A085005 -id:A085005 - cp's OEIS frontend

A085002 a(n) = floor(phin) - 2floor(phi*n/2) where phi is the golden ratio.

1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1

Offset: 1

Benoit Cloitre, Jun 17 2003

The lower Wythoff sequence (A000201) mod 2 (see formula section). - Michel Dekking, Feb 01 2021
A fractal sequence.
From Michel Dekking, Apr 24 2018: (Start)
Usually an integer sequence is called 'fractal' if it has the self-generating properties of a morphic sequence, i.e., the letter to letter projection of a fixed point of a morphism. Indeed, take the alphabet {1,2,...,8} and the morphism eta defined by
    eta:  1->5, 2->7, 3->8, 4->6, 5->53, 6->71, 7->82, 8->64.
Then eta has fixed point
    x = (5,3,8,6,4,7,1,6,8,2,5,71,6,4,7,5,3,...).
Let pi be the projection morphism
  pi(1)=1, pi(2)=0, pi(3)=1, pi(4)=0, pi(5)=1, pi(6)=0, pi(7)=1, pi(8)=0.
Then pi(x) = (a(n)).
To prove this, one may use my paper "Iteration of maps by an automaton".
The two maps are phi_a and phi_b defined by
    phi_a(0) = 1, phi_a(1) = 0, phi_b(0) = 0, phi_b(1) = 1.
The substitution is the Fibonacci substitution sigma given by
    sigma(a) = b, sigma(b) = ba.
Since the first differences of the lower Wythoff sequence are given by the Fibonacci substitution 1->2, 2-> 21, it follows from replacing 1 with a and 2 with b that (a(n)) is generated by iterating the two maps phi_a and phi_b according to the fixed point babba...of sigma. The two maps phi_a and phi_b are commuting bijections on {a,b}, exactly as in the Example on page 85 of "Iteration of maps by an automaton". It follows as in that example that (a(n)) is generated by the projection of a fixed point of a substitution on an alphabet of 8 letters, and a simple computation as on that page yields the morphism eta. (End)

Antti Karttunen, Table of n, a(n) for n = 1..10946
B. Cloitre, Graph of A085005(n) for n=1 up to 3874 [archive.org link]
Benoit Cloitre, Fractal walk starting at (0,0) with step of unit length turning 90° right if a(n)=0 left otherwise for n=1 up to 10^6
Michel Dekking, Iteration of maps by an automaton, Discrete Mathematics 126 (1994), 81-86.
M. Schaefer, E. Sedgwick, and D. Štefankovič, Spiraling and folding: the word view, Algorithmica 60 (2011), 609-626. See section 4.

Characteristic function of A283766.
Cf. A000035, A001622, A083035, A083036, A083037, A083038, A085003 (partial sums), A085004, A085005, A105774.
See also A171587.

Table[Floor[GoldenRatio n] - 2 Floor[GoldenRatio n/2], {n, 110}] (* Harvey P. Dale, Dec 11 2012 *)

a(n)=(n+sqrtint(5*n^2))%4>1 \\ Charles R Greathouse IV, Feb 07 2013

from math import isqrt
def A085002(n): return ((n+isqrt(5*n**2))&2)>>1 # Chai Wah Wu, Aug 10 2022

(define (A085002 n) (A000035 (A105774 n))) ;; Antti Karttunen, Mar 17 2017

a(n) = A105774(n) mod 2 = A000201(n) mod 2. - Benoit Cloitre, May 10 2005
From Michel Dekking, Apr 24 2018: (Start)
Proof that this sequence is the parity sequence of the lower Wythoff sequence:
if n*phi/2 = M + e, with 0 < e < 1, then 2*floor(phi*n/2) = 2M, and
floor(phi*n) = floor(2M+2e) = 2M or 2M+1.
So floor(phi*n) - 2*floor(phi*n/2) = 0 if floor(phi*n) is even, and equals 1 if floor(phi*n) is odd. (End)

A085003 Partial sums of A085002.

Original entry on oeis.org

1, 2, 2, 2, 2, 3, 4, 4, 4, 4, 5, 6, 7, 7, 7, 8, 9, 10, 10, 10, 11, 12, 13, 13, 13, 13, 14, 15, 15, 15, 15, 16, 17, 18, 18, 18, 19, 20, 21, 21, 21, 22, 23, 24, 24, 24, 24, 25, 26, 26, 26, 26, 27, 28, 28, 28, 28, 29, 30, 31, 31, 31, 32, 33, 34, 34, 34, 34, 35, 36, 36, 36, 36, 37, 38

Offset: 1

Benoit Cloitre, Jun 17 2003

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10946
B. Cloitre, Graph of A085005(n) for n=1 up to 3874 [archive.org link]

Partial sums of A085002. A left inverse of A283766.

Cf. A083035, A083036, A083037, A083038, A085004, A085005.

