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3 Identified the two other $w$-images as linear combinations of $\varphi$ and $\phi$; also some minor copy editing

[More a comment than an answer, but too long for the comment space]

Call this form $$\varphi := \frac{\eta(q^3)^2 \eta(q^6)^3 \eta(q^9)^2}{\eta(q^{18})} = q - 2q^4 - 4q^7 + 6q^{10} + 8q^{13} \cdots.$$ The listing of coefficients in the OEIS is correct as far as it goes (checked with copy-and-paste to gp). The form is not CM: the coefficients are supported on $q^n$ with $n \equiv 1 \bmod 3$ but do not vanish even for $n$ such as $10$ and $22$ that are $1 \bmod 3$ but not norms from ${\bf Q}(\sqrt{-3})$. Indeed In particular the coefficients aren't multiplicative, so $\varphi$ isn't quite an eigenform. It seems that the relevant eigenforms are obtained as follows. Apply $w_{18}$ to get (within a multiplicative factor) $$\phi := \frac{\eta(q^6)^2 \eta(q^3)^3 \eta(q^2)^2}{\eta(q)} = q + q^2 - 2q^4 - 3q^5 - 4q^7 - 2q^8 + 6q^{10} + 12q^{11} + 8q^{13} - 4q^{14} \cdots,$$ whose $q^n$ coefficient is 0 if $n \equiv 0 \bmod 3$, and coincides with the $q^n$ coefficient of $\varphi$ also when $n \equiv 1 \bmod 3$, but need not vanish for $n \equiv 2 \bmod 3$. Then "experimentally" if $m,n$ are relatively prime then the $q^{mn}$ coefficient of $\phi$ equals the product of the $q^m$ and $q^n$ coefficients, except when unless both $m$ and $n$ are $2 \bmod 3$, when the $q^{mn}$ coefficient is $-2$ times that product. Hence we obtain an eigenform by choosing a square root of $-2$ and multiplying the $q^n$ coefficient of $\phi$ by that square root for each $n \equiv 2 \bmod 3$.

As Dror Speiser notes, the Edixhoven program promises to compute the $q^n$ coefficient of such a form in time $\log^{O(1)}n$ for $n$ prime, and thus for all $n$ given the factorization of $n$; but I don't think this has been implemented yet to the point that one could actually carry out the computation this way. For specific forms there can be shortcuts that make a $\log^{O(1)}n$ computation practical (still assuming $n$ is factored), but here I've tried a few things and not yet(?) found such a shortcut.

[added later] Curiously the images of $\phi$ under the other two $w$ operators are in the linear span of $\varphi$ and $\phi$: if we write $$\psi = \frac{\eta(q^3)^3 \eta(q^6)^2 \eta(q^{18})^2}{\eta(q^9)} = q^2 - 3q^5 - 2q^8 + 12q^{11} - 4q^{14} \cdots$$ for (a multiple of) the $w_2$ image, then $\phi = \varphi + \psi$, while $\varphi - 2 \psi$ is the multiple $$\frac{\eta(q)^2 \eta(q^3)^2 \eta(q^6)^3}{\eta(q^2)} = q - 2q^2 - 2q^4 + 6q^5 - 4q^7 + 4q^8 + 6q^{10} - 24q^{11} + 8q^{13} + 8q^{14} \ldots$$ of the $w_9$ image.

2 Correction at the end of penultimate paragraph (as noted by D.Hansen)

[More a comment than an answer, but too long for the comment space]

