%I #14 Nov 29 2017 06:39:36
%S 0,1,1,2,3,1,2,3,5,2,5,3,4,1,3,4,7,3,8,5,7,2,7,5,8,3,7,4,5,1,4,5,9,4,
%T 11,7,10,3,11,8,13,5,12,7,9,2,9,7,12,5,13,8,11,3,10,7,11,4,9,5,6,1,5,
%U 6,11,5,14,9,13,4,15,11,18,7,17,10,13,3,14,11,19,8,21,13,18,5,17,12,19,7,16
%N Numerators of the L-tree, left-to-right enumeration.
%C a(n) is a subsequence of A174980. a(n)/A002487(n+2) enumerates all the reduced nonnegative rational numbers exactly once (L-tree).
%H Edsger Dijkstra, Selected Writings on Computing, Springer, 1982, p. 232. <a href="http://www.cs.utexas.edu/users/EWD/ewd05xx/EWD578.PDF"> EWD 578: More about the function fusc</a>.
%H Peter Luschny, <a href="http://www.oeis.org/wiki/User:Peter_Luschny/SternsDiatomic"> Rational Trees and Binary Partitions</a>.
%H Moritz A. Stern, Über eine zahlentheoretische Funktion, <a href="http://www.digizeitschriften.de/resolveppn/GDZPPN002150301"> J. Reine Angew. Math., 55 (1858), 193-220</a>.
%e The sequence splits into rows of length 2^k:
%e 0,
%e 1, 1,
%e 2, 3, 1, 2,
%e 3, 5, 2, 5, 3, 4, 1, 3,
%e 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5, 1, 4,
%e ...
%e The fractions are
%e 0/1,
%e 1/2, 1/1,
%e 2/3, 3/2, 1/3, 2/1,
%e 3/4, 5/3, 2/5, 5/2, 3/5, 4/3, 1/4, 3/1,
%e 4/5, 7/4, 3/7, 8/3, 5/8, 7/5, 2/7, 7/2, 5/7, 8/5, 3/8, 7/3, 4/7, 5/4, 1/5, 4/1,
%e ...
%p SternDijkstra := proc(L, p, n) local k, i, len, M; len := nops(L); M := L; k := n; while k > 0 do M[1+(k mod len)] := add(M[i], i = 1..len); k := iquo(k, len); od; op(p, M) end:
%p Ltree := proc(n) 5*2^ilog2(n+1); SternDijkstra([0,1], 1, n + 2 + %) / SternDijkstra([1,0], 2, n + 2) end:
%p a := proc(n) 5*2^ilog2(n+1); SternDijkstra([0,1], 1, n + 2 + %) end:
%p seq(a(n), n=0..90);
%t SternDijkstra[L_, p_, n_] := Module[{k, i, len, M}, len := Length[L]; M = L; k = n; While[k > 0, M[[1 + Mod[k, len]]] = Sum[M[[i]], {i, 1, len}]; k = Quotient[k, len]]; M[[p]]]; Ltree[n_] := With[{k = 5*2^Simplify[ Floor[ Log[2, n + 1]]]}, SternDijkstra[{0, 1}, 1, n + 2 + k]/ SternDijkstra[{1, 0}, 2, n + 2]]; a[0] = 0; a[n_] := With[{k = 5*2^Simplify[ Floor[ Log[2, n + 1]]]}, SternDijkstra[{1, 0}, 1, n + 2 + k]]; row[0] = {a[0]}; row[n_] := Table[a[k], {k, 2^n - 3, 2^(n+1) - 4}] // Reverse; Table[row[n], {n, 0, 6}] // Flatten (* _Jean-François Alcover_, Jul 26 2013, after Maple *)
%Y Cf. A002487, A070879, A047679, A007306, A174980.
%K easy,nonn,frac,tabf
%O 0,4
%A _Peter Luschny_, Apr 03 2010