A006667 -id:A006667 - cp's OEIS frontend

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

Michel Lagneau, Feb 13 2013

nonn

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.

			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.

Cf. A006666, A006667.

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:

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 *)

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

Original entry on oeis.org

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

A287798 Least k such that A006667(k)/A006577(k) = 1/n.

Original entry on oeis.org

159, 6, 5, 10, 20, 40, 80, 160, 320, 640, 1280, 2560, 5120, 10240, 20480, 40960, 81920, 163840, 327680, 655360, 1310720, 2621440, 5242880, 10485760, 20971520, 41943040, 83886080, 167772160, 335544320, 671088640, 1342177280, 2684354560, 5368709120, 10737418240

Offset: 3

Michel Lagneau, Jun 01 2017

A006667: number of tripling steps to reach 1 in '3x+1' problem.
A006577: number of halving and tripling steps to reach 1 in '3x+1' problem.
a(n) = {159, 6} union {A020714}.

			a(3) = 159 because A006667(159)/A006577(159) = 18/54 = 1/3.

Colin Barker, Table of n, a(n) for n = 3..1000
Index entries for linear recurrences with constant coefficients, signature (2).

Cf. A006577, A006666, A006667. Essentially the same as A020714, A084215, A146523 and A257113.

nn:=10^12:
for n from 3 to 35 do:
ii:=0:
for k from 2 to 10^6 while(ii=0) do:
  m:=k:s1:=0:s2:=0:
   for i from 1 to nn while(m<>1) do:
    if irem(m,2)=0
     then
     s2:=s2+1:m:=m/2:
     else
     s1:=s1+1:m:=3*m+1:
    fi:
   od:
    if n*s1=s1+s2
     then
     ii:=1: printf(`%d, `,k):
     else
    fi:
od:od:

f[u_]:=Module[{a=u,k=0},While[a!=1,k++;If[EvenQ[a],a=a/2,a=a*3+1]];k];Table[f[u],{u,10^7}];g[v_]:=Count[Differences[NestWhileList[If[EvenQ[#],#/2,3#+1]&,v,#>1&]],_?Positive];Table[g[v],{v,10^7}];Do[k=3;While[g[k]/f[k]!=1/n,k++];Print[n," ",k],{n,3,35}]

a(n) = if(n < 5, [0,0,159,6][n], 5<<(n-5)) \\ David A. Corneth, Jun 01 2017

Vec(x^3*(159 - 312*x - 7*x^2) / (1 - 2*x) + O(x^50)) \\ Colin Barker, Jun 01 2017

For n >= 5, a(n) = 5*2^n/32. - David A. Corneth, Jun 01 2017
From Colin Barker, Jun 01 2017: (Start)
G.f.: x^3*(159 - 312*x - 7*x^2) / (1 - 2*x).
a(n) = 2*a(n-1) for n>5.
(End)

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]

A277367 a(n) = gcd(A006666(n), A006667(n)) where A006666 and A006667 are respectively the number of halving and tripling steps in the '3x+1' problem.

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 1, 3, 1, 1, 2, 1, 1, 1, 1, 4, 3, 2, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 5, 2, 1, 1, 3, 3, 3, 1, 1, 1, 1, 1, 4, 4, 4, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 1, 2, 10, 10, 1, 1, 3, 3

Offset: 1

Michel Lagneau, Oct 11 2016

nonn

			a(17) = 3 because gcd(A006666(17), A006667(17)) = gcd(9, 3) = 3.

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

Cf. A006666, A006667, A277068.

nn:=100:
for n from 1 to nn do:
m:=n:it0:=0:it1:=0:
   for j from 1 to 1000 while(m<>1) do:
    if irem(m,2)=0
     then
     m:=m/2:it0:=it0+1:
     else
     m:=3*m+1:it1:=it1+1:
    fi:
   od:
   q:=gcd(it0,it1):printf(`%d, `,q):
  od:

Table[GCD[Count[NestWhileList[If[OddQ@ #, 3 # + 1, #/2] &, n, # > 1 &], ?EvenQ], Count[Differences[NestWhileList[If[EvenQ@ #, #/2, 3 # + 1] &, n, # > 1 &]], ?Positive]], {n, 87}] (* Michael De Vlieger, Oct 13 2016, after Harvey P. Dale at A006666 and A006667 *)

a(n) = {my(se = 0); my(so = 0); while (n!=1, if (n % 2, so++; n = 3*n+1, se++; n = n/2);); gcd(se, so);} \\ Michel Marcus, Oct 13 2016

a(2^m) = m.

