Two-cycle count conjectures for the discrete logarithm map

At least 21 years old · documented by

Let pp be prime. Write Tg A,h B(p)T_{g\,A,h\,B}(p) for the number of two-cycles with the indicated restrictions on gg and hh, and Ch A,a B(p)C_{h\,A,a\,B}(p) for the corresponding count in the auxiliary formulation; the symbols ANY\mathsf{ANY}, PR\mathsf{PR}, RP\mathsf{RP}, RPPR\mathsf{RPPR}, ORD h\mathsf{ORD}\,h, and ∙\bullet have the meanings fixed earlier in the paper. Let ϕ\phi be Euler's totient function.

Two-cycle counting conjectures.

Tg ANY,h RP(p)=Ch RP,a ANY(p)≈2ϕ(p−1),T_{g\,\mathsf{ANY},h\,\mathsf{RP}}(p)=C_{h\,\mathsf{RP},a\,\mathsf{ANY}}(p)\approx 2\phi(p-1), Th RP,g ORD h(p)=Ch RP,a RP(p)≈ϕ(p−1)+ϕ(p−1)2p−1,T_{h\,\mathsf{RP},g\,\mathsf{ORD}\,h}(p)=C_{h\,\mathsf{RP},a\,\mathsf{RP}}(p)\approx \phi(p-1)+\frac{\phi(p-1)^2}{p-1}, Tg PR,h RP(p)=Ch RP,a PR(p)≈2ϕ(p−1)2p−1,T_{g\,\mathsf{PR},h\,\mathsf{RP}}(p)=C_{h\,\mathsf{RP},a\,\mathsf{PR}}(p)\approx \frac{2\phi(p-1)^2}{p-1}, Tg PR,h RPPR(p)=Tg ANY,h RPPR(p)=Ch RPPR,a ∙(p)=Ch ∙,a RPPR(p)≈ϕ(p−1)2p−1+ϕ(p−1)3(p−1)2,T_{g\,\mathsf{PR},h\,\mathsf{RPPR}}(p)=T_{g\,\mathsf{ANY},h\,\mathsf{RPPR}}(p)=C_{h\,\mathsf{RPPR},a\,\bullet}(p)=C_{h\,\bullet,a\,\mathsf{RPPR}}(p)\approx \frac{\phi(p-1)^2}{p-1}+\frac{\phi(p-1)^3}{(p-1)^2}, Th ANY,g ORD h(p)=Ch ANY,a RP(p)≈2ϕ(p−1),T_{h\,\mathsf{ANY},g\,\mathsf{ORD}\,h}(p)=C_{h\,\mathsf{ANY},a\,\mathsf{RP}}(p)\approx 2\phi(p-1), Tg PR,h PR(p)=Tg ANY,h PR(p)=Ch PR,a RP(p)≈2ϕ(p−1)2p−1.T_{g\,\mathsf{PR},h\,\mathsf{PR}}(p)=T_{g\,\mathsf{ANY},h\,\mathsf{PR}}(p)=C_{h\,\mathsf{PR},a\,\mathsf{RP}}(p)\approx \frac{2\phi(p-1)^2}{p-1}.

The equalities in the first, fifth, third, and sixth relations are noted in the source to be exact by symmetry. The asymptotic predictions arise from a random-map heuristic and are not proved in the supplied text.

References

Primary source

Joshua Holden and Pieter Moree, “Some Heuristics and Results for Small Cycles of the Discrete Logarithm”, arXiv:math/0401013 (2004).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.