Displaying 1-10 of 12 results found.
Wythoff AB-numbers: floor(floor(n*phi^2)*phi), where phi = (1+sqrt(5))/2.
(Formerly M2715)
+10
25
3, 8, 11, 16, 21, 24, 29, 32, 37, 42, 45, 50, 55, 58, 63, 66, 71, 76, 79, 84, 87, 92, 97, 100, 105, 110, 113, 118, 121, 126, 131, 134, 139, 144, 147, 152, 155, 160, 165, 168, 173, 176, 181, 186, 189, 194, 199, 202, 207, 210, 215, 220, 223, 228, 231, 236, 241, 244, 249
COMMENTS
Previous name was: "From a 3-way splitting of positive integers: [[n*phi^2]*phi]."
A005614(a(n)-1)=1 and A005614(a(n))=1, n>=1. Because Wythoff AB-numbers (see the formula section) mark the first entry of pairs of 1s in the rabbit sequence A005614(n-1), n>=1. - Wolfdieter Lang, Jun 28 2011
REFERENCES
J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 10.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Aviezri S. Fraenkel, The Raleigh game, INTEGERS: Electronic Journal of Combinatorial Number Theory 7.2 (2007): A13, 10 pages. See Table 1.
FORMULA
a(n) = 2*floor(n*phi) + n = 2* A000201(n) + n. a(n) = A(B(n)) with A(k):= A000201(k) and B(k):= A001950(k), k>=1 (Wythoff AB-numbers).
MATHEMATICA
f[n_] := Floor[ GoldenRatio * Floor[ n * GoldenRatio^2]]; Array[f, 47]
(* another *) Table[n+2Floor[n*GoldenRatio], {n, 1, 100}]
PROG
(Python)
from sympy import floor
from mpmath import phi
def a(n): return floor(n*phi) + floor(n*phi**2) # Indranil Ghosh, Jun 10 2017 (Python)
from math import isqrt
Characteristic sequence for Wythoff AB-numbers A003623.
+10
9
0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1
COMMENTS
Left shifted sequence is the characteristic function of A035336, and also the second lowest digit of the Zeckendorf expansion of n. - Franklin T. Adams-Watters, Jun 30 2009 This is a morphic sequence, i.e., the letter-to-letter image of the fixed point of a morphism. The fixed point is the unique fixed point A270788 of the three symbol Fibonacci morphism. The letter-to-letter map is 1->0, 2->0, 3->1. - Michel Dekking, May 02 2019
FORMULA
a(n) = 1 if n=A(B(k)) for some k>=1, else 0, with A(k):= A000201(k) and B(k):= A001950(k), k>=1. a(n) = 1-(1-h(n))-(1-h(n+1)) = h(n)-(1-h(n+1))= h(n)*h(n+1) with h(n):= A005614(n-1), n>=1, the rabbit sequence.
MATHEMATICA
a[_] = 0; s = Table[n + 2 Floor[n*GoldenRatio], {n, 24}]; Map[Set[a[#], 1] &, s]; Array[a, Max[s]] (* Michael De Vlieger, Mar 29 2023 *)
PROG
(Python)
from math import isqrt
def A123740(n): return (n+2+isqrt(m:=5*(n+2)**2)>>1)-(n+isqrt(m-20*(n+1))>>1)-3 # Chai Wah Wu, Aug 29 2022
Indices n at which blocks (1;1) occur in infinite Fibonacci word, i.e., such that A005614(n-1) = A005614(n-2) = 1.
