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Numbers n for which number of iterations to reach the largest equals number of iterations to reach 1 from the largest in Collatz (3x+1) trajectory of n.
2

%I #11 Apr 14 2013 21:16:13

%S 1,6,120,334,335,804,1249,2008,2010,2012,2013,6556,6557,6558,6801,

%T 6802,6803,7496,7498,7500,7501,7505,10219,22633,25182,25183,27074,

%U 27075,27864,27866,27868,31838,31839,32078,36630,36633,36690,36691,36914,39126,39344

%N Numbers n for which number of iterations to reach the largest equals number of iterations to reach 1 from the largest in Collatz (3x+1) trajectory of n.

%C For each n here we have A220421(n) = A222641(n).

%H T. D. Noe, <a href="/A224303/b224303.txt">Table of n, a(n) for n = 1..1000</a>

%e 6 is in the list because Collatz trajectory of 6 is {6, 3, 10, 5, 16, 8, 4, 2, 1} and number of steps to reach largest = number of steps to reach 1 from largest = 4.

%t Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3*# + 1] &, n, # > 1 &]; Select[Range[50000],Position[Collatz[#],Max[Collatz[#]]] == {{(Length[Collatz[#]] + 1)/2}} &]

%Y Cf. A220421, A222641, A224533 (odd n).

%K nonn

%O 1,2

%A _Jayanta Basu_, Apr 03 2013