Call this form $$\varphi := \frac{\eta(q^3)^2 \eta(q^6)^3 \eta(q^9)^2}{\eta(q^{18})} = q - 2q^4 - 4q^7 + 6q^{10} + 8q^{13} \cdots.$$ The listing of coefficients in the OEIS is correct as far as it goes (checked with copy-and-paste to gp). The form is not CM: the coefficients are supported on $q^n$ with $n \equiv 1 \bmod 3$ but do not vanish even for $n$ such as $10$ and $22$ that are $1 \bmod 3$ but not norms from ${\bf Q}(\sqrt{-3})$. Indeed $\varphi$ isn't quite an eigenform. It seems that the relevant eigenforms are obtained as follows. Apply $w_{18}$ to get $$\phi := \frac{\eta(q^6)^2 \eta(q^3)^3 \eta(q^2)^2}{\eta(q)} = q + q^2 - 2q^4 - 3q^5 - 4q^7 - 2q^8 + 6q^{10} + 12q^{11} + 8q^{13} - 4q^{14} \cdots,$$ whose $q^n$ coefficient is 0 if $n \equiv 0 \bmod 3$, and coincides with the $q^n$ coefficient of $\varphi$ also when $n \equiv 1 \bmod 3$, but need not vanish for $n \equiv 2 \bmod 3$. Then "experimentally" if $m,n$ are relatively prime then the $q^{mn}$ coefficient of $\phi$ equals the product of the $q^m$ and $q^n$ coefficients, except when both $m$ and $n$ are $2 \bmod 3$ when the $q^{mn}$ coefficient is $-2$ times that product. Hence we obtain an eigenform by choosing a square root of $-2$ and multiplying each the $q^n$ coefficient of $\phi$ by that square root for each $n \equiv 2 \bmod 3$.

As Dror Speiser notes, the Edixhoven program promises to compute the $q^n$ coefficient of such a form in time $\log^{O(1)}n$ for $n$ prime, and thus for all $n$ given the factorization of $n$; but I don't think this has been implemented yet to the point that one could actually carry out the computation this way. For specific forms there can be shortcuts that make a $\log^{O(1)}n$ computation practical (still assuming $n$ is factored), but here I've tried a few things and not yet(?) found such a shortcut.

1

[More a comment than an answer, but too long for the comment space]

Call this form $$\varphi := \frac{\eta(q^3)^2 \eta(q^6)^3 \eta(q^9)^2}{\eta(q^{18})} = q - 2q^4 - 4q^7 + 6q^{10} + 8q^{13} \cdots.$$ The listing of coefficients in the OEIS is correct as far as it goes (checked with copy-and-paste to gp). The form is not CM: the coefficients are supported on $q^n$ with $n \equiv 1 \bmod 3$ but do not vanish even for $n$ such as $10$ and $22$ that are $1 \bmod 3$ but not norms from ${\bf Q}(\sqrt{-3})$. Indeed $\varphi$ isn't quite an eigenform. It seems that the relevant eigenforms are obtained as follows. Apply $w_{18}$ to get $$\phi := \frac{\eta(q^6)^2 \eta(q^3)^3 \eta(q^2)^2}{\eta(q)} = q + q^2 - 2q^4 - 3q^5 - 4q^7 - 2q^8 + 6q^{10} + 12q^{11} + 8q^{13} - 4q^{14} \cdots,$$ whose $q^n$ coefficient is 0 if $n \equiv 0 \bmod 3$, and coincides with the $q^n$ coefficient of $\varphi$ also when $n \equiv 1 \bmod 3$, but need not vanish for $n \equiv 2 \bmod 3$. Then "experimentally" if $m,n$ are relatively prime then the $q^{mn}$ coefficient of $\phi$ equals the product of the $q^m$ and $q^n$ coefficients, except when both $m$ and $n$ are $2 \bmod 3$ when the $q^{mn}$ coefficient is $-2$ times that product. Hence we obtain an eigenform by choosing a square root of $-2$ and multiplying each coefficient of $\phi$ by that square root.

As Dror Speiser notes, the Edixhoven program promises to compute the $q^n$ coefficient of such a form in time $\log^{O(1)}n$ for $n$ prime, and thus for all $n$ given the factorization of $n$; but I don't think this has been implemented yet to the point that one could actually carry out the computation this way. For specific forms there can be shortcuts that make a $\log^{O(1)}n$ computation practical (still assuming $n$ is factored), but here I've tried a few things and not yet(?) found such a shortcut.