+10
6
4, 9, 12, 17, 22, 25, 30, 33, 38, 43, 46, 51, 56, 59, 64, 67, 72, 77, 80, 85, 88, 93, 98, 101, 106, 111, 114, 119, 122, 127, 132, 135, 140, 145, 148, 153, 156, 161, 166, 169, 174, 177, 182, 187, 190, 195, 200, 203, 208, 211, 216, 221, 224, 229, 232, 237, 242, 245
COMMENTS
a(n) is the number k such that floor(k/r) = floor(n*r^2), where r = golden ratio. - Clark Kimberling, May 03 2015
FORMULA
a(n) = floor((2+sqrt(5))*n) + 0 or 1;
floor(n*(2+sqrt(5))) + b(a(n)) - a(n) = 0 where b(x) = A078588(x) = x + 1 + ceiling(x*sqrt(5)) - 2*ceiling(x*(1+sqrt(5))/2). a(n) = A003623(n) + 1 = A(B(n)) + 1, where A(B(n)) are the Wythoff AB-numbers. - Michel Dekking, Sep 15 2016
MAPLE
phi:=(1+sqrt(5))/2: seq(floor(phi*floor(n*phi^2))+1, n=1..80); # Michel Dekking, Sep 15 2016
MATHEMATICA
r = GoldenRatio; u = Flatten[Table[Select[Range[Floor[(r^2 + r) n], Floor[(r^2 + r) n + 1]], Floor[#/r] == Floor[n*r^2] &], {n, 1, 100}]] (* Clark Kimberling, May 03 2015 *)
PROG
(Python)
from math import isqrt
1, 2, 3, 1, 2, 3, 1, 1, 2, 3, 1, 2, 3, 1, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 1, 2, 3, 1, 2, 3, 1, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 1, 2, 3, 1, 2, 3, 1, 1, 2, 3, 1, 2, 3, 1, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 1, 2, 3, 1, 2, 3, 1, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 1, 2, 3, 1, 2, 3, 1, 1, 2, 3, 1, 2, 3, 1, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 1, 2
COMMENTS
In the Pell word A171588 = 0, 0, 1, 0, 0, 1, 0, 0, 0, ..., group the letters in overlapping blocks of length two: [0,0],[0,1],[1,0],[0,0],[0,1],[1,0],... Then code [0,0]->1, [0,1]->2, [1,0]->3. This gives (a(n)). (a(n)) is the unique fixed point of the 3-symbol Pell morphism
1 -> 123, 2 ->123, 3 -> 1.
The morphism and the fixed point are in standard form.
Modulo a change of alphabet (1->0, 2->1, 3->2), this sequence is equal to A263844. The positions of 1 in (a(n)) are given by
Why is this true? First, the Pell word b is given by
b(n) = [(n+1)(1-r)]-[n(1-r)], where r =1/sqrt(2).
This can rewritten as
b(n) = [nr]-[(n+1)r]+1.
Second,
1 occurs at n in (a(n)) <=>
00 occurs at n in (b(n)) <=>
b(n)+b(n+1) = 0 <=>
[nr]-[(n+2)r]+2 = 0 <=>
[(n+2)r]-[nr]-1 = 1 <=>
The positions of 2 in (a(n)) are given by A001952 - 1 = 2,5,9,12,16,..., since 2 occurs at n in (a(n))) if and only if 3 occurs at n+1 in (a(n)). The positions of 3 in (a(n)) are given by A001952 = 3,6,10,13,17,..., since 3 occurs at n in (a(n)) if and only if 1 occurs at n in (b(n)). The sequence of positions of 3 in (a(n)) is equal to the sequence b in Carlitz et al. The sequence of positions of 1 in (a(n)) seems to be equal to the sequence ad' in Carlitz et al. (End)
See the comments of A188376 for a proof of the observation on the positions of 1 in (a(n)). - Michel Dekking, Feb 27 2018
LINKS
L. Carlitz, R. Scoville and V. E. Hoggatt, Jr., Pellian Representations, Fib. Quart. Vol. 10, No. 5, (1972), pp. 449-488.
FORMULA
a(n) = floor((n+2)r)+floor((n+1)r)-2*floor(nr)+1, where r = 1-1/sqrt(2).
