Jonathan. Frech’s WebBlog

Factoids #1 (#222)

Jonathan Frech,

IV) Com­mu­ta­tive, non-associative operations

For any natural num­ber 𝑛, let $\mathrm{Op}_n:=\left\{\star:\mathbb{Z}^2_n\to\mathbb{Z}_n\right\}$$\mathrm{Op}_n:=\left\{\star:\mathbb{Z}^2_n\to\mathbb{Z}_n\right\}$$\mathrm{Op}_n:=\left\{\star:\mathbb{Z}^2_n\to\mathbb{Z}_n\right\}$ denote the set of all operations on a set of that order. An operation ★ shall be called com­mu­ta­tive iff $\mathrm{commut}(\star):\Leftrightarrow\forall\,x,y\in\mathbb{Z}_n:x\star y=y\star x$$\mathrm{commut}(\star):\Leftrightarrow\forall\,x,y\in\mathbb{Z}_n:x\star y=y\star x$$\mathrm{commut}(\star):\Leftrightarrow\forall\,x,y\in\mathbb{Z}_n:x\star y=y\star x$ and be called associative iff $\mathrm{assoc}(\star):\Leftrightarrow\forall\,x,y,z\in\mathbb{Z}_n:x\star(y\star z)=(x\star y)\star z$$\mathrm{assoc}(\star):\Leftrightarrow\forall\,x,y,z\in\mathbb{Z}_n:x\star(y\star z)=(x\star y)\star z$$\mathrm{assoc}(\star):\Leftrightarrow\forall\,x,y,z\in\mathbb{Z}_n:x\star(y\star z)=(x\star y)\star z$ holds.

With the above defined, one may study $\mathrm{CnA}_n:=\{\star\in\mathrm{Op}_n:\mathrm{commut}(\star)\land\lnot\mathrm{assoc}(\star)\}$$\mathrm{CnA}_n:=\{\star\in\mathrm{Op}_n:\mathrm{commut}(\star)\land\lnot\mathrm{assoc}(\star)\}$$\mathrm{CnA}_n:=\{\star\in\mathrm{Op}_n:\mathrm{commut}(\star)\land\lnot\mathrm{assoc}(\star)\}$. For 𝑛 = 2, this set is nonempty for the first time, containing a manageable two elements, by name

$$\mathrm{CnA}_2=\Big\{\mathrm{nor}:(x,y)\mapsto 1+xy,\quad\mathrm{nand}:(x,y)\mapsto (1+x)\cdot(1+y)\Big\}.$$$$\mathrm{CnA}_2=\Big\{\mathrm{nor}:(x,y)\mapsto 1+xy,\quad\mathrm{nand}:(x,y)\mapsto (1+x)\cdot(1+y)\Big\}.$$$$\mathrm{CnA}_2=\Big\{\mathrm{nor}:(x,y)\mapsto 1+xy,\quad\mathrm{nand}:(x,y)\mapsto (1+x)\cdot(1+y)\Big\}.$$

However, based on the superexponential nature of $\#\mathrm{Op}_n=\#\mathbb{Z}_n^{\mathbb{Z}_n^2}=\#\mathbb{Z}_n^{{\#\mathbb{Z}_n}^2}=n^{n^2}$$\#\mathrm{Op}_n=\#\mathbb{Z}_n^{\mathbb{Z}_n^2}=\#\mathbb{Z}_n^{{\#\mathbb{Z}_n}^2}=n^{n^2}$$\#\mathrm{Op}_n=\#\mathbb{Z}_n^{\mathbb{Z}_n^2}=\#\mathbb{Z}_n^{{\#\mathbb{Z}_n}^2}=n^{n^2}$, the sequence $\mathrm{A079195}_n:=\#\mathrm{CnA}_n$$\mathrm{A079195}_n:=\#\mathrm{CnA}_n$$\mathrm{A079195}_n:=\#\mathrm{CnA}_n$ likely also grows rather quickly, OEIS on­ly listing four members;

$$\mathrm{A079195}=(0,2,666,1\,047\,436,\dots).$$$$\mathrm{A079195}=(0,2,666,1\,047\,436,\dots).$$$$\mathrm{A079195}=(0,2,666,1\,047\,436,\dots).$$

Based on this limited numerical evidence, I would suspect the com­mu­ta­tive yet non-associative operations to be rather sparse, i. e.

$$\lim\limits_{n\to\infty}\mathrm{A079195}_n\cdot\left(\#\mathrm{Op_n}\right)^{-1}=0\mod\square.$$$$\lim\limits_{n\to\infty}\mathrm{A079195}_n\cdot\left(\#\mathrm{Op_n}\right)^{-1}=0\mod\square.$$$$\lim\limits_{n\to\infty}\mathrm{A079195}_n\cdot\left(\#\mathrm{Op_n}\right)^{-1}=0\mod\square.$$

Analysis source: factoids-1_operations.hs

(Non-)com­mu­ta­tive and (non-)associative operations have also been studied nearly twenty years ago by Christian van den Bosch, author of OEIS sequence A079195. Un­for­tu­nate­ly, their site appears to be down, which is where they hosted closed binary operations on small sets (resource found on web.archive.org).

V) Arbitrary polynomial extremum difference

Let 𝜀 > 0 be an arbitrary distance, define $g:=-4x^4-x^3+8x^2+3x-4$$g:=-4x^4-x^3+8x^2+3x-4$$g:=-4x^4-x^3+8x^2+3x-4$. Then $f:=\sfrac{\epsilon}{4}\cdot g$$f:=\sfrac{\epsilon}{4}\cdot g$$f:=\sfrac{\epsilon}{4}\cdot g$ has two local maxima at - 1 and 1, whose vertical distance is 𝜀.

VI) Digit sum roots

It holds that $\mathrm{ds}_{10}(108^{12})=108$$\mathrm{ds}_{10}(108^{12})=108$$\mathrm{ds}_{10}(108^{12})=108$.