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A265687 Numbers n such that the concave polygon formed by the trajectory of n in the "3n+1" problem and the straight line between the coordinate points (0,n) and (r,1) where r is the number of iterations needed to reach 1 is not a self-intersecting polygon. 2
3, 5, 7, 11, 27, 31, 47, 63, 71 (list; graph; refs; listen; history; text; internal format)



This sequence is included in A266796 and considers only the polygons located above the line connecting (0,n) and (r,1).

Consider the 3n+1 function iterates (n, T(n), T(T(n)),...,4,2,1) plotted on standard vertical and horizontal scales. Each coordinate point (0,n), (1,T(n)), (2, T(T(n)),...,(r,1), where r is the number of iterations needed to reach 1, is connected by a straight line, and the two points (0,n) and (r,1) are also connected by a straight line in order to form a polygon. The sequence lists the numbers n such that all the points of the polygon are located above the line connecting (0,n) and (r,1).

In other words, the sequence lists the numbers n such that the concave polygons are entirely contained in a half-plane defined by the line passing through the points (0,n) and (r,1), of equation f(x)= x*(1-n)/r + n. We must have f(0)= n, f(1)<T(n), f(2)<T(T(n)),...,f(r-1)<2, f(r)=1.

Conjecture: the sequence is finite.

From David Consiglio, Jr., Jan 07 2016: (Start)

Any additional terms must be congruent to either 3 or 7 mod 8 and larger than 10^8.

Even terms are reduced in just one step, and so immediately fall below the line for any initial number > 2.

Odd terms that are 1 mod 8 or 5 mod 8 reduce to just over 3/4ths their original size in 3 steps.  This eliminates any term of this type that requires more than 12 steps from consideration (since 9/12 = 3/4). Since the last number which requires 12 steps or fewer to reduce is 341, and all numbers up to that have been checked, all numbers 1 mod 8 cannot be members of this sequence.

It may be able to further disqualify groups of numbers using this modulus approach.


Table of n, a(n) for n=1..9.

Michel Lagneau, Examples


See the links.



for n from 3 to 10000 do:


    for i from 1 to nn while(m<>1) do:

     if irem(m, 2)=0








    for j from 1 to it+1 do:


      if z>T[j] then jj:=1:




      if jj=0


       printf(`%d, `, n):





Cf. A006577, A266796.

Sequence in context: A227240 A226017 A303705 * A023368 A295973 A142247

Adjacent sequences:  A265684 A265685 A265686 * A265688 A265689 A265690




Michel Lagneau, Dec 20 2015



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