A162330 -id:A162330 - cp's OEIS frontend

A285869 a(n) is the number of zeros of the Chebyshev S(n, x) polynomial in the open interval (-sqrt(2), +sqrt(2)).

0, 1, 2, 1, 2, 3, 4, 3, 4, 5, 6, 5, 6, 7, 8, 7, 8, 9, 10, 9, 10, 11, 12, 11, 12, 13, 14, 13, 14, 15, 16, 15, 16, 17, 18, 17, 18, 19, 20, 19, 20, 21, 22, 21, 22, 23, 24, 23, 24, 25, 26, 25, 26, 27, 28, 27, 28, 29, 30, 29, 30, 31, 32, 31, 32, 33, 34, 33, 34

Offset: 0

Wolfdieter Lang, May 10 2017

See a May 06 2017 comment on A049310 where these problems are considered which originated in a conjecture by Michel Lagneau (see A008611) on Fibonacci polynomials.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A008611, A016627, A049310, A059169, A162330, A183041.

Table[2 (Floor[n/2] - Floor[(n + 1)/4]) + Boole[OddQ@ n], {n, 0, 52}] (* Michael De Vlieger, May 10 2017 *)

concat(0, Vec(x*(1 + x - x^2 + x^3) / ((1 - x)^2*(1 + x)*(1 + x^2)) + O(x^100))) \\ Colin Barker, May 18 2017

a(n) = 2*b(n) if n is even, else a(n) = 1 + 2*b(n), with b(n) = floor(n/2) - floor((n + 1)/4) = A059169(n+1).
G.f. for {b(n)}: Sum_{n>=0} b(n)*x^n = x^2*(1 - x + x^2)/((1 - x)*(1 - x^4)) (see A059169).
From Colin Barker, May 18 2017: (Start)
G.f.: x*(1 + x - x^2 + x^3) / ((1 - x)^2*(1 + x)*(1 + x^2)).
a(n) = a(n-1) + a(n-4) - a(n-5) for n>4.
(End)
a(n) = A162330(n-1) for n >= 2. - Michel Marcus, Nov 01 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(2) (A016627). - Amiram Eldar, Sep 17 2023
a(n) = A183041(n-1) for n>=2. - R. J. Mathar, May 13 2025

A083219 a(n) = n - 2*floor(n/4).

Original entry on oeis.org

0, 1, 2, 3, 2, 3, 4, 5, 4, 5, 6, 7, 6, 7, 8, 9, 8, 9, 10, 11, 10, 11, 12, 13, 12, 13, 14, 15, 14, 15, 16, 17, 16, 17, 18, 19, 18, 19, 20, 21, 20, 21, 22, 23, 22, 23, 24, 25, 24, 25, 26, 27, 26, 27, 28, 29, 28, 29, 30, 31, 30, 31, 32, 33, 32, 33, 34, 35, 34, 35, 36, 37, 36, 37, 38

Offset: 0

Reinhard Zumkeller, Apr 22 2003

Conjecture: number of roots of P(x) = x^n - x^(n-1) - x^(n-2) - ... - x - 1 in the left half-plane. - Michel Lagneau, Apr 09 2013

a(n) is n+2 with its second least significant bit removed (see A021913(n+2) for that bit). - Kevin Ryde, Dec 13 2019

Altug Alkan, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A083220, A129756, A162751 (second highest bit removed).
Cf. A021913, A162330, A285869.
Essentially the same as A018837.

a:= n-> n - 2*floor(n/4):
seq(a(n), n=0..74);  # Alois P. Heinz, Jan 24 2021

Array[# - 2 Floor[#/4] &, 75, 0] (* Michael De Vlieger, Oct 02 2017 *)
LinearRecurrence[{1,0,0,1,-1},{0,1,2,3,2},80] (* Harvey P. Dale, Oct 01 2018 *)

