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 A171870 For odd numbers x, a(x) is the number of complex numbers z in the zx + 1 problem giving the same number of iterations as the 3x + 1 problem requires to reach 1. 1

%I #13 Oct 20 2019 22:03:18

%S 0,1,0,4,5,3,1,4,2,5,0,3,6,40,4,38,7,2,5,10,39,8,3,37,6,6,1,40,9,9,4,

%T 38,7,7,2,36,41,2,5,10,5,39,0,8,8,32,3,37,42,6,6,30,11,35,40,23,1,9,4,

%U 9,33,14,38,14,43,7,7,12,31,12,2,36,41,41,5,2,10,29,10,17,34,5,39,22,15,44,8

%N For odd numbers x, a(x) is the number of complex numbers z in the zx + 1 problem giving the same number of iterations as the 3x + 1 problem requires to reach 1.

%C This sequence appears easy because a(n) = A075680(n) - 1, but its true object is the introduction of polynomials f(z) with interesting properties, for instance the study of the roots of f(z)= 0.

%C The 3x+1 problem is an exceptional case of the zx + 1 problem (for z real or complex). The sequence gives the number of z <> 3 which gives the same trajectory as the 3x + 1 problem. We associate each number x with a polynomial f(z) whose roots have the same behavior as the integer 3 in the 3x + 1 problem. The sequence gives the degree of each polynomial. Example: with n = 17, the trajectory is (17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1) and we obtain, for z = 3, the following steps:

%C 17;

%C 17z + 1 = 52;

%C (17z + 1)/4 = 13;

%C z(17z + 1)/4 + 1 = 40;

%C (z(17z + 1)/4 + 1)/8 = 5;

%C z(z(17z + 1)/4 + 1)/8 + 1 = 16;

%C (z(z(17z + 1)/4 + 1)/8 + 1)/16 = 1 => f(z) = 17z^3 + z^2 + 4z - 480 = (z-3)(17z^2 + 52z + 160)= 0.

%C The polynomial g(z) = 17z^2 + 52z + 160 of degree 2 is connected with the number 17. The two roots z1 and z2 have the same behavior as the integer 3, and the 3*x + 1 problem, z1*x + 1 problem and z2*x + 1 problem are identical for x = 17. The following polynomials are given for x = 1, 3, 5, ..., 21.

%C 1 -> z-3

%C 3 -> (z-3)*(3z + 10)

%C 5 -> 5*(z-3)

%C 7 -> (z-3)*(7z^4 + 22z^3 + 68z^2 + 208z + 640)

%C 9 -> (z-3)*(9z^5 + 28z^4 + 88z^3 + 272z^2 + 832z + 2560)

%C 11 -> (z-3)*(11z^3 + 34z^2 + 104z + 320)

%C 13 -> (z-3)*(13z + 40)

%C 15 -> (z-3)*(15z^4 + 46z^3 + 140z^2 + 424z + 1280)

%C 17 -> (z-3)*(17z^2 + 52z + 160)

%C 19 -> (z-3)*(19z^5 + 58z^4 + 176z^3 + 544z^2 + 1664z + 5120)

%C 21 -> 21*(z-3)

%C 23 -> (z-3)*(23z^3 + 70z^2 + 212z + 640)

%C ...

%C In the general case and for each number n, if the Collatz conjecture is true, the polynomial is of the form:

%C f(z) = (z-3)*g(z) = n*z^p + z^(p-1) + 2^a*z^(n-2) + 2^b*z^(n-3) + ... + 2^w*z + 2^r - 2^s

%C where

%C s is the number of divisions by 2 at the last step;

%C r is the number of divisions by 2 at before the last step;

%C a is the number of divisions by 2 at the first step;

%C b is the number of divisions by 2 at the second step.

%H Jeffrey C. Lagarias, <a href="http://arXiv.org/abs/math/0309224">The 3x+1 problem: An annotated bibliography</a> arXiv:math/0309224

%F a(n) = A075680(n) - 1.

%e For x=17, a(9)=2 is in the sequence because the associated polynomial of 17 is 17z^2 + 52z + 160 with degree 2.

%t nextOddK[n_] := Module[{m=3n+1}, While[EvenQ[m], m=m/2]; m]; (* assumes odd n *) Table[m=n; cnt=0; If[n>1, While[m=nextOddK[m]; cnt++; m!=1]]; cnt-1, {n, 1, 200, 2}]

%K nonn

%O 1,4

%A _Michel Lagneau_, Dec 30 2009, May 18 2010

%E Partially edited by _N. J. A. Sloane_, May 22 2010

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