MAPLE
a:=[seq(floor((n+2)*(1-1/sqrt(2)))+floor((n+1)*(1-1/sqrt(2)))-2*floor(n*(1-1/sqrt(2)))+1, n=1..130)];
MATHEMATICA
With[{r = 1 - 1/Sqrt[2]}, Table[Inner[Times, Map[Floor[(n + #) r] &, Range[0, 2]], {-2, 1, 1}, Plus] + 1, {n, 108}]] (* Michael De Vlieger, Feb 15 2018 *)
PROG
(Magma) [Floor((n+2)*r)+Floor((n+1)*r)-2*Floor(n*r)+1 where r is 1-1/Sqrt(2): n in [1..90]]; // Vincenzo Librandi, Feb 23 2018
Define strings S(0)=0, S(1)=11, S(n) = S(n-1)S(n-2); iterate.
+10
5
1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1
COMMENTS
a(n) = 1 - A123740(n). This can be seen as follows. Define words T(0)=0, T(1)=1, T(n) = T(n-1)T(n-2). Then T(infinity) is the binary complement of the infinite Fibonacci word A003849. Obviously S(n) is the [1->11] transform of T(n). The claim now follows from the observation (see Comments of A123740) that doubling the 0's in the infinite Fibonacci word A003849 gives A123740. - Michel Dekking, Oct 21 2018 Here is a proof of Cloitre's (corrected) formula
Since abs(-1) = abs(1) = 1, one has to prove that A014677(k)=0 if and only if there is an n such that AB(n) = k (using that a(n) = 1 - A123740(n)). Now A014677 is the sequences of first differences of A001468, and the 0's in A014677 occur if and only if there occurs a block 22 in A001468, which is given by A001468(n) = floor((n+1)*phi) - floor(n*phi), n >= 0.
But then
The sequence A014675 is fixed point of the morphism 1->2, 2->21, which is alphabet equivalent to the morphism 1->12, 2->1, the classical Fibonacci morphism in standard form. This implies that the 22 blocks in A001468 occur at position n+1 in if and only if 3 occurs in the fixed point A270788 of the 3-symbol Fibonacci morphism at k, which happens if and only if there is an n such that AB(n)=k (see Formula of A270788). (End)
FORMULA
For n > 1, a(n-1) = floor(phi*ceiling(n/phi)) - ceiling(phi*floor(n/phi)) where phi=(1+sqrt(5))/2.
For n >= 0, a(n) = abs( A014677(n+1)). (End)
EXAMPLE
The infinite word is S(infinity) = 110111101101111011110110...
MATHEMATICA
s[0] = {0}; s[1] = {1, 1}; s[n_] := s[n] = Join[s[n-1], s[n-2]]; s[10] (* Jean-François Alcover, May 15 2013 *) nxt[{a_, b_}]:={b, Join[a, b]}; Drop[Nest[nxt, {{0}, {1, 1}}, 10][[1]], 3] (* Harvey P. Dale, Jan 31 2019 *)
PROG
(PARI) a(n, f1, f2)=local(f3); for(i=3, n, f3=concat(f2, f1); f1=f2; f2=f3); f2
(PARI) printp(a(10, [ 0 ], [ 1, 1 ])) \\ Would give S(10). Sequence is S(infinity).
(Haskell)
a005713 n = a005713_list !! n
a005713_list = 1 : 1 : concat (sibb [0] [1, 1]) where
sibb xs ys = zs : sibb ys zs where zs = xs ++ ys
a(n) = g(n+1) - g(n) where g(k) = floor(phi*floor(k/phi)) and phi = (1+sqrt(5))/2.
+10
3
1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 0, 2, 1, 0, 2, 1, 0, 2, 0
COMMENTS
Here is a proof that (a(n)) is fixed point of the morphism 0->102, 1->102, 2->02.
Let alpha:=phi-1. Then alpha*phi = 1. So
g(k) = floor(phi*floor(k*alpha)).
Write k*alpha = floor(k*alpha) + {k*alpha}, i.e., {k*alpha} is the fractional part of k*alpha. Then
g(k) = floor(phi*(k*alpha-{k*alpha})) = k + floor(-phi*{k*alpha}).
