I. Let $n=2$ and $x = \frac1\phi$ with golden ratio $\phi$. Then it satisfies, $$x(1 - x^{12})^2(1 - x^{24})^2 = (1 - x^6)^7(1 - x^8)^4\tag1$$ $$x(1 - x^{12})^2 = (1 - x^4)^\color{blue}2(1 - x^6)^3\tag2$$

II. Let $n=3$ and $y = \frac1T$ with tribonacci constant $T$. Then,

$$y (1 - y^8)^2 = (1 - y^2)(1 - y^4)^\color{blue}3\tag3$$

III. Let $n=5$ and $z = \frac1P$ with pentanacci constant $P$. Then,

$$z(1 - z^8)^2(1 - z^{12})^4 = (1 - z^2)^2(1 - z^6)^\color{blue}5(1 - z^{24})\tag4$$

And this is where the pattern apparently stops.

The first $(1)$ was mentioned by D. Broadhurst in Multiple Landen values and the tribonacci numbers. (I corrected his typo involving the first factor $x$ of the LHS.) The rest were found by the OP using Mathematica.

Q: Is it possible to find $n=7$?

I expanded my search parameters using $1-q^{m}$ for all even $m\leq32$ but couldn't find anything, nor the next prime $n=7$, so I'm wondering what so special about orders $n=2,3,5$.

P.S. See this post for a related family for all $n$.

  • $\begingroup$ For part III isn't $z(1-z^6)^2=(1-z)(1-z^{12})$ a simpler choice? What is the pattern? $\endgroup$ – Somos Aug 31 '17 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.