, since we have $$n$$ features. Why does it make sense to divide that sum by $$m$$ and shouldn't we divide it by $$n$$?

This way the same $$\lambda$$ value has better chances for working with the whole dataset as well as with a small part of it without adjusting.