Thus
a(n) = n+1 +floor(-phi*{(n+1)*alpha})-n -floor(-phi*{n*alpha}).
It follows that
a(n) = 1 - floor(phi*{(n+1)*alpha}) + floor(phi*{n*alpha}).
The difference -floor(phi*{(n+1)*alpha}) + floor(phi*{n*alpha}) is equal to -1, 0 or 1, since floor(phi*{n*alpha}) is equal to 0 or 1.
In fact, phi*{n*alpha} can only take values between 0 and 1.619, and floor(phi*{n*alpha}) = 0 if and only if
{n*alpha} < 1/phi = alpha.
This is the same (putting rho:=1-alpha) as requiring
{n*alpha+rho} < 1-alpha.
Via the rotation description of Sturmian sequences (see, e.g., Lothaire), one sees that this sequence is the inhomogeneous Sturmian sequence s(alpha, rho), but with offset 1, and with 0 and 1 exchanged. Since rho+alpha=1, it follows that s(alpha, rho) with offset 2 equals s(1-alpha, 1-alpha), the classical Fibonacci sequence xF:= A003849, fixed point of 0->01, 1->0. We have found that a(n+1)=0 iff xF(n)=0, xF(n+1)=1,
a(n+1)=1 iff xF(n)=0, xF(n+1)=0,
a(n+1)=2 iff xF(n)=1, xF(n+1)=0.
This means that (a(n+1)) equals the 3-symbol Fibonacci sequence A270788 on the alphabet {0,2,1}. Then Proposition 5 in "Morphisms, Symbolic Sequences, and Their Standard Forms" yields that (a(n)) is fixed point of the morphism 0->102, 1->102, 2->02. (End)
FORMULA
a(floor(k*phi)+k+1)=0; a(floor(k*phi)+k+2)=2, if n is not in {floor(k*phi)+k+1}U{floor(k*phi)+k+2}_{k>=1} a(n)=1.
(a(n)) is a fixed point of the morphism 02-->10202 and 102-->10210202. [Corrected by Michel Dekking, Oct 29 2018] Fixed point of the morphism 0->102, 1->102, 2->02. - Michel Dekking, Oct 21 2018
MAPLE
g:=k->floor((1+sqrt(5))/2*floor(k/((1+sqrt(5))/2))): seq(g(n+1)-g(n), n=1..110); # Muniru A Asiru, Oct 21 2018
MATHEMATICA
#[[2]]-#[[1]]&/@Partition[Table[Floor[GoldenRatio*Floor[n/GoldenRatio]], {n, 0, 110}], 2, 1] (* Harvey P. Dale, Dec 14 2012 *)
PROG
(PARI) {phi=(1+sqrt(5))/2; g(k)=floor(phi*floor(k/phi))};
(Magma) [Floor((1+Sqrt(5))*Floor(2*(k+1)/(1+Sqrt(5)))/2) -
Floor((1+Sqrt(5))*Floor(2*k/(1+Sqrt(5)))/2): k in [1..100]]; // G. C. Greubel, Oct 23 2018
Constant term in expansion of n in Fraenkel's exotic ternary representation.
+10
3
0, 1, 2, 0, 1, 2, 0, 0, 1, 2, 0, 1, 2, 0, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 0, 1, 2, 0, 1, 2, 0, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 0, 1, 2, 0, 1, 2, 0, 0, 1, 2, 0, 1, 2, 0, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 0, 1, 2, 0, 1, 2, 0, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 0, 1, 2, 0, 1, 2, 0, 0, 1, 2, 0, 1, 2, 0, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 0, 1, 2
COMMENTS
Let {p_i, i >= 0} = {1,3,7,17,41,99,...} denote the numerators of successive convergents to sqrt(2) (see A001333). Then any n >= 0 has a unique representation as n = Sum_{i >= 0} d_i*p_i, with 0 <= d_i <= 2, d_{i+1}=2 => d_i=0. Sequence gives a(n+1) = d_0. (a(n)) is the unique fixed point of the morphism alpha given by
alpha: 0 -> 012, 1 -> 012, 2 -> 0.
