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A210838
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Coordinates (x,y) of the endpoint of a structure (or curve) formed by Q-toothpicks of size = 1..n. The inflection points are the n-th nodes if n is a triangular number A000217.
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7
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0, 0, 1, 1, 3, 3, 0, 6, -4, 10, 1, 15, 7, 9, 14, 2, 22, 10, 13, 19, 3, 9, -8, -2, -20, 10, -7, 23, 7, 9, -8, -6, -24, -22, -7, -39, 11, -21, -8, -2, -28, -22, -7, -43, 15, -65, -8, -88, -32, -64, -7, -39, 19, -65, -8, -92, -36, -64, -65, -35, -95, -65, -64, -96
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OFFSET
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0,5
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COMMENTS
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It appears there is an infinite family of this type of curves or structures in which the terms of a sequence of positive integers are represented as inflection points and the gaps between them are essentially represented as nodes of spirals. For example: consider a structure formed by Q-toothpicks of size = Axxxxxa connected by their endpoints in which the inflection points are the exposed endpoints at stage Axxxxxb(n), where both Axxxxxa and Axxxxxb are sequences with positive integers. Also instead of Q-toothpicks we can use semicircumferences or also 3/4 of circumferences. For the definition of Q-toothpicks see A187210.
We start at stage 0 with no Q-toothpicks.
At stage 1 we place a Q-toothpick of size 1 centered at (1,0) with its endpoints at (0,0) and (1,1). Since 1 is a positive triangular number we have that the end of the curve is also an inflection point.
At stage 2 we place a Q-toothpick of size 2 centered at (1,3) with its endpoints at (1,1) and (3,3).
At stage 3 we place a Q-toothpick of size 3 centered at (0,3) with its endpoints at (3,3) and (0,6). Since 3 is a positive triangular number we have that the end of the curve is also an inflection point.
At stage 4 we place a Q-toothpick of size 4 centered at (0,10) with its endpoints at (0,6) and (-4,10).
And so on...
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LINKS
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EXAMPLE
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-------------------------------------
Stage n also The end as
the size of Pair inflection
Q-toothpick (x y) point
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. 0 0, 0, -
. 1 1, 1, Yes
. 2 3, 3, -
. 3 0, 6, Yes
. 4 -4, 10, -
. 5 1, 15, -
. 6 7, 9, Yes
. 7 14, 2, -
. 8 22, 10, -
. 9 13, 19, -
. 10 3, 9, Yes
. 11 -8, -2, -
. 12 -20, 10, -
. 13 -7, 23, -
. 14 7, 9, -
. 15 -8, -6, Yes
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MATHEMATICA
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A210838[nmax_]:=Module[{ep={0, 0}, angle=3/4Pi, turn=Pi/2, infl=0}, Join[{ep}, Table[If[n>1&&IntegerQ[Sqrt[8(n-1)+1]], infl++, If[Mod[infl, 2]==1, turn*=-1]; angle-=turn; infl=0]; ep=AngleVector[ep, {Sqrt[2]n, angle}], {n, nmax}]]];
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PROG
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(PARI)
A210838(nmax) = my(ep=vector(nmax+1), turn=1, infl=0, ep1, ep2); ep[1]=[0, 0]; if(nmax==0, return(ep)); ep[2]=[1, 1]; for(n=2, nmax, ep1=ep[n-1]; ep2=ep[n]; if(issquare((n-1)<<3+1), infl++; ep[n+1]=[ep2[1]+n*sign(ep2[1]-ep1[1]), ep2[2]+n*sign(ep2[2]-ep1[2])], if(infl%2, turn*=-1); infl=0; ep[n+1]=[ep2[1]-turn*n*sign(ep1[2]-ep2[2]), ep2[2]+turn*n*sign(ep1[1]-ep2[1])])); ep;
(Python)
from numpy import sign
from sympy import integer_nthroot
ep, turn, infl = [(0, 0), (1, 1)], 1, 0
for n in range(2, nmax + 1):
ep1, ep2 = ep[-2], ep[-1]
if integer_nthroot(((n - 1) << 3) + 1, 2)[1]: # Continue straight
infl += 1
dx = n * sign(ep2[0] - ep1[0])
dy = n * sign(ep2[1] - ep1[1])
else: # Turn
if infl % 2: turn *= -1
infl = 0
dx = turn * n * sign(ep2[1] - ep1[1])
dy = turn * n * sign(ep1[0] - ep2[0])
ep.append((ep2[0] + dx, ep2[1] + dy))
return ep[:nmax+1]
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CROSSREFS
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Cf. A210841 (the same idea for primes).
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KEYWORD
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AUTHOR
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EXTENSIONS
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a(30)-a(33) corrected and more terms by Paolo Xausa, Jan 12 2023
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STATUS
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approved
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