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Define the Fibonacci terms $t_{n}(x,y)$ for all $n\geq 1$ by letting $t_{1}(x,y)=y,t_{2}(x,y)=x,t_{n+2}(x,y)=t_{n+1}(x,y)*t_{n}(x,y)$.

We say that an algebra $(X,*)$ is $N$-uniformly partially permutative if it satisfies the identities

  1. $x*(y*z)=(x*y)*(x*z)$ and

  2. $t_{N}(x,y)=t_{N+2}(x,y)$

  3. $t_{N+1}(x,y)=t_{N+3}(x,y)$

The collection $V_{N}$ of all $N$-uniformly partially permutative algebras is a variety, so whenever I come across any variety, I have to ask myself the following sorts of questions.

  1. Are the free algebras in $V_{N}$ on finitely many generators finite? I conjecture that the answer is no for 2 or more generators and for large enough $N$. For 1 generator, I am a bit more skeptical about the truth value of this conjecture.

  2. In the equational theory of the variety of $N$-uniformly partially permutative algebras decidable?

  3. Is the variety $V_{N}$ generated by its finite members? Can the free algebra in $V_{N}$ on one generator be embedded into the inverse limits of finite algebras in $V_{N}$?

  4. Do the algebras in $V_{N}$ generated by a single element satisfy any identities not satified by all algebras in $V_{N}$?

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  • $\begingroup$ In other words $t_n(x,y)=x^{F_{n-1}}y^{F_{n-2}}$ ? $\endgroup$
    – Max Alekseyev
    Commented Mar 22, 2019 at 4:51
  • $\begingroup$ The operation $*$ is not commutative nor associative, so $t_{5}(x,y)=((xy)x)(xy)$. The word algebra is used in a universal algebraic sense, and not a ring theoretic sense. $\endgroup$
    – Joseph Van Name
    Commented Mar 22, 2019 at 11:59
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    $\begingroup$ Is there any reason why this question is downvoted? $\endgroup$
    – Joseph Van Name
    Commented Mar 23, 2019 at 16:57
  • $\begingroup$ I doubt that the downvoters have much relevant knowledge about self-distributivity, the Fibonacci terms, or any topic relevant to this question. $\endgroup$
    – Joseph Van Name
    Commented Mar 23, 2019 at 17:15

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