a(n) = n - n\4*2 \\ Charles R Greathouse IV, Jun 11 2015

a(n) = A083220(n)/2.
a(n) = a(n-1) + n mod 2 + (n mod 4 - 1)*(1 - n mod 2), a(0) = 0.
G.f.: x*(1+x+x^2-x^3)/((1-x)^2*(1+x)*(1+x^2)). - R. J. Mathar, Aug 28 2008
a(n) = n - A129756(n). - Michel Lagneau, Apr 09 2013
Bisection: a(2*k) = 2*floor((n+2)/4), a(2*k+1) = a(2*k) + 1, k >= 0. - Wolfdieter Lang, May 08 2017
a(n) = (2*n + 3 - 2*cos(n*Pi/2) - cos(n*Pi) - 2*sin(n*Pi/2))/4. - Wesley Ivan Hurt, Oct 02 2017
a(n) = A162330(n+2) - 1 = A285869(n+3) - 1. - Kevin Ryde, Dec 13 2019
E.g.f.: ((1 + x)*cosh(x) - cos(x) + (2 + x)*sinh(x) - sin(x))/2. - Stefano Spezia, May 27 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(2) - 1. - Amiram Eldar, Aug 21 2023

A183041 Least number of knight's moves from (0,0) to (n,1) on infinite chessboard.

Original entry on oeis.org

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

Offset: 0

Clark Kimberling, Dec 20 2010

nonn

Row 2 of the array at A065775.

Apparently a(n)=A162330(n), n>0. - R. J. Mathar, Jan 29 2011

			a(0)=3 counts (0,0) to (2,1) to (1,3) to (0,1).

Cf. A065775, A018837, A183042-A183053.

def a(n):
  if n < 2: return [3, 2][n]
  m, r = divmod(n, 4)
  return [2*m+1, 2*m+2][r%2]
print([a(n) for n in range(70)]) # Michael S. Branicky, Mar 02 2021

T(0,1)=3, T(1,1)=2, and for m>=1,
T(4m-2,1)=2m-1, T(4m-1,1)=2m, T(4m,1)=2m+1, T(4m+1,1)=2m+2.
G.f.: (2*x^5-2*x^4+x^3-x^2-x+3) / ((x-1)^2*(x+1)*(x^2+1)). - Colin Barker, Feb 19 2014

A226279 a(4n) = a(4n+2) = 2n , a(4n+1) = a(4n+3) = 2n-1.

Original entry on oeis.org

0, -1, 0, -1, 2, 1, 2, 1, 4, 3, 4, 3, 6, 5, 6, 5, 8, 7, 8, 7, 10, 9, 10, 9, 12, 11, 12, 11, 14, 13, 14, 13, 16, 15, 16, 15, 18, 17, 18, 17, 20, 19, 20, 19, 22, 21, 22, 21, 24, 23, 24, 23, 26, 25, 26, 25, 28, 27, 28, 27, 30, 29, 30, 29

Offset: 0

Paul Curtz, Jun 02 2013

a(n)=c(n) in A214297(n).
In A214297 d(n)=-1,1,1,3,1,3,3,... = mix (-A186422(2n) , A186422(2n+1)).
A214297 is the (reduced) numerator of 1/4 - 1/A061038(n).
(i.e. (1/4 -(1/0, 1/4, 1, 1/36, 1/16,...)) = -1/0, 0/1, -3/4, 2/9, 3/16,... )
1/0 is a convention.
n^2=(a(n+1)+d(n+1))^2 are the denominators.

Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A134967, A162330, A103889, A000290.

Table[{0, -1} + 2*Floor[n/2], {n, 0, 31}] // Flatten (* Jean-François Alcover, Jun 03 2013 *)

a(n)=n\4*2-n%2 \\ Charles R Greathouse IV, Sep 15 2013

a(0) = a(2)=0, a(1)=a(3)=-1, a(4)=2.
a(n) = a(n-4) + 2, n > 3.
a(n) = a(n-1) + a(n-4) - a(n-5), n > 4.
A214297(n) = a(n+1) * d(n+1).
G.f.: x*(3*x^3-x^2+x-1) / ((x-1)^2*(x+1)*(x^2+1)). - Colin Barker, Sep 22 2013

A293990 a(n) = (3*n + ((n-2) mod 4))/2.

