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%I #5 Aug 09 2018 05:27:15
%S 1,1,1,5,7,5,5,5,13,11,1,5,19,23,11,23,25,5,7,19,31,37,17,23,37,29,5,
%T 47,43,19,23,61,49,19,13,61,55,1,29,19,61,47,1,37,67,61,35,91,73,7,19,
%U 61,79,91,41,61,85,65,11,101,91,59,47,77,97,37,25,43,103
%N a(n) is the first term less than the initial 2n+1 in the reduced Collatz trajectory.
%e a(3)= 5 because, starting with 7, the iteration produces 11,17,13,5 and 5 is the first term less than 7.
%t f[n_]:=NestWhileList[(3*#+1)/2^IntegerExponent[3*#+1,2]&,2*n+1,#>1&];
%t nextOddK[n_]:=Module[{m=3n+1},While[EvenQ[m],m=m/2];m];dt[n_]:=Module[{m=n,cnt=0},If[n>1,While[m=nextOddK[m];cnt++;m>n]];cnt];
%t Table[Part[f[i],dt[2i+1]+1],{i,0,70}]
%Y Cf. A122458, A256598.
%K nonn
%O 0,4
%A _Michel Lagneau_, Jul 21 2018
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