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a(n) = least number k such that 2^n is highest in Collatz(3x+1) trajectory of k.
3

%I #10 Dec 02 2013 14:32:22

%S 1,2,4,8,3,32,21,128,85,512,151,2048,1365,8192,5461,32768,14563,

%T 131072,87381,524288,349525,2097152,932067,8388608,5592405,33554432,

%U 22369621,134217728,26512143,536870912,357913941

%N a(n) = least number k such that 2^n is highest in Collatz(3x+1) trajectory of k.

%e a(4)=3 since 3 is the least number such that largest member of Collatz(3 x + 1) trajectory of 3 is 2^4 = 16.

%t Coll[n_]:=NestWhileList[If[EvenQ[#],#/2,3*#+1]&,n,#>1 &]; t={}; Do[i=1; While[Max[Coll[i]] != 2^n, i++]; AppendTo[t, i], {n,0,25}]; t

%o (Haskell)

%o a225124 = (+ 1) . fromJust . (`elemIndex` a025586_list) . a000079

%o -- _Reinhard Zumkeller_, Apr 30 2013

%Y Cf. A000079, A025586, A095381, A231610.

%K nonn

%O 0,2

%A _Jayanta Basu_, Apr 29 2013