$\sf V=HOD$ is stated as:


$\forall X \, \exists \theta \, \exists \alpha < \theta \, \exists \varphi : X=\{y \in V\_\theta\mid V\_\theta\models\varphi(y,\alpha)\}$


This use two ordinal parameters (other than the code for $\varphi$) $``\theta; \alpha"$.


Can we do with just ONE parameter (other than the code for $\varphi$)? that is:


$\forall X \, \exists \alpha \, \exists \varphi: X= \{y \in V\_\alpha \mid V\_\alpha \models \varphi(y)\}$



Can we state $\sf V=HOD$ using a single ordinal parameter(other than the formula code)?
Yes, because there is a definable ordinal pairing function.


Specifically, if you want to get the set $\{y\in V\_\theta\mid V\_\theta\models\varphi(y,\alpha)\}$, then let $\beta=\langle\theta,\alpha\rangle$ be the ordinal coding the pair, and then look at $V\_{\beta+1}$. Inside this structure, we have $\beta$ as the largest ordinal, and the model can decode $\beta$ as the pair $\langle\theta,\alpha\rangle$, and then get the set $\{y\in V\_\theta\mid V\_\theta\models\varphi(y,\alpha)\}$. So this set is definable inside $V\_{\beta+1}$ without any extra parameter.


Ultimately one should view $\varphi$ also as a parameter (and formulas can be coded by finite numbers, hence by ordinals, a fact that is important when proving that HOD as defined internally is the same as the external concept of definable-from-ordinal parameters, since the internal definition might use nonstandard $\varphi$, but this is still OK since the code of a nonstandard formula is still a number, hence an ordinal). But this is fine, since we can consider $\beta=\langle\theta,\alpha,\ulcorner\varphi\urcorner\rangle$ and then proceed as I explained.


