A211981 Numbers n such that floor(2^A006666(n)/3^A006667(n)) = n.

1, 2, 3, 4, 5, 8, 10, 16, 21, 32, 42, 64, 75, 85, 113, 128, 151, 170, 227, 256, 341, 512, 682, 1024, 1365, 2048, 2730, 4096, 5461, 7281, 8192, 10922, 14563, 16384, 21845, 32768, 43690, 65536, 87381, 131072, 174762, 262144, 349525, 466033, 524288, 699050, 932067

Offset: 1

Author

Michel Lagneau, Feb 13 2013

Keywords

• nonn

Comments

A006666 and A006667 give the number of halving and tripling steps to reach 1 in 3x+1 problem.

Properties of this sequence:

A006667(a(n)) <= 3, and if a(n) is even then a(n)/2 is in the sequence.

The sequence A000079(n) (power of 2) is included in this sequence.

{a(n)} = E1 union E2 where E1 = {A000079(n)} union {5, 10, 21, 85, 170, 227, 341, 682, 1365, 2730, 5461, ...} and E2 = {75, 113, 151, 7281, ...}. If an element k of E1 generates the Collatz sequence of iterates k -> T_1(k) -> T_2(k) -> T_3(k) -> ... then any T_i(k) is an element of E1 of the form [2^a /3^b] where a = A006666(n), or A006666(n)-1, or ... and b = A006667(n), or A006667(n)-1, or ... But if k is an element of E2, there exists at least an element T_i(k) that is not in the sequence a(n). For example 75 -> 226 ->113 -> 340 -> ... and 226 is not in the sequence because, if [x] = [2^a /3^b] = [ x. x0 x1 x2 ...], the rational number 0.x0 x1 x2 ... > 0.666666.... => [2^a /3^(b-1)] of the form [(3x+2).y0 y1 y2 ...], and this integer is different from T_(i+1)(k) = [(3x+1).y0 y1 y2 ...] = 3x+1.

Example: T_2(75) = floor(2^10 /3^2) = 113 => floor(2^10/3^1) = 341 instead T_3(75) = 340.

Examples

			227 is in the sequence because A006666(227) = 11, A006667(227) = 2 => floor(2^11/3^2) = 227.
The Collatz trajectory of 227 is 227 -> 682 -> 341 -> 1024 -> 512 -> ... -> 2 -> 1, and 227 is in the subset E1 implies the following Collatz iterates:
227 = floor(2^11/3^2);
682 = floor(2^11/3^1);
341 = floor(2^10/3^1);
1024 = floor(2^10/3^0);
512 = floor(2^9/3^0);
256 = floor(2^8/3^0);
128 = floor(2^7/3^0);
...
2 = floor(2^1/3^0);
1 = floor(2^0/3^0);
With the numbers of E1, we obtain another formulation of the Collatz problem.
		

Crossrefs

Cf. A006666, A006667.

Programs

Maple

A:= proc(n) if type(n, 'even') then n/2; else 3*n+1 ; end if; end proc:
B:= proc(n) a := 0 ; x := n ; while x > 1 do x := A(x) ; a := a+1 ; end do; a ; end proc:
C:= proc(n) a := 0 ; x := n ; while x > 1 do if type(x, 'even') then x := x/2 ; else x := 3*x+1 ; a := a+1 ; end if; end do; a ; end proc:
D:= proc(n) C(n) ; end proc:
A006666:= proc(n) B(n)- C(n) ; end:
A006667:= proc(n) C(n)- D(n) ; end:
G:= proc(n) floor(2^ A006666 (n)/3^ A006667 (n)) ; end:
for i from 1 to 1000000 do: if G(i) =i then printf(`%d, `,i):else fi:od:

Mathematica

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; nn = 30; t = {}; n = 0; While[Length[t] < nn, n++; c = Collatz[n]; ev = Length[Select[c, EvenQ]]; od = Length[c] - ev - 1; If[Floor[2^ev/3^od] == n, AppendTo[t, n]]]; t (* T. D. Noe, Feb 13 2013 *)