%I #15 Dec 13 2014 03:01:58
%S 0,2,1,2,0,1,2,4,0,4,1,3,0,1,3,4,0,2,1,2,0,1,2,3,0,3,1,7,0,1,4,6,0,2,
%T 1,2,0,1,2,5,0,6,1,3,0,1,3,5,0,2,1,2,0,1,2,3,0,3,1,4,0,1,5,6,0,2,1,2,
%U 0,1,2,4,0,4,1,3,0,1,3,4,0,2,1,2,0,1,2,3,0,3,1,5,0,1,4,5,0,2,1,2,0,1,2,7,0
%N Number of steps to reach an integer == 1 (mod 4) when iterating the map n -> 3n/2 if n even or (3n+1)/2 if n odd.
%D L. Flatto, Z-numbers and beta-transformations, in Symbolic dynamics and its applications (New Haven, CT, 1991), 181-201, Contemp. Math., 135, Amer. Math. Soc., Providence, RI, 1992.
%H Reinhard Zumkeller, <a href="/A063574/b063574.txt">Table of n, a(n) for n = 1..10000</a>
%H K. Mahler, <a href="http://carmaweb.newcastle.edu.au/mahler/docs/167.pdf">An unsolved problem on the powers of 3/2</a>, J. Austral. Math. Soc. 8 (1968), pp. 313-321.
%F For odd n: a(n)=A007814(n+1), for even n: A007814(n) steps until an odd number is reached, which leads directly to the formula: with b(n)=A007814(n) (binary carry sequence) a(n)=b(n)+b((3^b(n)*n/2^b(n)+1)/2) - Lambert Herrgesell (zero815(AT)googlemail.com) and Lambert Klasen (lambert.klasen(AT)gmx.net), Apr 24 2006. Hence in particular, a(n) is well-defined.
%e 8 -> 12 -> 18 -> 27 -> 41 takes 4 steps so a(8) = 4.
%t Table[Length[NestWhileList[If[EvenQ[#],(3#)/2,(3#+1)/2]&,n, Mod[#,4]!= 1&]]-1,{n,110}] (* _Harvey P. Dale_, Jul 06 2011 *)
%o (PARI) {stop=1000; for(n=1,105,c=0; k=n; while((k%4)!=1&&c<stop,k=if(k%2==0,3*k/2,(3*k+1)/2); c++); print1(if(c<stop,c,-1),","))}
%o (PARI) b(n)=valuation(n,2); a(n)=b(n)+b((3^b(n)*n/2^b(n)+1)/2) - Lambert Herrgesell (zero815(AT)googlemail.com) and Lambert Klasen (lambert.klasen(AT)gmx.net), Apr 24 2006
%o (Haskell)
%o a063574 n = fst $ until ((== 1) . flip mod 4 . snd)
%o (\(u, v) -> (u + 1, a007494 v)) (0, n)
%o -- _Reinhard Zumkeller_, Dec 13 2014
%Y Cf. A007494.
%K easy,nice,nonn
%O 1,2
%A _N. J. A. Sloane_, Sep 23 2002
%E Extended by _Klaus Brockhaus_, Sep 23 2002