A281665 Numbers m such that A006667(m)/A006577(m) = 1/3.

Original entry on oeis.org

159, 283, 377, 502, 503, 603, 615, 668, 669, 670, 799, 807, 888, 890, 892, 893, 1063, 1065, 1095, 1186, 1187, 1188, 1189, 1190, 1417, 1435, 1580, 1581, 1582, 1585, 1586, 1587, 1889, 1913, 1947, 1959, 1963, 2104, 2106, 2108, 2109, 2113, 2114, 2115, 2119, 2518

Offset: 1

Michel Lagneau, Jan 31 2017

nonn

A006667: number of tripling steps to reach 1 in '3x+1' problem.
A006577: number of halving and tripling steps to reach 1 in '3x+1' problem.
The corresponding number of iterations A006577(a(n)) is given by the sequence 54, 60, 63, 66, 66, 69, 69, 69, 69, 69, 72, 72, 72, 72, 72, 72, 75, 75, ... and the set of the distinct values of this sequence is {b(n)} = {54, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, ...}. We observe that {b(k)} = {54} union {60 + 3*k} for k = 1, 2, ...

			159 is in the sequence because A006667(159)/A006577(159) = 18/54 = 1/3.

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

Cf. A006577, A006666, A006667.

nn:=10000:
for n from 2 to 3000 do:
  m:=n:s1:=0:s2:=0:
   for i from 1 to nn while(m<>1) do:
    if irem(m,2)=0
     then
     s2:=s2+1:m:=m/2:
     else
     s1:=s1+1:m:=3*m+1:
    fi:
   od:
   s:=s1/(s1+s2):
    if s=1/3
     then
     printf(`%d, `,n):
     else
    fi:
od:

A281938 a(n) is the least k such that gcd(A006666(k), A006667(k)) = n.

Original entry on oeis.org

2, 4, 8, 16, 32, 64, 128, 256, 512, 82, 129, 4096, 327, 16384, 32768, 1249, 35655, 159, 4926, 283, 377, 502, 603, 799, 1063, 1417, 1889, 2518, 3356, 4472, 5960, 7944, 10594, 14124, 18833, 25110, 33481, 44641, 59521, 79361, 105814, 141084, 188113, 250817, 334422

Offset: 1

Michel Lagneau, Feb 02 2017

nonn

A006666: Number of halving steps to reach 1 in '3x+1' problem.
A006667: number of tripling steps to reach 1 in '3x+1' problem.
a(n) = 2^n for n = 1, 2, 3, 3, 4, 5, 6, 7, 8, 9, 12, 14, 15.
The primes in the sequence are 2, 283, 1063, 1249, 1889, 44641, ...

			a(10) = 82 because gcd(A006666(82), A006667(82)) = gcd(70, 40) = 10, and there is no k < 82 such that gcd(A006666(k), A006667(k)) = 10.

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

Cf. A006577, A006666, A006667.

for n from 1 to 45 do:
ii:=0:
for k from 2 to 10^7 while(ii=0) do:
  m:=k:s1:=0:s2:=0:
   for i from 1 to nn while(m<>1) do:
    if irem(m,2)=0
     then
     s2:=s2+1:m:=m/2:
     else
     s1:=s1+1:m:=3*m+1:
    fi:
   od:
    if gcd(s1,s2)=n
     then
     ii:=1:printf(`%d %d \n`,n,k):
     else
    fi:
od:
od:

Function[w, First /@ Lookup[w, Function[k, If[k == {}, #, Take[#, First@ k]]]@ Complement[Range@ Max@ #, #]] &@ Keys@ w]@ KeySort@ PositionIndex@ Table[GCD[Count[NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 1 &], ?(EvenQ[#] &)], Count[Differences[NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]], ?Positive]], {n, 2^16}] (* Michael De Vlieger, Feb 02 2017, Version 10, after Harvey P. Dale at A006666 and A006667 *)

A304119 Numerators of record low values of the ratio n*3^A006667(n)/2^A006666(n).

