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%I #8 Jun 24 2014 00:23:13
%S 1,1,1,1,1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,1,2,3,4,5,6,1,2,3,4,5,6,7,3,1,
%T 2,3,4,5,6,7,3,8,5,5,1,2,3,4,5,6,7,3,8,5,5,9,7,8,7,1,2,3,4,5,6,7,3,8,
%U 5,5,9,7,8,7,10,9,11,11,9,1,2,3,4,5,6
%N Irregular triangular array of denominators of the positive rational numbers ordered as in Comments.
%C Suppose that m >= 3, and define sets h(n) of positive rational numbers as follows: h(n) = {n} for n = 1..m, and thereafter, h(n) = Union({x+1: x in h(n-1), {x/(x+1) : x in h(n-m)}), with the numbers in h(n) arranged in decreasing order. Every positive rational lies in exactly one of the sets h(n). For the present array, put m = 5 and (row n) = h(n); the number of numbers in h(n) is A003520(n-1). (For m = 3, see A243712.)
%H Clark Kimberling, <a href="/A243732/b243732.txt">Table of n, a(n) for n = 1..1000</a>
%e First 9 rows of the array:
%e 1/1
%e 2/1
%e 3/1
%e 4/1
%e 5/1
%e 6/1 ... 1/2
%e 7/1 ... 3/2 ... 2/3
%e 8/1 ... 5/2 ... 5/3 ... 3/4
%e 9/1 ... 7/2 ... 8/3 ... 7/4 ... 4/5
%e 10/1 .. 9/2 ... 11/3 .. 11/4 .. 9/5 ... 5/6
%e 11/1 .. 11/2 .. 14/3 .. 15/4 .. 14/5 .. 11/6 .. 6/7 .. 1/3
%e The denominators, by rows: 1,1,1,1,1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,6,1,2,3,4,5,6,7,3,...
%t z = 23; g[1] = {1}; g[2] = {2}; g[3] = {3}; g[4] = {4}; g[5] = {5};
%t g[n_] := Reverse[Union[1 + g[n - 1], g[n - 5]/(1 + g[n - 5])]]
%t Table[g[n], {n, 1, 9}]
%t v = Flatten[Table[g[n], {n, 1, z}]];
%t v1 = Denominator[v]; (* A243732 *)
%t v2 = Numerator[v]; (* A243733 *)
%Y Cf. A243731, A243712, A243733, A003269.
%K nonn,easy,tabf,frac
%O 1,7
%A _Clark Kimberling_, Jun 09 2014
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