A382962 - OEIS

%I #18 Apr 25 2025 20:41:06

%S 1,5,48,831,21320,772422

%N Number of symmetric ternary maps f : S X S X S -> S on a set S of n elements which can be represented as a superposition of binary maps * : S X S -> S.

%C Sequence corresponds to Tcomm(n) from the Zusmanovich paper.

%C We consider ternary maps that can be represented as

%C   f(x, y, z) = (x * y) * z.

%C Symmetry is defined here as independence from the order of arguments, hence

%C   f(x, y, z) = f(x, z, y) = f(y, x, z) = f(y, z, x) = f(z, x, y) = f(z, y, x).

%C Question 15 from the Zusmanovich paper remains unanswered: at least for n <= 6, every f satisfying the above can be represented by a superposition of some commutative '*', but no proof that this is the case for all n.

%H Pasha Zusmanovich, <a href="https://arxiv.org/abs/1608.05863">Lie algebras and around: selected questions</a>, arXiv:1608.05863 [math.RA], 2016.

%e For N=2, 6 of the N^(N^2) binary maps '*' define a symmetric ternary map f(x,y,z)

%e   *:                  f:

%e   ((0, 0), (0, 0))    (((0, 0), (0, 0)), ((0, 0), (0, 0)))

%e   ((0, 0), (0, 1))    (((0, 0), (0, 0)), ((0, 0), (0, 1)))

%e   ((0, 1), (1, 0))    (((0, 1), (1, 0)), ((1, 0), (0, 1)))

%e   ((0, 1), (1, 1))    (((0, 1), (1, 1)), ((1, 1), (1, 1)))

%e   ((1, 0), (0, 1))    (((0, 1), (1, 0)), ((1, 0), (0, 1)))

%e   ((1, 1), (1, 1))    (((1, 1), (1, 1)), ((1, 1), (1, 1)))

%e The 3rd and 5th ternary maps are identical and are counted only once, hence a(2) = 5.

%Y Cf. A283840, A283841.

%K nonn,hard,more

%O 1,2

%A _Bert Dobbelaere_, Apr 10 2025