A108946 -id:A108946 - cp's OEIS frontend

A117808 Primes of the form ((2 + sqrt(3))^(2k+1) + (2 - sqrt(3))^(2k+1))/4.

3, 13, 181, 2521, 489061, 6811741, 1321442641, 18405321661, 381765135195632792959100810331957408101589361

Offset: 1

Roger L. Bagula, Apr 29 2006

nonn

Primes in A001570. - Joerg Arndt, Dec 30 2023
Primes among absolute values of A108946.
Also the Cosgrave-Dilcher primes that are a subset of the nontrivial cyclotomic lambda invariant for Q(sqrt{-3}) (or a subset of the 1-exceptional primes for M=3). - Christopher M. Stokes, Aug 04 2022

J. B. Cosgrave and K. Dilcher, The multiplicative orders of certain Gauss factorials, Intl. J. Number Theory 7 (1) (2011) 145-171.
John B. Cosgrave and Karl Dilcher, The multiplicative orders of certain Gauss factorials II, Funct. Approx. Comment. Math. Volume 54, Number 1 (2016), 73-93.
Christopher Stokes, On Gauss factorials and their application to Iwasawa theory for imaginary quadratic fields, arXiv:2207.07804 [math.NT], 2022.
Eric Weisstein's World of Mathematics, Prime-Generating Polynomial

Cf. A108946.

b(n)=my(w=quadgen(12)); ((w+2)^n+(2-w)^n)\4
for(n=2,800, if(isprime(p=b(n)), print1(p", "))) \\ Charles R Greathouse IV, Aug 22 2022

Definition and terms corrected by N. J. A. Sloane, May 21 2010

Edited by Joerg Arndt, Dec 30 2023

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.

Original entry on oeis.org

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

Thomas Scheuerle, May 28 2023

The curve can be drawn using turtle graphics rules. Each term of the sequence encodes an angle of rotation in units of (1/6)*Pi. For example, a(k) = 3 would mean a turn of 90 degrees to the left, a(k) = -2 a turn of 60 degrees to the right. To draw the tiling we draw a line of length l and then take a term of the sequence to determine the direction of further drawing by rotation relative to the current drawing orientation. The length of the line segments between terms of the sequence is either sqrt(3) or 1 units. We start by drawing with sqrt(3) units of length; every time we reach a term with 3 or -3 in the sequence we toggle the selected line length from sqrt(3) to 1, or back again from 1 to sqrt(3).

The drawing process works by recursion into the H8 metatile and its supertiles; this means a(1..14) draws a single "hat" monotile. Then the terms a(1..140) draw the H8 metatile and a(1..1588) and so forth (see formula section) draw the next larger supertile of the H8 metatile. (For details regarding H8 see page 18 in arXiv:2303.10798.) The number of "hat" tiles visible after k recursions is Fibonacci(4*k + 2) (A033890); however, tiles and line segments will be overdrawn multiple times in this process.

			We start by drawing a line of length sqrt(3):
___
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.
___|
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.
    /
___|
(In this ASCII representation, angles and length units are only symbolically represented and do not match the exact values in the description.)

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.

Cf. A363445 describes a curve around the perimeter of this tiling.

Cf. A003500, A003699, A033890, A052530, A061278, A108946.

MATLAB
```
% See Scheuerle link.
```

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); }

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.

A363445 Turn sequence of a fractal-like curve which is also the perimeter around an aperiodic tiling based on the "hat" monotile. See the comments section for details.

Original entry on oeis.org

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, 0, 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, 0, 2, -3, 2, 3, 2, -3, -2, 3, -2, 0, 2, -3, -2

Offset: 1

Thomas Scheuerle, Jul 09 2023

sign

The curve can be drawn by turtle graphics rules. Each term of the sequence encodes an angle of rotation in units of (1/6)*Pi. For example, a(k) = 3 would mean a turn of 90 degrees to the left, a(k) = -2 a turn of 60 degrees to the right. To draw the tiling we draw a line of length l and then take a term of the sequence to determine the direction of further drawing by rotation relative to the current drawing orientation. The length of the line segments between each term of the sequence is either sqrt(3) or 1 units. We start by drawing with sqrt(3) units of length; every time we reach a term with 3 or -3 in the sequence we toggle the selected line length from sqrt(3) in 1, or back again from 1 in sqrt(3).
The curve is defined by recursion; this means a(1..14) draws a single "hat" monotile. Then the interval a(15..56) draws the perimeter around the H8 metatile and a(57..202) will be the perimeter around the next higher composition of these tiles. (For details regarding H8 see page 18 in arXiv:2303.10798.) The number of "hat" tiles enclosed by this curve after k recursions is Fibonacci(4*k + 2) (A033890).
The number of new terms added after each iteration can be calculated as m(k) = 5*m(k-1) - 5*m(k-2) + m(k-3) with m(1..3) = {14, 42, 146, ...}. After each such iteration the curve will be closed with an enclosed area equivalent to A033890(k+1) "hat" tiles.

