A233293 -id:A233293 - cp's OEIS frontend

A033496 Numbers m that are the largest number in their Collatz (3x+1) trajectory.

1, 2, 4, 8, 16, 20, 24, 32, 40, 48, 52, 56, 64, 68, 72, 80, 84, 88, 96, 100, 104, 112, 116, 128, 132, 136, 144, 148, 152, 160, 168, 176, 180, 184, 192, 196, 200, 208, 212, 224, 228, 232, 240, 244, 256, 260, 264, 272, 276, 280, 288, 296, 304, 308, 312, 320, 324

Offset: 1

Jeff Burch

Or, possible peak values in 3x+1 trajectories: 1,2 and m=16k+4,16k+8,16k but not for all k; those 4k numbers [like m=16k+12 and others] which cannot be such peaks are listed in A087252.

Possible values of A025586(m) in increasing order. See A275109 (number of times each value of a(n) occurs in A025586). - Jaroslav Krizek, Jul 17 2016

			These peak values occur in 1, 3, 6, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 21, 22, 27, 30, 39, 44, 71, 75, 1579 [3x+1]-iteration trajectories started with different initial values. This list most probably is incomplete.
From _Hartmut F. W. Hoft_, Jun 24 2016: (Start)
Let n be the maximum in some Collatz trajectory and let F(n), the initial fan of n, be the set of all initial values less than or equal to n whose Collatz trajectories lead to n as their maximum. Then the size of F(n) never equals 2, 4, 5, 7 or 10 (see the link).
Conjecture: Every number k > 10 occurs as the size of F(n) for some n.
Fans F(n) of size k, for all 10 < k < 355, exist for 4 <= n <= 50,000,000. The largest fan in this range, F(41163712), has size 7450.
(End)

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 (first 2000 terms from T. D. Noe)
Hartmut F. W. Hoft, initial Collatz fans
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A025586, A087251-A087256, A015999, A275109.
Cf. A006370, A008908, A070165.
Cf. A095384 (contains a definition of Collatz[]).
Cf. A105730, A233293.

a033496 n = a033496_list !! (n-1)
a033496_list = 1 : filter f [2, 4 ..] where
   f x = x == maximum (takeWhile (/= 1) $ iterate a006370 x)
-- Reinhard Zumkeller, Oct 22 2015

Set(Sort([Max([k eq 1 select n else IsOdd(Self(k-1)) and not IsOne(Self(k-1)) select 3*Self(k-1)+1 else Self(k-1) div 2: k in [1..5*n]]): n in [1..2^10] | Max([k eq 1 select n else IsOdd(Self(k-1)) and not IsOne(Self(k-1)) select 3*Self(k-1)+1 else Self(k-1) div 2: k in [1..5*n]]) le 2^10])) // Jaroslav Krizek, Jul 17 2016

Collatz[a0_Integer, maxits_:1000] := NestWhileList[If[EvenQ[ # ], #/2, 3# + 1] &, a0, Unequal[ #, 1, -1, -10, -34] &, 1, maxits]; (* Collatz[n] function definition by Eric Weisstein *)
Select[Range[324], Max[Collatz[#]] == # &] (* T. D. Noe, Feb 28 2013 *)

def a(n):
    if n<2: return [1]
    l=[n, ]
    while True:
        if n%2==0: n//=2
        else: n = 3*n + 1
        if n not in l:
            l.append(n)
            if n<2: break
        else: break
    return l
print([n for n in range(1, 501) if max(a(n)) == n]) # Indranil Ghosh, Apr 14 2017

A008908(a(n)) = A159999(a(n)). - Reinhard Zumkeller, May 04 2009

Max(A070165(a(n),k): k=1..A008908(a(n))) = A070165(a(n),1) = a(n). - Reinhard Zumkeller, Oct 22 2015

A087256 Number of different initial values for 3x+1 trajectories in which the largest term appearing in the iteration is 2^n.

