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COMMENTS
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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.
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.
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MAPLE
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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:
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