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 A125711 In the "3x+1" problem, let 1 denote a halving step and 0 denote an x->3x+1 step. Then a(n) is obtained by writing the sequence of steps needed to reach 1 from 2n and reading it as a decimal number. 5

%I

%S 1,3,175,7,47,431,87791,15,743151,111,22255,943,751,218863,175087,31,

%T 5871,1791727,1431279,239,191,55023,44015,1967,11917039,1775,

%U 3515647479163389605506303638875119,481007,382703,437231

%N In the "3x+1" problem, let 1 denote a halving step and 0 denote an x->3x+1 step. Then a(n) is obtained by writing the sequence of steps needed to reach 1 from 2n and reading it as a decimal number.

%e 6 -> 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1, so a(3) is the decimal equivalent of 10101111, which is 175.

%t f[x_] := If[EvenQ[x], x/2, 3x + 1];g[n_] := FromDigits[Mod[Most[NestWhileList[f, 2n, # > 1 &]], 2, 1] - 1, 2];Table[g[n], {n, 40}] (*Chandler*)

%Y Cf. A125710, A125626.

%K nonn

%O 1,2

%A _N. J. A. Sloane_, Feb 01 2007

%E Extended by _Ray Chandler_, Feb 02 2007

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