%N In repeated iterations of function m -> m/2 if m even, m -> 3m+1 if m odd, a(n) is maximum value achieved if starting from n.
%C If a(n) exists (which is the essence of the "3x+1" problem) then a(n) must be a multiple of 4, since if a(n) was odd then the next iteration 3*a(n)+1 would be greater than a(n), while if a(n) was twice an odd number then the next-but-one iteration (3/2)*a(n)+1 would be greater.
%C The variant A025586 considers the trajectory ending in 1, by definition. Therefore the two sequences differ for a(1) and a(2). - _M. F. Hasler_, Oct 20 2019
%H <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>
%e a(6)=16 since iteration starts: 6, 3, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, ... and 16 is highest value
%o (PARI) a(n)=my(r=max(4,n));while(n>2,if(n%2,n=3*n+1;if(n>r,r=n),n/=2));r \\ _Charles R Greathouse IV_, Jul 19 2011
%Y Cf. A006370, A056957, A056958.
%Y Essentially the same as A025586.
%A _Henry Bottomley_, Jul 18 2000