Original entry on oeis.org

1, 1, 1, 6, 1, 3, 1, 3, 1, 12, 1, 3, 1, 3, 1, 8, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 13, 1, 3, 1, 3, 1, 8, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 9, 1, 3, 1, 3, 1, 11, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 21, 1, 3, 1, 3, 1, 8, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 78, 1, 3, 1, 3, 1, 8, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 9, 1, 3, 1

Offset: 1

Labos Elemer, Sep 08 2003

nonn

It would be interesting to know whether the ...1,3,1,3,1,x,1,3,1,3,1,... pattern persists. - John W. Layman, Jun 09 2004
The observed pattern should persist. Proof: [1] a(odd)=1 because -1+2^odd is not divisible by 3, so in Collatz-algorithm 2^odd is preceded by increasing inverse step. Thus 2^odd is the only suitable initial value; [2] a[2k]>=3 for k>1 because 2^(2k)-1=-1+4^k=3A so {b=2^2k, (b-1)/3 and (2a-2)/3} are three relevant initial values. No more case arises unless condition-[3] (see below) was satisfied; [3] a[6k+4]>=5 for k>=1, ..iv=c=2^(6k+4); here {c, (c-1)/3, 2(c-1)/3, (2c-5)/9, (4c-10)/9} is 5 suitable initial values, iff (2c-5)/9 is an integer; e.g. at 6k+4=10, {1024<-341<-682<-227<-454} back-tracking the iteration. - Labos Elemer, Jun 17 2004
From Hartmut F. W. Hoft, Jun 24 2016: (Start)
Except for a(2)=1 the sequence has the 6-element quasiperiod 1, 3, 1, x, 1, 3  where x>=6, but unequal to 7 and 10 (see links below and in A033496). Observe that for n=2^(6k+4)=16*2^(6k), n mod 9 = 7 so that (2n-5)/9 is an integer and a(n)>=6.
Conjecture: All numbers m > 10 occur as values in A087256 (see A233293).
The conjecture has been verified for all 10 < k < 133 for Collatz trajectories with maximum value through 2^(36000*6 + 4). The largest fan of initial values in this range, F(6*1993+4), has maximum 2^11962 and size 3958.
(End)

			n = 10: 2^10 = 1024 = peak for trajectories started with initial value taken from the list: {151, 201, 227, 302, 341, 402, 454, 604, 682, 804, 908, 1024};
a trajectory with peak=1024: {201, 604, 302, 151, 454, 227, 682, 341, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1}

Hartmut F. W. Hoft, proof of quasi period 6

Cf. A025586, A087251, A087252, A087253, A087254, A105730, A233293.

c[x_]:=c[x]=(1-Mod[x, 2])*(x/2)+Mod[x, 2]*(3*x+1);c[1]=1; fpl[x_]:=FixedPointList[c, x]; {$RecursionLimit=1000;m=0}; Table[Print[{xm-1, m}];m=0; Do[If[Equal[Max[fpl[n]], 2^xm], m=m+1], {n, 1, 2^xm}], {xm, 1, 30}]

f(n, m) = 1 + if(2*n <= m, f(2*n, m), 0) + if (n%6 == 4, f(n\3, m), 0);
a(n) = f(2^n, 2^n); \\ David Wasserman, Apr 18 2005

a(6n+4) = A105730(n). - David Wasserman, Apr 18 2005

Terms a(19)-a(21) from John W. Layman, Jun 09 2004

More terms from David Wasserman, Apr 18 2005

A105730 Number of different initial values for 3x+1 trajectories in which the largest term appearing in the iteration is 2^(6n+4).

Original entry on oeis.org

6, 12, 8, 6, 13, 8, 6, 9, 11, 6, 21, 8, 6, 78, 8, 6, 9, 13, 6, 15, 8, 6, 16, 8, 6, 9, 20, 6, 12, 8, 6, 13, 8, 6, 9, 11, 6, 14, 8, 6, 32, 8, 6, 9, 32, 6, 23, 8, 6, 24, 8, 6, 9, 14, 6, 12, 8, 6, 13, 8, 6, 9, 11, 6, 14, 8, 6, 19, 8, 6, 9, 13, 6, 80, 8, 6, 29, 8, 6, 9, 18, 6, 12, 8, 6, 13, 8, 6, 9, 11

