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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.

%I A171870 #13 Oct 20 2019 22:03:18
%S A171870 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 A171870 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 A171870 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 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.
%C A171870 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 A171870 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 A171870 17;
%C A171870 17z + 1 = 52;
%C A171870 (17z + 1)/4 = 13;
%C A171870 z(17z + 1)/4 + 1 = 40;
%C A171870 (z(17z + 1)/4 + 1)/8 = 5;
%C A171870 z(z(17z + 1)/4 + 1)/8 + 1 = 16;
%C A171870 (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 A171870 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 A171870    1 -> z-3
%C A171870    3 -> (z-3)*(3z + 10)
%C A171870    5 -> 5*(z-3)
%C A171870    7 -> (z-3)*(7z^4 + 22z^3 + 68z^2 + 208z + 640)
%C A171870    9 -> (z-3)*(9z^5 + 28z^4 + 88z^3 + 272z^2 + 832z + 2560)
%C A171870   11 -> (z-3)*(11z^3 + 34z^2 + 104z + 320)
%C A171870   13 -> (z-3)*(13z + 40)
%C A171870   15 -> (z-3)*(15z^4 + 46z^3 + 140z^2 + 424z + 1280)
%C A171870   17 -> (z-3)*(17z^2 + 52z + 160)
%C A171870   19 -> (z-3)*(19z^5 + 58z^4 + 176z^3 + 544z^2 + 1664z + 5120)
%C A171870   21 -> 21*(z-3)
%C A171870   23 -> (z-3)*(23z^3 + 70z^2 + 212z + 640)
%C A171870   ...
%C A171870 In the general case and for each number n, if the Collatz conjecture is true, the polynomial is of the form:
%C A171870 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 A171870 where
%C A171870   s is the number of divisions by 2 at the last step;
%C A171870   r is the number of divisions by 2 at before the last step;
%C A171870   a is the number of divisions by 2 at the first step;
%C A171870   b is the number of divisions by 2 at the second step.
%H A171870 Jeffrey C. Lagarias, <a href="http://arXiv.org/abs/math/0309224">The 3x+1 problem: An annotated bibliography</a> arXiv:math/0309224
%F A171870 a(n) = A075680(n) - 1.
%e A171870 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 A171870 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 A171870 nonn
%O A171870 1,4
%A A171870 _Michel Lagneau_, Dec 30 2009, May 18 2010
%E A171870 Partially edited by _N. J. A. Sloane_, May 22 2010