A033890 -id:A033890 - cp's OEIS frontend

A134501 a(n) = Fibonacci(7n + 3).

2, 55, 1597, 46368, 1346269, 39088169, 1134903170, 32951280099, 956722026041, 27777890035288, 806515533049393, 23416728348467685, 679891637638612258, 19740274219868223167, 573147844013817084101, 16641027750620563662096

Offset: 0

Artur Jasinski, Oct 28 2007

Index entries for linear recurrences with constant coefficients, signature (29,1).

Cf. A000045, A001906, A001519, A033887, A015448, A014445, A033888, A033889, A033890, A033891, A102312, A099100, A134490-A134495, A103134, A134497 - A134504.

[Fibonacci(7*n+3): n in [0..100]]; // Vincenzo Librandi, Apr 16 2011

Mathematica
```
Table[Fibonacci[7n+3], {n, 0, 30}]
```

a(n)=fibonacci(7*n+3) \\ Charles R Greathouse IV, Jun 11 2015

From R. J. Mathar, Jul 04 2011: (Start)
G.f.: (-2+3*x) / (-1 + 29*x + x^2).
a(n) = 2*A049667(n+1) - 3*A049667(n). (End)
a(n) = A000045(A017017(n)). - Michel Marcus, Nov 07 2013

Offset changed to 0 by Vincenzo Librandi, Apr 16 2011

A134502 a(n) = Fibonacci(7n + 4).

Original entry on oeis.org

3, 89, 2584, 75025, 2178309, 63245986, 1836311903, 53316291173, 1548008755920, 44945570212853, 1304969544928657, 37889062373143906, 1100087778366101931, 31940434634990099905, 927372692193078999176, 26925748508234281076009

Offset: 0

Artur Jasinski, Oct 28 2007

Index entries for linear recurrences with constant coefficients, signature (29,1)

Cf. A000045, A001906, A001519, A033887, A015448, A014445, A033888, A033889, A033890, A033891, A102312, A099100, A134490-A134495, A103134, A134497-A134504.

[Fibonacci(7*n +4): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

Mathematica
```
Table[Fibonacci[7n+4], {n, 0, 30}]
```

a(n)=fibonacci(7*n+4) \\ Charles R Greathouse IV, Jun 11 2015

From R. J. Mathar, Jul 04 2011: (Start)
G.f.: (-3-2*x) / (-1 + 29*x + x^2).
a(n) = 2*A049667(n) + 3*A049667(n+1). (End)
a(n) = A000045(A017029(n)). - Michel Marcus, Nov 07 2013

Offset changed from 1 to 0 by Vincenzo Librandi, Apr 17 2011

A180033 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 5x - 5x^2).

Original entry on oeis.org

1, 6, 35, 205, 1200, 7025, 41125, 240750, 1409375, 8250625, 48300000, 282753125, 1655265625, 9690093750, 56726796875, 332084453125, 1944056250000, 11380703515625, 66623798828125, 390022511718750, 2283231552734375

Offset: 0

Johannes W. Meijer, Aug 09 2010

The a(n) represent the number of n-move routes of a fairy chess piece starting in the corner and side squares (m = 1, 3, 7, 9; 2, 4, 6, 8) on a 3 X 3 chessboard. This fairy chess piece behaves like a white queen on the eight side and corner squares but on the central square the queen explodes with fury and turns into a red queen, see A180032.
The sequence above corresponds to 56 red queen vectors, i.e., A[5] vector, with decimal values between 47 and 488. The central squares lead for these vectors to A057088.
Inverse binomial transform of A004187 (without the leading 0).
Equals the INVERT transform of A086347 and the INVERTi transform of A180167. - Gary W. Adamson, Aug 14 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Jean-Paul Allouche, Jeffrey Shallit, and Manon Stipulanti, Combinatorics on words and generating Dirichlet series of automatic sequences, arXiv:2401.13524 [math.CO], 2025. See p. 14.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, Example 7.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Index entries for linear recurrences with constant coefficients, signature (5,5).

Cf. A057088, A180032.

