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%I #5 Apr 04 2013 16:18:32
%S 1,2,4,8,16,32,64,20,40,80,160,52,96,208,68,136,280,88,184,56,112,232,
%T 72,148,304,100,196,400,808,260,520,1048,13120,4372,9232,472,916,1888,
%U 628,1264,2536,4912,1672,3256,39364,2248,740,1480,2968,5992,1972,3844
%N The maximum value of the Collatz (3x+1) iteration beginning at A223699(n).
%H T. D. Noe, <a href="/A223700/b223700.txt">Table of n, a(n) for n = 1..800</a>
%e The Collatz iteration of 15 is {15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1}. The maximum is 160, which occurs at position 11, counting from the right. Hence, a(11) = 160 because no number smaller than 15 has its maximum value at the 11 position.
%t Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; t= Table[c = Reverse[Collatz[n]]; Position[c, Max[c]][[1, 1]], {n, 10000}]; t3 = {}; n = 0; While[n++; p = Position[t, n, 1, 1]; p != {}, c = Collatz[p[[1,1]]]; AppendTo[t3, Max[c]]]; t3
%Y Cf. A070165 (Collatz iteration of n), A223699.
%K nonn
%O 1,2
%A _T. D. Noe_, Mar 26 2013