A348007 Starting value of the longest run of halving steps in the trajectory from n to 1 in the Collatz map (or 3x+1 problem), or -1 if no such trajectory exists.

2, 16, 4, 16, 16, 16, 8, 16, 16, 16, 16, 16, 16, 160, 16, 16, 16, 16, 16, 64, 16, 160, 16, 16, 16, 160, 16, 16, 160, 160, 32, 16, 16, 160, 16, 112, 16, 304, 16, 160, 64, 112, 16, 16, 160, 160, 48, 112, 16, 16, 16, 160, 160, 160, 16, 112, 16, 304, 160, 160, 160

Offset: 2

Author

Paolo Xausa, Sep 24 2021

Keywords

• nonn

Comments

In case of ties (two or more longest runs of same length), the highest starting value is picked. The first n for which the longest run of halving steps occurs at two different subtrajectories is 37, where the Collatz map contains the 4-step subtrajectories 112 -> 56 -> 28 -> 14 > 7 and 16 -> 8 -> 4 -> 2 -> 1. a(37) is therefore 112 (highest starting value).

If the Collatz conjecture (i.e., all trajectories reach 1) is true then, except for n = 2, 4 and 8, a(n) mod 16 = 0, since all trajectories contain (at least) 4 consecutive halvings.

Examples

			a(2) = 2 because the Collatz trajectory from 2 to 1 is simply 2 -> 1 (one halving step, starting at 2).
a(3) = 16 because the trajectory from 3 to 1 is 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1. Here, the longest halving run is the 4-step subtrajectory 16 -> 8 -> 4 -> 2 -> 1, which starts at 16.
a(15) = 160 because the longest halving run in the trajectory from 15 to 1 (the 5-step subtrajectory 160 -> 80 -> 40 -> 20 -> 10 -> 5) starts at 160.
		

Links

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

Crossrefs

Cf. A006370, A070165, A135282, A347409.

Programs

Mathematica

nterms=100;Table[c=n;cm=sm=0;While[c>1,If[OddQ[c],c=3c+1,s=IntegerExponent[c,2];If[s>sm,sm=s;cm=c,If[s==sm,cm=Max[cm,c]]];c/=2^s]];cm,{n,2,nterms+1}]

PARI

A348007(n) = { my(m2v=valuation(n,2), mx=n, t); while(n>1, if((t=valuation(n,2))>m2v, m2v=t; mx=n, if(t==m2v && n>mx, mx=n)); if(!(n%2),n/=2,n+=(n+n+1))); (mx); }; \\ Antti Karttunen, Oct 13 2021

Formula

a(2^k) = 2^k, for integers k >= 1.

a(n) mod 2^A347409(n) = 0.