Cf. A086347, A180167. - Gary W. Adamson, Aug 14 2010

I:=[1,6]; [n le 2 select I[n] else 5*Self(n-1)+5*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 15 2011

with(LinearAlgebra): nmax:=20; m:=1; A[5]:= [0,0,0,1,0,1,1,1,1]: A:=Matrix([[0,1,1,1,1,0,1,0,1], [1,0,1,1,1,1,0,1,0], [1,1,0,0,1,1,1,0,1], [1,1,0,0,1,1,1,1,0], A[5], [0,1,1,1,1,0,0,1,1], [1,0,1,1,1,0,0,1,1], [0,1,0,1,1,1,1,0,1], [1,0,1,0,1,1,1,1,0]]): for n from 0 to nmax do B(n):=A^n: a(n):= add(B(n)[m,k],k=1..9): od: seq(a(n), n=0..nmax);

LinearRecurrence[{5,5},{1,6}, 30] (* Vincenzo Librandi, Nov 15 2011 *)

my(x='x+O('x^30)); Vec((1+x)/(1-5*x-5*x^2)) \\ G. C. Greubel, Apr 07 2019

((1+x)/(1-5*x-5*x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Apr 07 2019

G.f.: (1+x)/(1 - 5*x - 5*x^2).
a(n) = 5*a(n-1) + 5*a(n-2) with a(0) = 1 and a(1) = 6.
a(n) = ((7+5*A)*A^(-n-1) + (7+5*B)*B^(-n-1))/45 with A = (-5+3*sqrt(5))/10 and B = (-5-3*sqrt(5))/10.
Limit_{k->oo} a(n+k)/a(k) = 2*5^(n/2)/(L(2*n) - F(2*n)*sqrt(5)) with L(n) = A000032(n) and F(n) = A000045(n).
Limit_{k->oo} a(2*n+k)/a(k) = 2*A000351(n)/(A056854(n) - 3*A004187(n)*sqrt(5)) for n >= 1.
Limit_{k->oo} a(2*n-1+k)/a(k) = 2*A000351(n)/(3*A049685(n-1)*sqrt(5) - 5*A033890(n-1)) for n >= 1.
a(n) = A057088(n+1)/5. a(2*n) = 5^n*F(4*(n+1))/3, a(2*n+1) = 5^n*L(2*(2*n+3))/3. - Ehren Metcalfe, Apr 04 2019
E.g.f.: exp(5*x/2)*(15*cosh(3*sqrt(5)*x/2) + 7*sqrt(5)*sinh(3*sqrt(5)*x/2))/15. - Stefano Spezia, Mar 17 2025

A342709 12-gonal (dodecagonal) square numbers.

Original entry on oeis.org

1, 64, 3025, 142129, 6677056, 313679521, 14736260449, 692290561600, 32522920134769, 1527884955772561, 71778070001175616, 3372041405099481409, 158414167969674450625, 7442093853169599697984, 349619996931001511354641, 16424697761903901433970161

Offset: 1

Bernard Schott, Mar 19 2021

The 12-gonal square numbers k correspond to the nonnegative integer solutions of the Diophantine equation k = d*(5*d-4) = c^2, equivalent to (5*d-2)^2 - 5*c^2 = 4. Substituting x = 5*d-2 and y = c gives the Pell-Fermat's equation x^2 - 5*y^2 = 4.
The solutions x are in A342710, while corresponding solutions y that are also the indices c of the squares which are 12-gonal are in A033890.
The indices d of the corresponding 12-gonal which are squares are in A081068.

			142129 = 169*(5*169-4) = 377^2, so 142129 is the 169th 12-gonal number and the 377th square, hence 142129 is a term.

Index entries for linear recurrences with constant coefficients, signature (48,-48,1).

Intersection of A000290 (squares) and A051624 (12-gonal numbers).

Similar for n-gonal squares: A001110 (triangular), A036353 (pentagonal), A046177 (hexagonal), A036354 (heptagonal), A036428 (octagonal), A036411 (9-gonal), A188896 (there are no 10-gonal squares > 1), A333641 (11-gonal), this sequence (12-gonal).

with(combinat):
seq(fibonacci(4*n-2)^2, n=1..16);

Table[Fibonacci[4*n - 2]^2, {n, 1, 16}] (* Amiram Eldar, Mar 19 2021 *)

a(n) = fibonacci(4*n-2)^2; \\ Michel Marcus, Mar 21 2021

G.f.: x*(1 + 16*x + x^2)/((1 - x)*(1 - 47*x + x^2)). - Stefano Spezia, Mar 20 2021
a(n) = 48*a(n-1) - 48*a(n-2) + a(n-3). - Kevin Ryde, Mar 20 2021
a(n) = 9*A161582(n) + 1. - Hugo Pfoertner, Mar 19 2021
a(n) = A033890(n-1)^2.

A221366 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = (1/2)(7 - 3sqrt(5)).

