cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

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

Views

Author

Labos Elemer, Sep 08 2003

Keywords

Comments

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)

Examples

			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}
		

Crossrefs

Programs

  • Mathematica
    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}]
  • PARI
    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

Formula

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

Extensions

Terms a(19)-a(21) from John W. Layman, Jun 09 2004
More terms from David Wasserman, Apr 18 2005

A087254 If we start the Collatz-iteration at these values, each divisible by 4, all subsequent terms in trajectory are smaller than the initial value.

Original entry on oeis.org

4, 8, 20, 24, 32, 48, 56, 68, 72, 80, 84, 96, 104, 116, 128, 132, 144, 152, 168, 176, 180, 192, 200, 212, 224, 228, 240, 260, 264, 272, 276, 288, 296, 308, 312, 320, 324, 336, 344, 356, 360, 368, 372, 384, 392, 404, 408, 416, 452, 456, 464, 468, 480, 488, 512
Offset: 1

Views

Author

Labos Elemer, Sep 08 2003

Keywords

Comments

Numbers that are not highest in any Collatz trajectory other than n. - Jayanta Basu, May 27 2013

Examples

			n=104: iteration list = {104,52,26,13,40,20,10,5,16,8,4,2,1}, where initial-value = largest-term.
		

Crossrefs

Cf. A025586, A087251-A087253, A222562 (with 1 and 2 prepended).

Programs

  • Mathematica
    mcoll[n_]:=Max@@NestWhileList[If[EvenQ[#],#/2,3#+1] &,n,#>1 &]; t={}; Do[c=i=0; While[c!=1 && ++iJayanta Basu, May 27 2013 *)
Showing 1-2 of 2 results.