Accumulate[Table[Floor[GoldenRatio*n]-2*Floor[GoldenRatio*n/2],{n,110}]] (* Harvey P. Dale, Dec 11 2012 *)

sum(k=1,n,(k+sqrtint(5*k^2))%4>1) \\ Charles R Greathouse IV, Feb 07 2013
(Scheme, with memoization-macro definec)
(definec (A085003 n) (if (= 1 n) n (+ (A085002 n) (A085003 (- n 1)))))
;; Antti Karttunen, Mar 17 2017

a(n)=sum(k=1, n, A085002(k)).

a(A283766(n)) = n for all n >= 1. - Antti Karttunen, Mar 17 2017

A085004 a(n)=2*A085003(n)-n.

Original entry on oeis.org

1, 2, 1, 0, -1, 0, 1, 0, -1, -2, -1, 0, 1, 0, -1, 0, 1, 2, 1, 0, 1, 2, 3, 2, 1, 0, 1, 2, 1, 0, -1, 0, 1, 2, 1, 0, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 2, 1, 0, 1, 2, 1, 0, -1, 0, 1, 2, 1, 0, 1, 2, 3, 2, 1, 0, 1, 2, 1, 0, -1, 0, 1, 0, -1, -2, -1, 0, 1, 0, -1, 0, 1, 2, 1, 0, -1, 0, 1, 0, -1, -2, -1, 0, -1, -2, -3, -2, -1, 0, -1, -2, -1, 0, 1, 0, -1, 0, 1, 2, 1, 0, -1

Offset: 1

Benoit Cloitre, Jun 17 2003

sign

Cf. A083035, A083036, A083037, A083038, A085003, A085005.

2*sum(k=1, n, (k+sqrtint(5*k^2))%4>1)-n \\ Charles R Greathouse IV, Feb 07 2013

|a(n+1)-a(n)| = 1. - Charles R Greathouse IV, Feb 07 2013

A094200 a(n) = 16n^4 + 32n^3 + 36n^2 + 20n + 3.

Original entry on oeis.org

3, 107, 699, 2547, 6803, 15003, 29067, 51299, 84387, 131403, 195803, 281427, 392499, 533627, 709803, 926403, 1189187, 1504299, 1878267, 2318003, 2830803, 3424347, 4106699, 4886307, 5772003, 6773003, 7898907, 9159699, 10565747

Offset: 0

Benoit Cloitre, May 25 2004

Let x(n) = (1/2)*(-(2*n + 1) + sqrt((2*n + 1)^2 + 4)) and f(n,k) = (-1)*Sum_{i=1..k} Sum_{j=1..i} (-1)^floor(j*x(n)). Then a(n) = k is the least integer k > 0 such that f(n, k) = 0. In other words, f(n, a(n)) = 0, and if f(n,k) = 0, then a(n) <= k. [Edited by Petros Hadjicostas, Jul 12 2020]

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A085005, A094201.

Table[16n^4+32n^3+36n^2+20n+3,{n,0,30}] (* or *) LinearRecurrence[ {5,-10,10,-5,1},{3,107,699,2547,6803},30] (* Harvey P. Dale, Jul 23 2013 *)

PARI
```
a(n) = 16*n^4+32*n^3+36*n^2+20*n+3
```

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n >= 5. - Harvey P. Dale, Jul 23 2013

A094201 a(n) = 4n^5 + 10n^4 + 13n^3 + 11n^2 + 5*n + 1.

Original entry on oeis.org

1, 44, 447, 2248, 7685, 20676, 47299, 96272, 179433, 312220, 514151, 809304, 1226797, 1801268, 2573355, 3590176, 4905809, 6581772, 8687503, 11300840, 14508501, 18406564, 23100947, 28707888, 35354425, 43178876, 52331319

Offset: 0

Benoit Cloitre, May 25 2004

Let x(n) = (1/2)*(-(2*n+1) + sqrt((2*n+1)^2 + 4)) and f(k) = (-1)*(Sum_{i=1..k} Sum_{j=1..i} (-1)^floor(j*x(n))), then a(n) = Max{f(k): 0 < k < A094200(n)}.

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A085005, A094200.

LinearRecurrence[{6,-15,20,-15,6,-1},{1,44,447,2248,7685,20676},30] (* Harvey P. Dale, Oct 23 2021 *)

PARI
```
a(n)=4*n^5+10*n^4+13*n^3+11*n^2+5*n+1
```

G.f.: (37*x^4 + 206*x^3 + 198*x^2 + 38*x + 1)/(x - 1)^6. - Jinyuan Wang, Apr 06 2020

Corrected by T. D. Noe, Nov 09 2006

cp's OEIS Frontend

A085002 a(n) = floor(phi*n) - 2*floor(phi*n/2) where phi is the golden ratio.

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A085003 Partial sums of A085002.

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A085004 a(n)=2*A085003(n)-n.

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A094200 a(n) = 16*n^4 + 32*n^3 + 36*n^2 + 20*n + 3.

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A094201 a(n) = 4*n^5 + 10*n^4 + 13*n^3 + 11*n^2 + 5*n + 1.

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A085002 a(n) = floor(phin) - 2floor(phi*n/2) where phi is the golden ratio.

A094200 a(n) = 16n^4 + 32n^3 + 36n^2 + 20n + 3.

A094201 a(n) = 4n^5 + 10n^4 + 13n^3 + 11n^2 + 5*n + 1.