Original entry on oeis.org

1, 5, 1, 45, 1, 320, 1, 2205, 1, 15125, 1, 103680, 1, 710645, 1, 4870845, 1, 33385280, 1, 228826125, 1, 1568397605, 1, 10749957120, 1, 73681302245, 1, 505019158605, 1, 3461452808000, 1, 23725150497405, 1

Offset: 0

Peter Bala, Jan 15 2013

The function F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) is analytic for |x| < 1. When x is a quadratic irrational of the form x = 1/2*(N - sqrt(N^2 - 4)), N an integer greater than 2, the real number F(x) has a predictable simple continued fraction expansion. The first examples of these expansions, for N = 2, 4, 6 and 8, are due to Hanna. See A174500 through A175503. The present sequence is the case N = 7. See also A221364 (N = 3), A221365 (N = 5) and A221367 (N = 9).

If we denote the present sequence by [1, c(1), 1, c(2), 1, c(3), ...] then for k = 1, 2, ..., the simple continued fraction expansion of F((1/2*(7 - sqrt(45)))^k) is given by the sequence [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...]. Examples are given below.

			F(1/2*(7 - sqrt(45))) = 1.16725 98258 10214 95210 ... = 1 + 1/(5 + 1/(1 + 1/(45 + 1/(1 + 1/(320 + 1/(1 + 1/(2205 + ...))))))).
F((1/2*(7 - sqrt(45)))^2) = 1.02173 93445 69104 86504 ... = 1 + 1/(45 + 1/(1 + 1/(2205 + 1/(1 + 1/(103680 + 1/(1 + 1/(4870845 + ...))))))).
F((1/2*(7 - sqrt(45)))^3) = 1.00311 52648 91110 10148 ... = 1 + 1/(320 + 1/(1 + 1/(103680 + 1/(1 + 1/(33385280 + 1/(1 + 1/(10749957120 + ...))))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,8,0,-8,0,1).

Cf. A004187, A033890, A049682, A081070.

Cf. A174500 (N = 4), A221364 (N = 3), A221365 (N = 5), A221369 (N = 9).

LinearRecurrence[{0,8,0,-8,0,1},{1,5,1,45,1,320},40] (* or *) Riffle[ LinearRecurrence[{8,-8,1},{5,45,320},20],1,{1,-1,2}] (* Harvey P. Dale, Jan 04 2018 *)

a(2*n-1) = (1/2*(7 + sqrt(45)))^n + (1/2*(7 - sqrt(45)))^n - 2 = A081070(n); a(2*n) = 1.
a(4*n-1) = 45*A049682(n) = 45*(A004187(n))^2;
a(4*n+1) = 5*(A033890(n))^2.
a(n) = 8*a(n-2)-8*a(n-4)+a(n-6). G.f.: -(x^4+5*x^3-7*x^2+5*x+1) / ((x-1)*(x+1)*(x^2-3*x+1)*(x^2+3*x+1)). - Colin Barker, Jan 20 2013

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)).

A159951 Fibonacci integral quotients associated with the dividends in A159950 and the divisors in A003481.

Original entry on oeis.org

12, 856800, 139890541190400, 50664770469826998541056000, 40527253814267058837705250384270510080000, 71554565901386985191123530075861409411081105273676595200000

Offset: 1

Enoch Haga, Apr 27 2009

The first example of an integral quotient in the Fibonacci sequence is 12 because 240/20=12. 240 is the product of terms through 8, and 20 the sum. Thereafter, with every other additional pair of terms in the Fibonacci sequence, another integral quotient occurs.

Let m be an even positive integer. Then the sequence defined by b_m(n) = Product_{k = 1..2*n+1} F(m*k) / Sum_{k = 1..2*n+1} F(m*k) appears to be integral. - Peter Bala, Nov 12 2021

			The first two integral quotients occur in the Fibonacci sequence as illustrated by the following: (1*1*2*3*5*8)/(1+1+2+3+5+8) = 240/20 = 12, integral; (1*1*2*3*5*8*13*21*34*55)/(1+1+2+3+5+8+13+21+34+55) = 122522400/143 = 856800, integral.

Cf. A000045, A001519, A001906, A003481, A033890, A159950, A175553.

with(combinat):
seq(mul(fibonacci(k), k = 1..4*n+2)/(fibonacci(4*n+4) - 1), n = 1..10); # Peter Bala, Nov 04 2021

10 'Fibo 20 'R=SUM:S=PRODUCT 30 'T integral every other pair 40 A=1:S=1:print A;:S=S*1 50 B=1:print B;:S=S*B 60 C=A+B:print C;:R=R+C:S=S*C 70 D=B+C:print D;:R=R+D:R=R+2:print R:S=S*D:print S 80 T=S/R:if T=int(S/R) then print T:stop 90 A=C:B=D:R=R-2:goto 60

a(n) = (Product_{k = 1..4*n+2} Fibonacci(k))/(Sum_{k = 1..4*n+2} Fibonacci(k)) = (Product_{k = 1..4*n+2} Fibonacci(k))/(Fibonacci(4*n+4) - 1) = Fibonacci(2*n+1)/Fibonacci(2*n+3) * Product_{k = 1..4*n+1} Fibonacci(k), which shows a(n) is integral. Cf. A175553. - Peter Bala, Nov 11 2021

A161582 The list of the k values in the common solutions to the 2 equations 5k+1=A^2, 9k+1=B^2.