To see this, note first that the p_i satisfy p_{i+2}=2p_{i+1}+p_i for all i=1,2,... Then define a sequence of words by
w(0) = 0, w(1) = 012, w(i+2) = w(i+1)w(i+1)w(i).
The length of w(i) is equal to p_i. In the numeration system, the representation of n = p_i is d = 10..0, and the representation of n = 2p_i is d = 20..0. By unicity of the representation, the numbers n = p_i +m have the representation d = 1c, where c is the representation of m for m = 1,...,p_{i-1}. Similarly, because the digit 2 is required to be followed by the digit 0, the numbers n = 2p_i + m have the representation d =20c, where c is the representation of m for m = 1,...,p_{i-2}. It follows from this that the d_0 digits in the range 0 to p_{i+2} have to satisfy the equation w(i+2) = w(i+1)w(i+1)w(i). But alpha(0)=alpha(1)=w(1), and alpha(2)=w(0), which implies by induction that w(i) = alpha^i(0):
w(i+1) = w(i)w(i)w(i-1) = alpha^i(0) alpha^i(0) alpha^{i-1}(0) =
alpha^i(0) alpha^i(1)alpha^i(2) = alpha^i(012) = alpha^{i+1}(0).
(a(n)) is modulo a change of alphabet (0->1, 1->2, 2->3) equal to A294180, the standard form of (a(n)). This combined with the fact that the Pell word A171588 is a Sturmian word leads to the formula for (a(n)) below. (End)
FORMULA
a(n) = floor((n+2)r) + floor((n+1)r) - 2*floor(nr), where r = 1 - 1/sqrt(2). - Michel Dekking, Feb 11 2018
EXAMPLE
See the link to Table 2 of Fraenkel (2000).
MATHEMATICA
Table[Floor[(n + 2) (1 - 1/Sqrt[2])] + Floor[(n + 1) (1 - 1/Sqrt[2])] - 2 Floor[n (1 - 1/Sqrt[2])], {n, 100}] (* Vincenzo Librandi, Feb 12 2018 *)
PROG
(Magma) [Floor((n+2)*(1-1/Sqrt(2)))+Floor((n+1)*(1-1/Sqrt(2)))- 2*Floor(n*(1-1/Sqrt(2))): n in [1..100]]; // Vincenzo Librandi, Feb 12 2018
1, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, 0, -1, 1, -1, 1, 0, -1, 1, 0, -1, 1, -1, 1
COMMENTS
c(n):=a(n-1), n >= 1, is -1 if n is a Wythoff B-number from A001950, it is 0 if n=A(B(m)+1) for some m >= 1, with A(k):= A000201(k) (Wythoff A-numbers) and it is +1 if n=A(A(m)+1)=B(m)+1 for some m >= 0, with B(0):=0. - Wolfdieter Lang, Oct 13 2006 This sequence is a symbolic sequence as discussed in the paper "Morphisms, Symbolic Sequences, and Their Standard Forms". It can be derived directly from the 2-block morphism induced by the morphism generating A001468. Since A001468 is the Fibonacci word A003849, but on the alphabet {2,1}, with an extra 1 in front, this 2-block morphism has 3-symbol Fibonacci as a fixed point: A270788. The 2-blocks in A001468 are 12, 21, and 22, yielding the differences a(n) = 1, a(n) = -1, and a(n) = 0. In 3-symbol Fibonacci these correspond to the letters 2, 1, and 3. Expressing this coding with pi given by pi(1)=-1, pi(2)=1, pi(3)=0, we obtain the formula below. Wolfdieter Lang's Wythoff description of (a(n)) follows from the corresponding Wythoff description in A270788. - Michel Dekking, Dec 30 2019
FORMULA
abs(a(n)) = floor(f*ceiling(n/f)) - ceiling(f*floor(n/f)) where f=phi=(1+sqrt(5))/2; for n > 1, abs(a(n)) = A005713(n-1). - Benoit Cloitre, Apr 21 2003 G.f. equals the continued fraction: A(x) = [0;1, 1/x, 1/x, 1/x^2, 1/x^3, 1/x^5, 1/x^8, ..., 1/x^Fibonacci(n), ...]. - Paul D. Hanna, Dec 17 2004 a(n) = b(n) - b(n-1) with b(n):= A005614(n), n >= 1. a(n) = pi( A270788(n)), n >= 1, where pi is the letter-to-letter map pi(1)=-1, pi(2)=1, pi(3)=0. - Michel Dekking, Dec 30 2019
PROG
(Python)
from math import isqrt
def A014677(n): return (n+isqrt(m:=5*(n+2)**2)>>1)-(n+1+isqrt(m-10*n-15)&-2)+(n+isqrt(m-20*n-20)>>1)+1 # Chai Wah Wu, Aug 25 2022
Infinite Fibonacci word on the alphabet {1,2,3,4}.