Offset: 0

David Wasserman, Apr 18 2005

From Hartmut F. W. Hoft, Jun 24 2016: (Start)
The sequence has the quasiperiod 6, x, 8, 6, y, 8, 6, 9, z of length 9 starting at index 0 where x, y, z > 10; in addition, a(3*9*n+1) = 12, a(3*9*n+4) = 13 and a(3*9*n+8) = 11 for all n>=0; proof by induction (see this link) as in the link in A087256 based on the modular identities in the link in A033496.
Conjecture: All numbers greater than 10 appear in the sequence (see also A033496 and A233293). (End)

			a(1) = 12, i.e. the number of initial values for 2^10, since 804 -> 402 -> 201 -> 604 -> 302 -> 151 -> 454 -> 227 -> 682 -> 341 -> 1024 and 908 -> (454 -> ... -> 1024) are the two maximal trajectories containing all 12 initial values. a(8) = 11 since 2^(6*8+4) has 11 different initial values for Collatz trajectories leading to it. - _Hartmut F. W. Hoft_, Jun 24 2016

Hartmut F. W. Hoft, identities to prove for quasi period 9

Cf. A025586, A033496, A087256, A233293.

trajectory[start_] := NestWhileList[If[OddQ[#], 3#+1, #/2]&, start, #!=1&]
fanSize[max_] := Module[{active={max}, fan={}, current}, While[active!={}, current=First[active];active=Rest[active]; AppendTo[fan, current]; If[2*current<=max, AppendTo[active, 2*current]]; If[Mod[current, 3]==1 && OddQ[(current-1)/3] && current>4, AppendTo[active, (current-1)/3]]]; Length[fan]]/;max==Max[trajectory[max]]
a105730[low_, high_] := Map[fanSize[2^(6#+4)]&, Range[low, high]]
a105730[0,89] (* Hartmut F. W. Hoft, Jun 24 2016 *)

a(n) = A087256(6n+4).

A232870 Numbers that are the largest value in the Collatz (3x+1) trajectories of exactly three initial values.

Original entry on oeis.org

40, 64, 100, 112, 136, 148, 184, 208, 244, 256, 280, 352, 400, 424, 472, 532, 544, 616, 640, 688, 712, 724, 784, 820, 832, 868, 904, 928, 964, 976, 1048, 1072, 1108, 1120, 1156, 1192, 1216, 1264, 1300, 1360, 1396, 1408, 1432, 1480, 1540, 1576, 1588, 1624, 1684

Offset: 1

Jon E. Schoenfield, Dec 01 2013

nonn

Numbers that appear exactly 3 times in A025586, which gives the largest value in the 3x + 1 trajectory of n.
For each term k in this sequence, the three initial values, that is, values of n at which A025586(n) = k, are (in ascending order) n1 = (k-1)/3, n2 = 2*n1 = 2*(k-1)/3, and n3 = k. n1 is the odd number from which an upward (that is, 3x + 1) step lands at k = 3*n1 + 1. It cannot be the case that n1 = 3 (mod 4), because we would then have k = 10 (mod 12), so k/2 would be odd, and its successor in the trajectory would be 3*k/2 + 1 > k, so k would not be the largest value in the trajectory. Thus, n1 = 1 (mod 4), so n2 = 2 (mod 8) and n3 = 4 (mod 12).
Numbers that are the largest value in the 3x + 1 trajectory of exactly one initial value (that is, numbers that appear exactly once in A025586) are in A222562.
Numbers that are not the largest value in the 3x + 1 trajectory of any initial value (that is, numbers that do not appear at all in A025586) are in A213199.

			40 is in the sequence because it is the largest value in the 3x + 1 trajectories of exactly three initial values: 13, 26, and 40 itself. The trajectories are as follows:
..... 13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1
26 -> 13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1
........... 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000

Cf. A025586, A213199, A222562, A233293.

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A033496 Numbers m that are the largest number in their Collatz (3x+1) trajectory.

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A087256 Number of different initial values for 3x+1 trajectories in which the largest term appearing in the iteration is 2^n.

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A105730 Number of different initial values for 3x+1 trajectories in which the largest term appearing in the iteration is 2^(6n+4).

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A232870 Numbers that are the largest value in the Collatz (3x+1) trajectories of exactly three initial values.

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