A211981 -id:A211981 - cp's OEIS frontend

A225089 a(n) = floor(2^A006666(m)/3^A006667(m)) - m, where m = 2n + 1.

0, 0, 0, 0, 0, 1, 1, 0, 2, 0, 1, 2, 1, 2, 1, 0, 1, 4, 3, 1, 0, 3, 2, 4, 4, 2, 5, 5, 4, 3, 5, 0, 6, 3, 2, 8, 7, 6, 8, 2, 7, 0, 10, 6, 5, 4, 7, 8, 10, 9, 8, 4, 3, 10, 9, 11, 14, 9, 12, 7, 6, 10, 9, 0, 14, 13, 12, 7, 6, 5, 10, 17, 13, 15, 0, 13, 12, 16, 15, 5, 8

Offset: 1

Michel Lagneau, Apr 27 2013

nonn

A006666 and A006667 are the number of halving and tripling steps to reach 1 in 3x+1 problem.
Properties of this sequence:
a(m) = 0 for m =  A211981(m).

			a(9) = 3 because floor(2^A006666(19)/3^A006667(19)) - 19 = floor(2^14 /3^6) - 19 = floor(22.474622) - 19 = 22 - 19 = 3.

Michel Lagneau, Table of n, a(n) for n = 1..10000

Cf. A006666, A006667, A211981, A075680.

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 100 do: printf(`%d, `, G(i)-i):od:

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; nn = 100; t = {}; n = 0; While[Length[t] < nn, n++; c = Collatz[n]; ev = Length[Select[c, EvenQ]]; od = Length[c] - ev - 1; AppendTo[t, Floor[2^ev/3^od]-n]]; t

A265099 Least k such that floor(2^A006666(k)/3^A006667(k)) - k = n.

Original entry on oeis.org

1, 6, 9, 19, 18, 27, 33, 37, 36, 50, 43, 56, 59, 66, 57, 74, 78, 72, 97, 87, 86, 98, 112, 119, 118, 134, 123, 115, 114, 130, 149, 148, 157, 135, 179, 144, 153, 187, 220, 174, 173, 172, 197, 196, 255, 224, 238, 219, 236, 203, 249, 268, 247, 246, 230, 229, 228

Offset: 0

Michel Lagneau, Dec 01 2015

nonn

A006666 and A006667 are the number of halving and tripling steps to reach 1 in 3x+1 problem.
Conjecture: k exists for all n.
In other words, given an integer n, there always exists at least an integer k and a pair of integers (a, b) such that n + k = 2^a/3^b where a is the number of halving steps to reach 1, and b is the number of tripling steps to reach 1, in the 3x+1 problem.

			a(0) = 1 because A006666(1) = 0 and A006667(1) = 0 => floor(2^0/3^0) - 1 = 1 - 1 = 0;
a(1) = 6 because A006666(6) = 6 and A006667(6) = 2 => floor(2^6/3^2) - 6 = floor(64/9) - 6 = 7 - 6 = 1.

Cf. A006666, A006667, A075680, A211981, A225089.

lst={};Do[Collatz[k_]:=NestWhileList[If[EvenQ[#],#/2,3 #+1]&,k,#>1&];nn=500;t={};k=0;While[Length[t]

A302175 a(n) = [2^A006666(n)/3^A006667(n)], where [x] = floor(x).

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 8, 11, 10, 12, 14, 14, 16, 16, 16, 18, 22, 22, 21, 21, 25, 25, 28, 29, 28, 32, 33, 33, 33, 36, 32, 39, 37, 37, 44, 44, 44, 47, 42, 48, 42, 53, 50, 50, 50, 54, 56, 59, 59, 59, 56, 56, 64, 64, 67, 71, 67, 71, 67, 67, 72, 72, 64, 79, 79, 79, 75

Offset: 1

Michel Lagneau, Apr 03 2018

The sequence contains A211981 and the powers of 2 (A000079).
There exists a subset E = { 1, 2, 3, 4, 5, 8, 10, 16, 21, 32, 42, 64, 85, 128, 170, 227, 256, 341, 512, 682, 1024, 2048, ...} in {a(n)} such that each element m of E generates the Collatz sequence of iterates m -> T_1(m) -> T_2(m) -> T_3(m) -> ... -> 1 where any T_i(m) is an element of E of the form [2^i /3^j] where i = A006666(m), or A006666(m)-1, or ... and j = A006667(m), or A006667(m)-1, or ..., but with A006667(m) <= 3. If m is even then m/2 is in E.
For example, the statement that "3 is an element of E" implies that each element of the trajectory 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 belongs to E. Thus the trajectory of the number 3 can be represented by [2^5/3^2] -> [2^5/3^1] -> [2^4/3^1] -> [2^4/3^0] -> [2^3/3^0] -> [2^2/3^0] -> [2^1/3^0] -> [2^0/3^0].

			a(39) = [2^A006666(39)/3^A006667(39)] = [2^23/3^11] = [47.353937...] = 47.

Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A000079, A006666, A006667, A211981, A225089, A265099.

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; nn = 70; t = {}; n = 0; While[Length[t] < nn, n++; c = Collatz[n]; ev = Length[Select[c, EvenQ]]; od = Length[c] - ev - 1; AppendTo[t, Floor[2^ev/3^od]]]; t

a(n) = my(t, h); while(n>1, if(n%2, n=3*n+1; t++, n>>=1; h++)); 2^h\3^t; \\ Michel Marcus, May 05 2018

cp's OEIS Frontend

A225089 a(n) = floor(2^A006666(m)/3^A006667(m)) - m, where m = 2n + 1.

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A265099 Least k such that floor(2^A006666(k)/3^A006667(k)) - k = n.

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A302175 a(n) = [2^A006666(n)/3^A006667(n)], where [x] = floor(x).

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