

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



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, 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, 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
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OFFSET

1,4


COMMENTS

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.
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:
17;
17z + 1 = 52;
(17z + 1)/4 = 13;
z(17z + 1)/4 + 1 = 40;
(z(17z + 1)/4 + 1)/8 = 5;
z(z(17z + 1)/4 + 1)/8 + 1 = 16;
(z(z(17z + 1)/4 + 1)/8 + 1)/16 = 1 => f(z) = 17z^3 + z^2 + 4z  480 = (z3)(17z^2 + 52z + 160)= 0.
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.
1 > z3
3 > (z3)*(3z + 10)
5 > 5*(z3)
7 > (z3)*(7z^4 + 22z^3 + 68z^2 + 208z + 640)
9 > (z3)*(9z^5 + 28z^4 + 88z^3 + 272z^2 + 832z + 2560)
11 > (z3)*(11z^3 + 34z^2 + 104z + 320)
13 > (z3)*(13z + 40)
15 > (z3)*(15z^4 + 46z^3 + 140z^2 + 424z + 1280)
17 > (z3)*(17z^2 + 52z + 160)
19 > (z3)*(19z^5 + 58z^4 + 176z^3 + 544z^2 + 1664z + 5120)
21 > 21*(z3)
23 > (z3)*(23z^3 + 70z^2 + 212z + 640)
...
In the general case and for each number n, if the Collatz conjecture is true, the polynomial is of the form:
f(z) = (z3)*g(z) = n*z^p + z^(p1) + 2^a*z^(n2) + 2^b*z^(n3) + ... + 2^w*z + 2^r  2^s
where
s is the number of divisions by 2 at the last step;
r is the number of divisions by 2 at before the last step;
a is the number of divisions by 2 at the first step;
b is the number of divisions by 2 at the second step.


LINKS



FORMULA



EXAMPLE

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


MATHEMATICA

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]]; cnt1, {n, 1, 200, 2}]


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