			We start by drawing a line of length sqrt(3):
___
We then take 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.
___|
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.
    /
___|
(In this ASCII representation, angles and length units are only symbolically represented and do not match the exact values in the description.)

Thomas Scheuerle, Table of n, a(n) for n = 1..2718
David Smith, Joseph Samuel Myers, Craig S. Kaplan, and Chaim Goodman-Strauss, An aperiodic monotile, arXiv:2303.10798 [math.CO], 2023.
Thomas Scheuerle, MATLAB program
Thomas Scheuerle, a(1..140480) drawn out by turtle graphics. This is the result of the seventh iteration.

Cf. A363348 describes how to draw this curve together with all "hat" monotiles enclosed by it.

Cf. A003500, A003699, A033890, A052530, A061278, A108946.

MATLAB
```
% See Scheuerle link.
```

L(k) = { my(v = [0, 14, 56, 202]); if(k > 3,return(6*L(k-1) - 10*L(k-2) + 6*L(k-3) - L(k-4)),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))
c1(k) = r2(k) + L(k-1)
c2(k) = r2(k) + r3(k) + L(k-1)
c3(k) = r2(k) + r5(k+1) + L(k-1)
c4(k) = r2(k) + r7(k) + L(k-1)
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[(L(k-1)+1)..(c1(k)-1)], -a[c1(k)], a[(c1(k)+1)..(c2(k)-1)], -a[c1(k)], a[(c1(k)+1)..(c2(k)-1)], -a[c1(k)], a[(c1(k)+1)..(c3(k)-1)], -a[c4(k)], a[(c4(k)+1)..(c2(k)-1)], -a[c1(k)], a[(c1(k)+1)..(c2(k)-1)], -a[c1(k)],  a[(c1(k)+1)..(c3(k)-1)], -a[c3(k)], a[(c3(k)+1)..length(a)] ]) ); 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);}

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-1)), a(L(k)+1..c1(k)-1), -a(c1(k)), a(c1(k)+1..c2(k)-1), -a(c1(k)), a(c1(k)+1..c2(k)-1), -a(c1(k)), a(c1(k)+1..c3(k)-1), -a(c4(k)), a(c4(k)+1..c2(k)-1), -a(c1(k)), a(c1(k)+1..c2(k)-1), -a(c1(k)),  a(c1(k)+1..c3(k)-1), -a(c3(k)), a(c3(k)+1..L(k))}. With:
L(k) = 6*L(k-1) - 10*L(k-2) + 6*L(k-3) - L(k-4), for k > 3 and L(0..3) = {0, 14, 56, 202}.
L(k) = L(k-1) + r1(k-1) + 3*r3(k-1) + 2*r4(k-1) + r6(k-1).
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) = r6(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).
c1(k) = r2(k) + L(k)         = {6, 24, 80, ...}.
c2(k) = r2(k) + r3(k) + L(k) = {12, 46, 162, ...}.
c3(k) = r2(k) + r4(k) + L(k) = {14, 54, 192, ...}.
c4(k) = r2(k) + r7(k) + L(k) = {10, 38, 132, ...}.
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.
For some nonnegative integers b and c:
  OrientationAngle(L(b)) = OrientationAngle(L(c)).
  Xcoordinate(L(b)) = Xcoordinate(L(c)).
  Ycoordinate(L(b)) = Ycoordinate(L(c)).

cp's OEIS Frontend

A097948 G.f.: -(1-3*x^2-x^3)/(1+4*x-4*x^3-x^4).

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A117808 Primes of the form ((2 + sqrt(3))^(2*k+1) + (2 - sqrt(3))^(2*k+1))/4.

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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.

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MATLAB

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A363445 Turn sequence of a fractal-like curve which is also the perimeter around an aperiodic tiling based on the "hat" monotile. See the comments section for details.

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MATLAB

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A097948 G.f.: -(1-3x^2-x^3)/(1+4x-4*x^3-x^4).

A117808 Primes of the form ((2 + sqrt(3))^(2k+1) + (2 - sqrt(3))^(2k+1))/4.