%I #11 May 01 2013 17:05:38
%S 1,3,5,21,13,53,67,141,93,61,37,149,101,65,261,173,693,461,305,209,
%T 813,541,2165,1445,961,657,2315,1709,1169,4557,3037,2021,3659,5389,
%U 3589,2413,5123,2291,4253,2755,7235,7557,5037,3357,10123,8949,5965,3973,15893
%N Least odd number d such that the Collatz (3x+1) iteration of d has the following property: if the length of the iteration is b and the maximum value occurs at c, the ratio c/b is 1/n.
%C This sequence is the first column of the irregular triangle shown in A224537.
%C For 45 < n <= 300, the length the Collatz sequence of a(n) is 2n. - _T. D. Noe_, May 01 2013
%H T. D. Noe, <a href="/A224994/b224994.txt">Table of n, a(n) for n = 1..300</a>
%e The Collatz iteration of 5 is {5, 16, 8, 4, 2, 1}, which has length 6. The maximum occurs at the second position. Note that 2/6 = 1/3. No number less than 5 has this property. Hence a(3) = 5.
%t Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; nn = 50; t = Table[0, {nn}]; t[[1]] = 1; n = 1; While[Times @@ t == 0, n = n + 2; c = Collatz[n]; frac = Position[c, Max[c]][[1, 1]]/Length[c]; numer = Numerator[frac]; denom = Denominator[frac]; If[numer == 1 && denom <= nn && t[[denom]] == 0, t[[denom]] = n]]; t
%Y Cf. A224537.
%K nonn
%O 1,2
%A _T. D. Noe_, Apr 23 2013