Polynomial cocycles for Alexander quandles were used for these calculations. The polynomial f(x,y)=(x-y)^(p^m) is a 2-cocycle for any Alexander quandle with Z_p coefficients (p denotes a prime), and f(x,y)=(x-y)^(p^m) y^(p^n) is a 2-cocycle for Z_p Alexander quandles modulo g(t) if g(t) divides (t^(p^m+p^n)-1).
We computed the invariants for knots in the table up to 9 crossings. We computed only those knots whose Alexander polynomials are not coprime mod p with g(t).(By Inoue's theorem only such knots are colored non-trivially.) Here the Alexander quandle we use is Z_p coefficients mod g(t).
When we found non-trivial values in our calculations, it is marked ``non-trivial.'' Note, however, that there are non-trivial (i.e., non-coboundary) cocycles that give trivial invariant values. It may be possible, for example, that it takes non-trivial values only for ``virtual knots'' that are non-classical.
Comments: It turned out that all of the following had trivial values (i.e., positive integers that represent the number of colorings). It may be conjectured that this cocycle is a coboundary for any Alexander quandle.
For Z_2 coefficients:
First we try cocycles of the form f(x,y)=(x-y)^(2^m) y. Alexander quandle modulo t^2+t+1 has non-trivial 2-cocycle f(x,y)=(x-y)^2 y.
Comments: This 4 element quandle is well- known, and for example, isomorphic to the quandle consisting of 120 degree rotations of a regular tetrahedron. It is known to have 2-dimensional cohomology group Z_2 with Z_2 coefficient. The outputs are all of the form k*[4 + 12*u^(t+1)], so we conjecture that it is always the case. It is also an interesting problem to characterize this invariant.
For f(x,y)= (x-y)^(2^2) y, the quandle must be mod g(t) where g(t) divides t^5-1, but t^5-1 is factored into prime polynomials (t+1)(t^4+t^3+t^2+t^2+1) mod 2, so we set g(t)=t^4+t^3+t^2+t^2+1.
Comments: We observe that the cocycle values are {0, t+1, t^3, t^3+t+1} and in fact, the invariant is 16+80*u^(t^3+t+1)+80*u^(t+1)+80*u^(t^3) for all knots with non-trivial values. Are there knots with different values?
For f(x,y)= (x-y)^(2^3) y, we factor t^9-1 mod 2 to (t+1)(t^2+t+1)(t^6+t^3+1), so we try Alexander quandle mod t^6+t^3+1.
Comments: Again we found only one non-trivial value for a few knots: 64+576*u^(t^4+t^3+1)+576*u^(t^4+t^3+t)+576*u^(t^5+t^4+t^3+1)+576*u^(t^5)+576*u^(t^5+t+1)+576*u^(t+1)+576*u^(t^5+t^4+t^3+t).
Higher powers seem too big to compute, even after prime decompositions.
Next we try cocycles of the form f(x,y) = (x-y)^(2^m) y^2. For f(x,y) = (x-y)^2 y^2 we compute t^4-1=(t+1)^2 (t^2+1) but t^2+1 gives a trivial quandle of 4 elements [Nel03]. For f(x,y) = (x-y)^(2^2) y^2 we compute t^6-1=(t+1)^2 (t^2+t+1)^2, so that we try the quandle modulo (t^2+t+1)^2.
f(x,y) = (x-y)^(2^3) y^2, t^(10)-1=(t+1)^2 (t^4+t^3+t^2+t+1)
f(x,y) = (x-y)^(2^2) y^(2^2), t^(8)-1=(t+1)^2 (t^2+1) (t^4+1)
f(x,y) = (x-y)^(2^3) y^(2^2), t^(12)-1=(t+1)^2 (t^2+t+1)^2 (t^2+1) (t^4+t^2+1)
f(x,y) = (x-y)^(2^4) y^2, t^(18)-1=(t+1)^2 (t^2+t+1)^2 (t^6+t^3+1)^2
f(x,y) = (x-y)^(2^3) y^(2^3), t^(16)-1=(t+1)^2 (t^2+1) (t^4+1) (t^8+1)
For f(x,y)=(x-y)^3 y, we have t^4-1=(t+1)(t+2)(t^2+1) mod 3, so we try the quandle mod t^2+1.
Comments: Up to 9 crossings, all non-trivial values are 9+36*u^(t+1)+36*u^(2*t+2), except 297++216*u^(t+1)+216*u^(2*t+2) for 9_40 !
For f(x,y)=(x-y)^(3^2) y, we have t^(10)-1=(t+1)(t+2)(t^4+t^3+t^2+t+1)(t^4+2t^3+t^2+2t+1) mod 3, and degree 4 is too large. For f(x,y)=(x-y)^3 y^3, we compute t^6-1=(t+1)(t+2)(t^2+t+1)(t^2+2t+1) mod 3, so we try two of them.
Comments: The polynomial t^2+t+1 of the quandle above was coprime with Alexander polynomials of all knots up to 9 crossings, so that the output is empty. There is no non-trivial values for the other case, up to 9 crossings.
Further possibilities seemed to have too many elements to compute at this time.