+10
2
1, 2, 3, 1, 4, 1, 2, 3, 1, 2, 3, 1, 4, 1, 2, 3, 1, 4, 1, 2, 3, 1, 2, 3, 1, 4, 1, 2, 3, 1, 2, 3, 1, 4, 1, 2, 3, 1, 4, 1, 2, 3, 1, 2, 3, 1, 4, 1, 2, 3, 1, 4, 1, 2, 3, 1, 2, 3, 1, 4, 1, 2, 3, 1, 2, 3, 1, 4, 1, 2, 3, 1, 4, 1, 2, 3, 1, 2, 3, 1, 4, 1, 2, 3, 1, 2, 3, 1, 4, 1, 2, 3, 1, 4, 1, 2, 3, 1, 2, 3, 1, 4, 1, 2, 3
COMMENTS
Start with the infinite Fibonacci word A003849, which is 0100101001001010010... and replace each 0 by 1,2,3 and each 1 by 1,4. (a(n)) is the unique fixed point of the morphism 1->12, 2->3, 3->14, 4->3, obtained by coding the overlapping 3-block morphism of the Fibonacci morphism according to 010<->1, 100<->2, 001<->3, 101<->4. - Michel Dekking, Sep 28 2017
FORMULA
a(n) = 3 for n = 3, 8, 21, 55, ..., F(2*k), where k>1.
a(n) = 4 for n = 5, 13, 34, 89, ..., F(2*k+1), where k>1.
Let A(n)=floor(n*tau), B(n)=n+floor(n*tau); i.e., A and B are the lower and upper Wythoff sequences, A= A000201, B= A001950. a(n)=1 if n=A(A(k)) for some k; a(n)=2 if n=B(A(k)) for some k; a(n)=3 if n=A(B(k)) for some k; a(n)=4 if n=B(B(k)) for some k.
Infinite Fibonacci word on the alphabet {1,2,3,4,5}.
+10
2
1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 1, 2
COMMENTS
Start with the infinite Fibonacci word A003849, which is 0100101001001010010... and replace each 0 by 1,2,3 and each 1 by 4,5. The unique fixed point of the 4-block Fibonacci substitution 1 -> 12, 2 -> 3, 3 -> 45, 4 -> 12, 5 -> 3. Here the 4-blocks are coded as 0100 <-> 1, 1001 <-> 2, 0010 <-> 3, 0101 <-> 4, 1010 <-> 5.
FORMULA
Let A(n) = floor(n*phi), B(n) = n + floor(n*phi), i.e., A and B are the lower and upper Wythoff sequences, A = A000201, B = A001950. Then a(n) = 1 if n = A(A(A(k))) for some k; a(n) = 2 if n = B(A(k)) for some k; a(n) = 3 if n = A(B(k)) for some k; a(n) = 4 if n = A(A(B(k))) for some k; a(n) = 5 if n = B(B(k)) for some k.
Search completed in 0.015 seconds
| 
