rwg>There's a lot of interesting math here. I believe n (3 - 1) t ---------- 2 2 i pi d(k) ==== ----------- \ k 3 > (- 1) e / ==== n 3 + 1 k = - ------ 2
f(t) := limit ----------------------------------- n -> oo n 2
is a continuous map of [0,1] onto the closure of the Sierpinski triangle with the unit interval of the real axis as base. This gives us explicit valuations of f where t has a simple ternary expansion, e.g., the rationals.
Based on self-similarity of the graph, exp(4 pi i/3) (f(1-3t) - 1)/2, 0 <= t <= 1/3; f(t) := { (exp(i pi/3) + f(3t-1))/2, 1/3 <= t <= 2/3; 1 + f(3-3t) exp(2 i pi/3)/2, 2/3 <= t <= 1; plus continuity defines f on [0,1]. E.g., (c282) makelist(funmake(f,[k/5])=f(k/5),k,1,4) 2 1 %i sqrt(3) + 1 f(-) - 1 f(-) + -------------- 1 5 2 5 2 (d282) [f(-) = --------------, f(-) = ---------------------, 5 %i sqrt(3) - 1 5 2 4 %i sqrt(3) + 1 %i sqrt(3) 1 3 f(-) + -------------- (---------- - -) f(-) + 2 3 5 2 4 2 2 5 f(-) = ---------------------, f(-) = -------------------------] 5 2 5 2 (c283) linsolve(%,map(lhs,%)) 1 %i sqrt(3) + 5 2 2 %i sqrt(3) + 3 (d283) [f(-) = --------------, f(-) = ----------------, 5 14 5 7 3 2 %i sqrt(3) + 4 4 %i sqrt(3) + 9 f(-) = ----------------, f(-) = --------------] 5 7 5 14 1 1 4 %i sqrt(3) + 2 5 %i sqrt(3) + 5 6 2 (d284) [f(-) = -, f(-) = --------------, f(-) = --------------, f(-) = -, 7 3 7 3 7 6 7 3 3 %i sqrt(3) + 1 2 %i sqrt(3) + 1 f(-) = --------------, f(-) = --------------] 7 3 7 6 Note f(1/7) and f(6/7) lie on the bottom edge despite period six ternary expansions. Only a set of t of measure 0 map onto an edge of a triangular cell. 1 4 %i sqrt(3) + 5 25 %i sqrt(3) + 56 (d285) [f(--) = ----------------, f(--) = ---------------, 28 73 28 73 9 16 %i sqrt(3) + 20 f(--) = ------------------] 28 73 1 541 %i sqrt(3) + 717 (d286) f(--) = -------------------- 29 10922 f(1/sqrt 2) and f(1/sqrt 3) do not appear to be algebraic of degree <= 12. --rwg