Original entry on oeis.org

1, 27, 1701, 6561, 1760826122505, 115093142840908791, 460166680231540515, 1840049047529878113

Offset: 1

Michel Lagneau, May 03 2018

This has been verified for n up to 10^7.
Conjecture: Consider A006666 and A006667, the sequences giving the number of halving and tripling steps to reach 1 in 3x+1 problem. There exists a rational constant c such that c <= n*3^A006667(n)/2^A006666(n) <= 1 where c = 1840049047529878113/2305843009213693952 is the last term in the sequence of the ratios.
Note that n*3^A006667(n)/2^A006666(n) = 1 if n is a power of 2 (A000079).
It seems that n*3^A006667(n)/2^A006666(n) = c for n = 993*2^k, k = 0, 1, 2, ... In this case, c = 993*2^k*3^32/2^(61+k), where 32 = A006667(993*2^k) and 61+k = A006666(993*2^k). For example, c = 993*3^32/2^61 = 1986*3^32/2^62 = 3972*3^32/2^63 = 7944*3^32/2^64 = ...

			For n=1 to 10 the ratios are: 1, 1, 27/32, 1, 15/16, 27/32, 1701/2048, 1, 6561/8192, 15/16, so the low records are 1, 27/32, 1701/2048, 6561/8192, ...

Cf. A006666, A006667, A127789 (for the indices where these records occur).

q=1; Collatz[n_]:=NestWhileList[If[EvenQ[#], #/2, 3 #+1]&, n, #>1&]; nn=5000; t={}; n=0; While[Length[t]

ht(n) = my(h, t); while(n>1, if(n%2, n=3*n+1; t++, n>>=1; h++)); return([h, t]);
lista(nn) = {m = 2; for (n=1, nn, v = ht(n); newm = n*3^v[2]/2^v[1]; if (newm < m, print1(numerator(newm), ", "); m = newm));} \\ Michel Marcus, May 06 2018

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

A303811 Least k such that A006666(k)/A006667(k) = prime(n).

Original entry on oeis.org

159, 6, 10, 40, 640, 2560, 40960, 163840, 2621440, 167772160, 671088640, 42949672960, 687194767360, 2748779069440, 43980465111040, 2814749767106560, 180143985094819840, 720575940379279360, 46116860184273879040, 737869762948382064640, 2951479051793528258560

Offset: 1

Michel Lagneau, Sep 10 2018

nonn

A006666 and A006667 are respectively the number of halving and tripling steps in the '3x+1' problem.

For n > 2, it seems that a(n) is of the form a(n) = 5*2^q with q = 1, 3, 7, 9, 13, 15, 19, 25, 27, 33, 37, 39, 43, 49, 55, 57, 63, 67, 69, ... (Numbers q such that q+4 is prime: A172367)

			a(4) = 40 because A006666(40)/A006667(40) = 7/1 = prime(4).

Cf. A006577, A006666, A006667, A172367, A287798.

nn:=10^20:
for n from 1 to 10 do:
ii:=0:
   for k from 1 to nn while(ii=0) do:
    it0:=0:it1:=0:m:=k:
      for i from 1 to nn while(m<>1) do:
        if irem(m, 2)=0
         then
         m:=m/2:it0:=it0+1:
         else
         m:=3*m+1:it1:=it1+1:
       fi:
     od:
      if it1<>0 and it0/it1 = ithprime(n)
       then
       ii:=1:printf(`%d %d \n`,n,k):
       else
     fi:
od:
od:

cp's OEIS Frontend

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

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

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A287798 Least k such that A006667(k)/A006577(k) = 1/n.

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

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A277367 a(n) = gcd(A006666(n), A006667(n)) where A006666 and A006667 are respectively the number of halving and tripling steps in the '3x+1' problem.

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A281665 Numbers m such that A006667(m)/A006577(m) = 1/3.

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A281938 a(n) is the least k such that gcd(A006666(k), A006667(k)) = n.

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A304119 Numerators of record low values of the ratio n*3^A006667(n)/2^A006666(n).

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