A363348 Turn sequence of a non-Eulerian path for drawing an infinite aperiodic tiling based on the "hat" monotile. See the comments section for details.
3, -2, 3, -2, 3, 2, 0, 2, -3, 2, 3, 2, -3, 2, 3, -2, 3, -2, 3, -2, 0, 2, -3, 2, 3, 2, -3, 2, 3, -2, 3, -2, 3, 2, 0, 2, -3, 2, 3, -2, 0, 2, -3, 2, 3, 2, -3, 2, 3, -2, 3, -2, 3, 2, 0, 2, -3, 2, 3, -2, 0, 2, -3, 2, 3, 2, -3, 2, 3, -2, 3, -2, 3, 2, 0, 2, -3, 2, 3, 2, -3, -2, 3, 2, -3, 2, 3, -2, 3, -2
Offset: 1
Examples
We start by drawing a line of length sqrt(3):
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We take then the first term of the sequence a(1) = 3 this means
we turn our drawing turtle 90 degrees to the left and also switch to a length unit of 1.
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We take the second term from the sequence a(2) = -2 this means
we turn our drawing turtle 60 degrees to the right, and we keep the selected line length of 1 unit.
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(In this ASCII representation, angles and length units are only symbolically represented and do not match the exact values in the description.)
Links
- Thomas Scheuerle, Table of n, a(n) for n = 1..1588
- Thomas Scheuerle, MATLAB program
- Thomas Scheuerle, 1st iteration: drawing of a(1..140) (results in 8 "hat" tiles).
- Thomas Scheuerle, 2nd iteration: drawing of a(1..1588) (results in 55 "hat" tiles).
- Thomas Scheuerle, 3rd iteration: drawing of a(1..206104) (results in 377 "hat" tiles).
- Thomas Scheuerle, 4th iteration: drawing of a(1..2462272) (results in 2584 "hat" tiles).
- David Smith, Joseph Samuel Myers, Craig S. Kaplan, and Chaim Goodman-Strauss, An aperiodic monotile, arXiv:2303.10798 [math.CO], 2023.
Crossrefs
Programs
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MATLAB
% See Scheuerle link.
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PARI
L(k) = { my(v = [0, 14, 140, 1588]); if(k > 3, return(12*L(k-1) - 7*L(k-2) + L(k-3)), return(v[k+1])) } r1(k) = if(k > 1, return(r5(k-1) + r1(k-1) + r7(k-1)), return(6)) r2(k) = if(k > 1, return(r2(k-1) + r7(k-1)), return(6)) r3(k) = if(k > 1, return(2*r5(k-1) + r3(k-1) + r5(k) + r7(k-1)), return(6)) r5(k) = if(k > 1, return(r5(k-1) + r3(k-1)), return(2)) r7(k) = if(k > 1, return(r5(k-1) + 2*r3(k-1)), return(4)) r8(k) = if(k > 1, return(r12(k-1) + r8(k-1) + r14(k-1)), return(1)) r9(k) = if(k > 1, return(r9(k-1) + r14(k-1)), return(1)) r10(k) = if(k > 1, return(2*r13(k-1) + r10(k-1) + r11(k-1) + r14(k-1)), return(1)) r11(k) = if(k > 1, return(2*r13(k-1) + 3*r10(k-1) + r11(k-1)), return(1)) r12(k) = if(k > 1, return(r13(k-1) + r10(k-1)), return(1)) r13(k) = if(k > 1, return(r12(k-1) + r13(k-1) + r14(k-1)), return(1)) r14(k) = if(k > 1, return(r13(k-1) + 2*r10(k-1)), return(1)) c1(k) = r2(k) + sum(m=1, k-1, r9(k+1-m)*L(m)) c2(k) = c1(k) - sum(m=1, k-1, L(m)) c3(k) = r2(k) + r3(k) + sum(m=1, k-1, (r9(k+1-m) + r10(k+1-m) - 1)*L(m)) c4(k) = r2(k) + r5(k+1) + sum(m=1, k-1, (r9(k+1-m) + r11(k+1-m) - 1)*L(m)) c5(k) = r2(k) + r7(k) + sum(m=1, k-1, (r9(k+1-m) + r14(k+1-m) - 2)*L(m)) c6(k) = c4(k) - sum(m=1, k-1, L(m)) a(NumIter) = { my(a = [3, -2, 3, -2, 3, 2, 0, 2, -3, 2, 3, 2, -3, 2]); for(k = 1, NumIter, a = concat([a, a[1..