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 A211981 Numbers n such that floor(2^A006666(n)/3^A006667(n)) = n. 3
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 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. LINKS EXAMPLE 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. 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 *) CROSSREFS Cf. A006666, A006667. Sequence in context: A206555 A110539 A222297 * A019997 A263875 A125157 Adjacent sequences:  A211978 A211979 A211980 * A211982 A211983 A211984 KEYWORD nonn AUTHOR Michel Lagneau, Feb 13 2013 STATUS approved

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Last modified December 3 02:56 EST 2020. Contains 338899 sequences. (Running on oeis4.)