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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)
 OFFSET 1,1 COMMENTS 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) 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. LINKS Michel Lagneau, Examples EXAMPLE See the links. MAPLE T:=array(1..1000):nn:=1000: for n from 3 to 10000 do:   kk:=1:m:=n:T[kk]:=n:it:=0:     for i from 1 to nn while(m<>1) do:      if irem(m, 2)=0       then       m:=m/2:kk:=kk+1:T[kk]:=m:it:=it+1:       else       m:=3*m+1:kk:=kk+1:T[kk]:=m:it:=it+1:      fi:     od:    jj:=0:     for j from 1 to it+1 do:      z:=((1-n)/it)*(j-1)+n:       if z>T[j] then jj:=1:       else       fi:     od:       if jj=0        then        printf(`%d, `, n):        else       fi: od: CROSSREFS Cf. A006577, A266796. Sequence in context: A227240 A226017 A303705 * A023368 A295973 A142247 Adjacent sequences:  A265684 A265685 A265686 * A265688 A265689 A265690 KEYWORD nonn AUTHOR Michel Lagneau, Dec 20 2015 STATUS approved

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Last modified April 24 18:03 EDT 2019. Contains 322430 sequences. (Running on oeis4.)