Big O notation where the "squeeze" starts at $0$?

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There is the standard notation : $$f = O(g) \iff \exists M > 0, \,\exists n_0 > 0, \,\forall n > n_0, \quad \left|f(n)\right| \leq M\cdot g(n) $$ But this is not adequate for a paper I'm writing because I need to force $n_0$ to be $0$, i.e. : $$ \exists M > 0, \,\forall n > 0, \quad \left|f(n)\right| \leq M\cdot g(n) $$

I'll be using this a lot, so a synthetic notation like $f = O(g)$ is required.

Is there a standard notation for this? Or a not so standard one?

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What is often used in this case is $$ |f(n)| \leq \operatorname{const} g(n) $$ or $$ |f| \leq \operatorname{const} g $$ or $$ |f| \leq C\, g. $$ where it is understood that the constant does not depend on $n$.

However, it sometimes depends on the context, if $C$ can depend on $f$ and $g$.

I do not think that this is a standard notation.