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%I #12 Oct 02 2019 16:02:59
%S 3,5,9,15,17,27,31,45,51,57,85,93,105,135,153,155,171,193,255,279,285,
%T 315,355,459,465,513,525,527,579,653,765,837,855,945,965,969,1065,
%U 1201,1395,1539,1575,1581,1737,1767,1775,1785,1959,2209,2295,2565,2635
%N Products of distinct tribonacci numbers (A000213).
%H Clark Kimberling, <a href="/A274432/b274432.txt">Table of n, a(n) for n = 1..10000</a>
%e The tribonacci numbers are 1,1,1,3,5,9,17,31,..., so that the sequence of all products of distinct members, in increasing order, is (3, 5, 9, 15, 17, 27, 31, 45,...).
%t r[1] := 1; r[2] := 1; r[3] = 1; r[n_] := r[n] = r[n - 1] + r[n - 2] + r[n - 3];
%t s = {1}; z = 60; f = Map[r, Range[z]]; Take[f, 20]
%t Do[s = Union[s, Select[s*f[[i]], # <= f[[z]] &]], {i, z}];
%t Take[s, 2 z] (*A274432*)
%t infQ[n_] := MemberQ[f, n];
%t ans = Table[#[[Flatten[Position[Map[Apply[Times, #] &, #], s[[n]]]][[1]]]] &[
%t Rest[Subsets[Map[#[[1]] &, Select[Map[{#, infQ[#]} &, Divisors[s[[n]]]], #[[2]] && #[[1]] > 1 &]]]]], {n, 2, 300}];
%t Map[Apply[Times, #] &, Select[ans, Length[#] == 2 &]] (* A274433 *)
%t Map[Apply[Times, #] &, Select[ans, Length[#] == 3 &]] (* A274434 *)
%t (* _Peter J. C. Moses_, Jun 17 2016 *)
%Y Cf. A160009, A274280, A274433 (binary products), A274434 (trinary products).
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Jun 22 2016
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