(c1(k)-1)], -a[c1(k)], a[(c2(k)+1)..L(k)], a[1..(c3(k)-1)], -a[c3(k)], a[(c2(k)+1)..L(k)], a[1..(c3(k)-1)], -a[c3(k)], a[(c2(k)+1)..L(k)], a[1..(c4(k)-1)], -a[c4(k)], a[(c5(k)+1)..L(k)], a[1..(c3(k)-1)], -a[c3(k)], a[(c2(k)+1)..L(k)], a[1..(c3(k)-1)], -a[c3(k)], a[(c2(k)+1)..L(k)], a[1..(c4(k)-1)], -a[c4(k)], a[(c6(k)+1)..L(k)]]) ); return(a) } draw(NumIter) = {my(p = [0, sqrt(3)]); my(dl = [1]); my(s = a(NumIter)); for(j=2, length(s), dl = concat(dl, ((dl[j-1]+(abs(s[j-1])==3))%2)); p = concat(p, p[j]+sqrt(1+2*dl[j])*exp(I*Pi*vecsum(s[1..j-1])*(1/6)) )); plothraw(apply(real, p), apply(imag, p), 1); }
Formula
a(1..14) = {3,-2, 3,-2, 3, 2, 0, 2, -3, 2, 3, 2,-3, 2} = a(1..L(1)) and for k > 0:
a(1..L(k+1)) = {a(1..L(k)), a(1..c1(k)-1), -a(c1(k)), a(c2(k)+1..L(k)), a(1..c3(k)-1), -a(c3(k)), a(c2(k)+1..L(k)), a(1..c3(k)-1), -a(c3(k)), a(c2(k)+1..L(k)), a(1..c4(k)-1), -a(c4(k)), a(c5(k)+1..L(k)), a(1..c3(k)-1), -a(c3(k)), a(c2(k)+1..L(k)), a(1..c3(k)-1), -a(c3(k)), a(c2(k)+1..L(k)), a(1..c4(k)-1), -a(c4(k)), a(c6(k)+1..L(k))}. With:
L(k) = 12*L(k-1) - 7*L(k-2) + L(k-3) for k > 3 with L(1..3) = {14, 140, 1588}.
r1(k) = r5(k-1) + r1(k-1) + r7(k-1), with r1(1) = 6.
r2(k) = r2(k-1) + r7(k-1), with r2(1) = 6.
r3(k) = 2*r6(k-1) + r3(k-1) + r4(k-1) + r7(k-1), with r3(1) = 6 (A003699).
r4(k) = r5(k+1) = 2*r5(k-1) + 3*r3(k-1) + r4(k-1), with r4(1) = 8 (A052530).
r5(k) = r5(k-1) + r3(k-1), with r5(1) = 2. r4, r5, r6 are in the case of this tiling accidentally essentially the same recurrence.
r6(k) = r5(k) = r5(k-1) + r6(k-1) + r7(k-1), with r6(1) = 2 (A052530).
r7(k) = r6(k-1) + 2*r3(k-1), with r7(1) = 4 (A003500).
r8(k) = r12(k-1) + r8(k-1) + r14(k-1), with r8(1) = 1
r9(k) = r9(k-1) + r14(k-1), with r9(1) = 1.
r10(k) = 2*r13(k-1) + r10(k-1) + r11(k-1) + r14(k-1), with r10(1) = 1 (A061278).
r11(k) = 2*r13(k-1) + 3*r10(k-1) + r11(k-1), with r11(1) = 1.
r12(k) = r13(k-1) + r10(k-1), with r12(1) = 1.
r13(k) = r12(k-1) + r13(k-1) + r14(k-1), with r13(1) = 1.
r14(k) = r13(k-1) + 2*r10(k-1), with r14(1) = 1 (A108946 unsigned).
c1(k) = r2(k) + Sum_{m=1..k-1} (r9(k+1-m)*L(m)) = {6, 38, 374, 4204, ...}.
c2(k) = c1(k) - Sum_{m=1..k-1} L(m) = {6, 24, 220, 2462, ...}.
c3(k) = r2(k) + r3(k) + Sum_{m=1..k-1} ((r9(k+1-m) + r10(k+1-m) - 1)*L(m)) = {12, 116, 1282, 14572, ...}.
c4(k) = r2(k) + r4(k) + Sum_{m=1..k-1} ((r9(k+1-m) + r11(k+1-m) - 1)*L(m)) = {14, 138, 1550, 17630, ...}.
c5(k) = r2(k) + r7(k) + Sum_{m=1..k-1} ((r9(k+1-m) + r14(k+1-m) - 2)*L(m)) = {10, 66, 720, 8170, ...}.
c6(k) = c4(k) - Sum_{m=1..k-1} L(m) = {14, 124, 1396, 15888, ...}.
Description of curve position:
OrientationAngle(n) = Sum_{k = 1..n-1} a(k)*Pi*(1/6).
Xcoordinate(n) = Sum_{k = 1..n} cos(OrientationAngle(n))*sqrt(1 + 2*((1 + Sum_{k = 1..n-1} [abs(a(k)) = 3]) mod 2)).
Ycoordinate(n) = Sum_{k = 1..n} sin(OrientationAngle(n))*sqrt(1 + 2*((1 + Sum_{k = 1..n-1} [abs(a(k)) = 3]) mod 2)). [] is the Iverson bracket here.
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