Original entry on oeis.org

0, 7, 336, 15792, 741895, 34853280, 1637362272, 76921173511, 3613657792752, 169764995085840, 7975341111241735, 374671267233275712, 17601574218852716736, 826899317018844410887, 38846666325666834594960, 1824966417989322381552240, 85734574979172485098360327

Offset: 1

Paul Weisenhorn, Jun 14 2009

The 2 equations are equivalent to the Pell equation x^2-45*y^2=1, with x=(45*k+7)/2 and y= A*B/2, case C=5 in A160682.

Index entries for linear recurrences with constant coefficients, signature (48,-48,1).

Cf. A160682, A049685 (sequence of A), A033890 (sequence of B).

t:=0: for n from 0 to 1000000 do a:=sqrt(5*n+1); b:=sqrt(9*n+1);
if (trunc(a)=a) and (trunc(b)=b) then t:=t+1; print(t,n,a,b): end if: end do:

LinearRecurrence[{48,-48,1},{0,7,336},30] (* or *) Rest[CoefficientList[ Series[ -7x^2/((x-1)(x^2-47x+1)),{x,0,30}],x]] (* Harvey P. Dale, Mar 21 2013 *)

k(t+3) = 48*(k(t+2)-k(t+1))+k(t).
With w = sqrt(5),
k(t) = ((7+3*w)*((47+21*w)/2)^(t-1)+(7-3*w)*((47-21*w)/2)^(t-1))/90.
k(t) = floor((7+3*w)*((47+21*w)/2)^(t-1)/90) = 7*|A156093(t-1)|.
G.f.: -7*x^2/((x-1)*(x^2-47*x+1)).
a(1)=0, a(2)=7, a(3)=336, a(n) = 48*a(n-1)-48*a(n-2)+a(n-3). - Harvey P. Dale, Mar 21 2013

Edited, extended by R. J. Mathar, Sep 02 2009

A342710 Solutions x to the Pell-Fermat equation x^2 - 5*y^2 = 4.

Original entry on oeis.org

3, 18, 123, 843, 5778, 39603, 271443, 1860498, 12752043, 87403803, 599074578, 4106118243, 28143753123, 192900153618, 1322157322203, 9062201101803, 62113250390418, 425730551631123, 2918000611027443, 20000273725560978, 137083915467899403, 939587134549734843

Offset: 0

Bernard Schott, Mar 19 2021

This Pell equation is used to find the 12-gonal square numbers (see A342709).
The corresponding solutions y are in A033890.
Essentially the same as A246453. - R. J. Mathar, Mar 24 2021

			a(1)^2 - 5 * A033890(1)^2 = 18^2 - 5 * 8^2 = 4.

Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A033890, A342709.

a(n) = 3*A049685(n). - Hugo Pfoertner, Mar 19 2021

LinearRecurrence[{7, -1}, {3, 18}, 20] (* Amiram Eldar, Mar 19 2021 *)
Table[2 ChebyshevT[2 n + 1, 3/2], {n, 0, 20}] (* Eric W. Weisstein, Sep 02 2025 *)
Table[2 Cos[(2 n + 1) ArcCos[3/2]], {n, 0, 20}] // FunctionExpand (* Eric W. Weisstein, Sep 02 2025 *)

a(n) = 7*a(n-1) - a(n-2).
a(n) = 2*T(2*n+1, 3/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Jul 02 2022
From Stefano Spezia, Apr 14 2025: (Start)
G.f.: 3*(1 - x)/(1 - 7*x + x^2).
E.g.f.: exp(7*x/2)*(3*cosh(3*sqrt(5)*x/2) + sqrt(5)*sinh(3*sqrt(5)*x/2)). (End)
a(n) = 2*cos((2*n+1)*arccos(3/2)). - Eric W. Weisstein, Sep 02 2025

cp's OEIS Frontend

A134501 a(n) = Fibonacci(7n + 3).

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A134502 a(n) = Fibonacci(7n + 4).

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A180033 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 5*x - 5*x^2).

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A342709 12-gonal (dodecagonal) square numbers.

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A221366 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) when x = (1/2)*(7 - 3*sqrt(5)).

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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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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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A180033 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 5x - 5x^2).

A221366 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = (1/2)(7 - 3sqrt(5)).

A161582 The list of the k values in the common solutions to the 2 equations 5k+1=A^2, 9k+1=B^2.