A211981 - OEIS

%I #18 Oct 26 2024 22:59:29

%S 1,2,3,4,5,8,10,16,21,32,42,64,75,85,113,128,151,170,227,256,341,512,

%T 682,1024,1365,2048,2730,4096,5461,7281,8192,10922,14563,16384,21845,

%U 32768,43690,65536,87381,131072,174762,262144,349525,466033,524288,699050,932067

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

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

%C Properties of this sequence:

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

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

%C {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.

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

%e 227 is in the sequence because A006666(227) = 11, A006667(227) = 2 => floor(2^11/3^2) = 227.

%e 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:

%e 227 = floor(2^11/3^2);

%e 682 = floor(2^11/3^1);

%e 341 = floor(2^10/3^1);

%e 1024 = floor(2^10/3^0);

%e 512 = floor(2^9/3^0);

%e 256 = floor(2^8/3^0);

%e 128 = floor(2^7/3^0);

%e ...

%e 2 = floor(2^1/3^0);

%e 1 = floor(2^0/3^0);

%e With the numbers of E1, we obtain another formulation of the Collatz problem.

%p A:= proc(n) if type(n, 'even') then n/2; else 3*n+1 ; end if; end proc:

%p B:= proc(n) a := 0 ; x := n ; while x > 1 do x := A(x) ; a := a+1 ; end do; a ; end proc:

%p 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:

%p D:= proc(n) C(n) ; end proc:

%p A006666:= proc(n) B(n)- C(n) ; end:

%p A006667:= proc(n) C(n)- D(n) ; end:

%p G:= proc(n) floor(2^ A006666 (n)/3^ A006667 (n)) ; end:

%p for i from 1 to 1000000 do: if G(i) =i then printf(`%d, `,i):else fi:od:

%t 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 *)

%Y Cf. A006666, A006667.

%K nonn

%O 1,2

%A _Michel Lagneau_, Feb 13 2013