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I. Degree 8

Assume the $j_i$ to be free parameters. The following octics in $x$ belong to $8T43,$ have group $\text{PGL}(2,7)$, and order $2\times168 = 336.$

\begin{align} {j_1}\; &=\frac{(x^2 + 5x + 1)^3(x^2 + 13x + 49)}x\\ {j_2}^2 &=\frac{j_2\,(-7x^4 - 196x^3 - 1666x^2 - 3860x + 49)+(x^2 + 14 x + 21)^4}x\\ {j_3}\; &=\frac{(x + 1)^6(x^2 + x + 7)}x\\ {j_4}^2 &=\frac{7 j_4\,(x + 1)^4+(x + 1)^7 (x - 7)}x \end{align}

Expanded out, some of the coefficients are quadratic in $j_i$. They also have nice discriminants $D_i$. (Note: $D_2$ has another square factor I've suppressed for brevity. I have a feeling the second octic may have a simpler version.)

\begin{align} D_1 &= -7^7(j_1-1728)^4\,{j_1}^4\\ D_2 &= -7^7(j_2-256)^4\,{j_2}^6 \\ D_3 &= -7^7(j_3-108)^3\,{j_3}^5\\ D_4 &= -7^7(j_4-64)^4\,{j_4}^{12} \end{align}

For general $j_i$, these octics are not solvable in radicals. However, if we substitute the following integers,

\begin{align} j_1 &= -640320^3\quad \\ j_2 &= -63^2,\, 396^4\quad \\ j_3 &= -300^3\quad \\ j_4 &= -2^9,\, 2^{12}\quad \end{align}

then they become solvable. The first two $j_i$ are famous being in the Chudnovsky and Ramanujan pi formulas, and the last two can be used in Ramanujan-Sato pi formulas as well.

II. Eta quotients

There are infinitely many such radical $j_i$ given by,

\begin{align} \quad j_1 &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{8}+2^8 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{16}\right)^3 \\ \quad j_{2} &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2 \\ \quad j_{3} &=\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2 \\ \quad j_{4} &=\left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4} + 4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24} \end{align}

where $\eta(\tau)$ is the Dedekind eta function and $j_1$ is just the j-function. (For examples of $\tau$, click on the link.)

Thus, the octics in $x$ are also formulas for the $j_i$ and where $x$ involve eta quotients (though I didn't include them to prevent clutter).

III. Degree 6

So far, I've only found two with group $\text{PGL}(2,5)$ hence $6T14$ and order $2\times60 = 120$. (Edit: For instant comparison, we add the other two $6T14$ from Joachim König's answer, in red.)

\begin{align} j_1 &=\frac{(x^2 + 10x + 5)^3}x\\ \color{red}{j_2} &=\frac{(x+1)^4(x^2+6x+25)}{x}\\ \color{red}{{j_3}^2} &= \frac{j_3\,(2 y^6 + 29y^5 + 85y^4 + 50y^3) + 5^4 y^6}{(2y - 1)} \\ j_4 &=\frac{(x + 1)^5 (x + 5)}x \end{align}

with discriminants,

\begin{align} D_1 &= 5^5\,(j_1-1728)^2\,{j_1}^4\\ D_2 &= 5^5\,(j_2-256)^3\,{j_2}^3 \\ D_3 &= 5^5(j_3-108)^3\,{j_3}^{11}\\ D_4 &= 5^5\,(j_4-64)^2\,{j_4}^4 \end{align}

It is preferred the constant factor is only $5^5$. Note: $D_3$ actually has the extra square factor $(32 j_3+84375)^2$, but I truncated it for aesthetics.

IV. Question:

So are there analogous sextic $6T14$ formulas in $x$ for $j_2, j_3$, where the $j_i$ are linear or quadratic, and discriminants similar to the others? (Note: Has been answered in the affirmative.)

V. Addendum:

(The addendum has been moved as an "answer" to prevent clutter and loading issues.)

Tito Piezas III
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  • For another example, $j_3 = -(2\sqrt3,r)^6$ such that, $$e^{\pi\sqrt{929/3}} \approx (2\sqrt3,r)^6 +41.99999999999999999999923\dots$$ and $r$ is a root of the simple cubic $r^3 - 2898r^2 + 2346r - 583 = 0.$ – Tito Piezas III May 06 '23 at 11:11
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  • There is a $d$ in the title that is never defined or referred to in the body of the question. 2) $j_1$ and $j_3$ are not polynomials, so it's not clear how they are octics. 3) The formula for $j_2^2$ expresses it in terms of $j_2$; similarly for $j_4^2$; what is going on there? 4) "for some integers $a,b,c,d$" but there is no other mention of those integers. This whole post is very confusing.
    • – Gerry Myerson May 07 '23 at 00:44
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    It may help to add a more combinatorial viewpoint: these four all define three-point ramified covers (i. e., essentially Belyi maps) of the line parametrized by $j_i$ with inertia groups of orders $(2,3,p)$, $(2,4,p)$, $(2,6,p)$ and $(2,p,p)$ in the four respective cases. That makes them essentially unique (in the respective groups $PGL_2(p)$), and changing from $p=7$ to $p=5$ gives the "analoga". Not all of them are genus zero in the action on $p+1$ points though, some are genus $1$ (as easily read off from the cycle types of the permutations involved), and then one needs degree 2 in $j_i$. – Joachim König May 07 '23 at 08:57
  • @JoachimKönig Thanks for the novel analysis. Do you suppose u can find an alternative octic formula for $j_2$? I find it overly complex compared to the others (and its discriminant has an extra polynomial factor I suppressed). I still have to re-edit the post anyway. – Tito Piezas III May 07 '23 at 09:09
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    @GerryMyerson I've edited the post to address the points you mentioned. And I've included an addendum to explain the role of $d$, among other things. – Tito Piezas III May 07 '23 at 15:23