

A304363


"Stationkeeping Collatz numbers": a(n) is the smallest even number whose Collatz ('3x+1') trajectory, after its initial step downward, is directed back toward its starting value at each of its next n steps.


0



2, 6, 6, 6, 6, 6, 22, 22, 54, 54, 118, 246, 246, 502, 502, 1014, 2038, 2038, 2038, 6134, 6134, 14326, 14326, 14326, 47094, 47094, 47094, 47094, 47094, 309238, 309238, 833526, 1882102, 1882102, 3979254, 3979254, 8173558, 8173558, 8173558, 24950774, 24950774
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OFFSET

0,1


COMMENTS

These trajectories may be called "stationkeeping" in that, like a satellite correcting perturbations to its orbit as they arise, these trajectories continue (for a time) to be directed toward their starting values. (Using even numbers as starting values results in a smaller relative initial step away from the starting value, i.e., from a(n) downward to a(n)/2 rather than from a(n) upward to 3*a(n)+1.)
If the Collatz conjecture is true, there are no cycles, i.e., there exists no number k > 1 whose trajectory ever returns to k, so for each starting value k, the pattern of being directed back toward k at every step after the initial downward step is eventually broken; this occurs when the trajectory reaches either an odd number greater than k (and thus moves even farther upward above k) or an even number less than k (and thus moves even farther downward below k). (For the case n=0, the trajectory from k=2, on its initial step downward, goes immediately to 1 and stops.)
It can be shown that the difference between distinct consecutive terms is always a power of 2.


LINKS

Table of n, a(n) for n=0..40.
Index entries for sequences related to 3x+1 (or Collatz) problem


EXAMPLE

The Collatz trajectory of 6, after its initial downward step (i.e., "step 0"), is directed back toward its starting value (6) at each of the next 5 steps, but its 6th step (from 4 to 2) is directed away from the starting value:
.
start of step 2 16 < start of step 4
 / \
starting v / \
value 10 / \
 / \ / \
v / \ / 8 < start of step 5
6/\/\
\ / \ / \
\ / 5 \
\ / ^ 4 < start of step 6
\ /  \
3 start of step 3 \
^ \
 2
start of step 1 \
...
The trajectory of 2 reaches 1 (and stops) on its initial downward step, and the trajectory of 4 continues downward (to 1) after its initial downward step, so 6 is the smallest even number whose trajectory, after its initial downward step, is directed back toward its starting value for even a single step; thus a(1) = 6. Since the trajectory of 6 continues to be directed back toward its initial value for the next 4 steps as well, a(1) = a(2) = a(3) = a(4) = a(5) = 6. However, since that trajectory's 6th step after the initial downward step is directed away from the starting value, a(6) > 6.
The smallest even number whose Collatz trajectory, after its initial step downward, is directed back toward its starting value at each of its next 65 steps is 1475648206838, so a(65) = 1475648206838; that trajectory continues to be directed back toward its starting value at each of the next 11 steps as well, so a(65) = a(66) = ... = a(76).


PROG

(Magma) nMax:=40; a:=[]; up:=0; dn:=1; kStart:=6; kAtN:=3; for n in [1..nMax] do if (3^up ge 2^dn) ne IsEven(kAtN) then kStart+:=2^dn; kAtN+:=3^up; end if; if IsEven(kAtN) then dn+:=1; kAtN:=kAtN div 2; else up+:=1; kAtN:=3*kAtN+1; end if; a[n]:=kStart; end for; [2] cat a;


CROSSREFS

Cf. A006370, A070165.
Sequence in context: A184408 A184409 A137479 * A247097 A258576 A278253
Adjacent sequences: A304360 A304361 A304362 * A304364 A304365 A304366


KEYWORD

nonn


AUTHOR

Jon E. Schoenfield, May 11 2018


STATUS

approved



