The question is pretty much in the title. It is a classical fact that a filtered dga gives rise to a multiplicative spectral sequence. It is claimed in Remark 4.1 of <https://arxiv.org/pdf/1410.6728.pdf> that this generalizes to filtered $A\_\infty$- algebras(see the page 27 of the same for this definition). One can define a notion of filtered $A\_N$-algebras for $N \in \mathbb{N}^{\geq 0}$, $N \geq 3$ where $m\_i$ are only defined for $i \leq N$ and the $A\_{\infty}$ equations hold for $i \leq N$ (so the multiplication is still associative on the level of homology).


Question: Is the natural spectral sequence associated to a filtered $A\_N$ algebra multiplicative?


I would guess the answer is yes, but could not find a reference for this fact and there seems to be some subtleties in the literature according to this earlier post:


[Multiplicative structure on spectral sequence](https://mathoverflow.net/questions/24025/multiplicative-structure-on-spectral-sequence)


Motivation: It seems like this might be an efficient way to prove that some spectral sequences that arise "in nature" are multiplicative, say by equipping the relevant (co) chain complex with an $A\_3$ structure. 



Does a filtered A_N algebra give rise to a multiplicative spectral sequence?
For the $d\_r$-differentials to be derivations, i.e., to satisfy the Leibniz rule $d\_r(x \cdot y) = d\_r(x) \cdot y \pm x \cdot d\_r(y)$ with $x \cdot y = m\_2(x \otimes y)$, it is enough to have a filtered differential graded $A\_2$-algebra. This follows from Sections 7 and 8 of Massey's 1954 paper "Products in Exact Couples", since the pairing $m\_2 \colon A \otimes A \to A$ is a filtration-preserving chain map.


Assuming the Leibniz rule, if the pairing of $E\_1$-terms (or $E\_2$-terms) is associative and unital, then so are the induced pairings of $E\_r$-terms for all greater $r$.


Aside: The "subtleties in the literature" concern exact couples that do not come from filtered chain complexes, e.g., those that arise from the homotopy groups of sequences of spectra. If a pairing of sequences is only defined in the stable homotopy category, then it is not generally clear that there will be a corresponding product in the spectral sequence.