Original entry on oeis.org

1, 3, 3, 5, 7, 9, 9, 11, 13, 15, 15, 17, 19, 21, 21, 23, 25, 27, 27, 29, 31, 33, 33, 35, 37, 39, 39, 41, 43, 45, 45, 47, 49, 51, 51, 53, 55, 57, 57, 59, 61, 63, 63, 65, 67, 69, 69, 71, 73, 75, 75, 77, 79, 81, 81, 83, 85, 87, 87, 89, 91, 93, 93

Offset: 0

Dimitris Valianatos, Oct 21 2017

The product (2/3) * (4/3) * (6/5) * (6/7) * (8/9) * (10/9) * (12/11) * (12/13) * ... = Pi/(2*sqrt(3)). The denominators are a(n) for n >= 1 and numerators are a(n-1) + A093148(n) for n >= 1 -> [2, 4, 6, 6, 8, 10, 12, 12, ...].
Let r(n) = (a(n)-1)/(a(n)+1) if a(n) mod 4 = 1, (a(n)+1)/(a(n)-1) otherwise; then Product_{n>=1} r(n) = (2/1) * (2/1) * (2/3) * (4/3) * (4/5) * (4/5) * (6/5) * (6/7) * ... = Pi*sqrt(3)/2 = 2.72069904635132677...
The odd numbers of partial sums this sequence, are identified with the A003215 sequence. Also the prime numbers that appear in partial sums in this sequence, are identified with the A002407 sequence.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A007310, A047298, A063305, A093148, A134967, A157932, A162330, A168329, A214549, A285869, A003215, A002407.

[(3*n+((n-2) mod 4))/2 : n in [0..100]]; // Wesley Ivan Hurt, Oct 29 2017

A293990:=n->(3*n+((n-2) mod 4))/2: seq(A293990(n), n=0..100); # Wesley Ivan Hurt, Oct 29 2017

Table[(3*n + Mod[(n - 2), 4])/2, {n, 0, 100}] (* Wesley Ivan Hurt, Oct 29 2017 *)
f[n_] := (3n + Mod[n - 2, 4])/2; Array[f, 65, 0] (* or *)
LinearRecurrence[{1, 0, 0, 1, -1}, {1, 3, 3, 5, 7}, 65] (* or *)
CoefficientList[ Series[(x^4 + 2x^3 + 2x + 1)/((x - 1)^2 (x^3 + x^2 + x + 1)), {x, 0, 64}], x] (* Robert G. Wilson v, Nov 28 2017 *)

PARI
```
a(n) = (3*n + (n-2)%4) / 2
```

Vec(x*(1 + 2*x + 2*x^3 + x^4) / ((1 - x)^2*(1 + x)*(1 + x^2)) + O(x^30)) \\ Colin Barker, Oct 21 2017

first(n) = my(start=[1,3,3,5,7,9,9,11]); if(n<=8, return(start)); my(res=vector(n)); for (i=1, 8, res[i] = start[i]); for(i = 1, n-8 ,res[i+8] = res[i] + 12); res \\ David A. Corneth, Oct 21 2017

Sum_{n>=0} 1/a(n)^2 = 5*Pi^2/36 = 1.3707783890401886970... = 10*A086729.
(a(n) - n) * (-1)^(n+1) = A134967(n) for n >= 0.
a(n) - n = A162330(n) for n >= 0.
a(n) - n = A285869(n+1) for n >= 0.
a(n) + a(n+1) = A157932(n+2) for n >= 0.
a(n) + (2*n+1) = A047298(n+1) for n >= 0.
From Colin Barker, Oct 21 2017: (Start)
G.f.: x*(1 + 2*x + 2*x^3 + x^4) / ((1 - x)^2*(1 + x)*(1 + x^2)).
a(n) = a(n-1) + a(n-4) - a(n-5) for n > 5.
(End)
a(n + 8) = a(n) + 12. - David A. Corneth, Oct 21 2017
a(4*k+4) * a(4*k+3) - a(4*k+2) * a(4*k+1) = 2*A063305(k+3) for k >= 0.
Sum_{n>=0} 1/(a(n) + a(n+2))^2 = (4*Pi^2 - 27) / 108 = (A214549 - 1) / 4.

A330239 Minimum circular (strong) similarity of a length-n binary word.

Original entry on oeis.org

0, 0, 1, 2, 3, 4, 3, 4, 5, 6, 5, 6, 7, 8, 7, 8, 9, 10, 9, 10, 11, 12, 11, 12, 13, 14, 15, 14, 15, 16, 15, 16, 17, 18, 17, 18

Offset: 1

Jeffrey Shallit, Dec 06 2019

The circular (strong) similarity of a word w is the maximum, over all nontrivial cyclic shifts x of w, of the number of positions where x and w agree.
The plausible identification of this sequence with A285869, A162330, A183041 is just illusory because a(27) = 15.
Circular (strong) similarity is basically a one-sided version of autocorrelation, where we only care about agreement of terms, not the difference between agreement and disagreement.

			For n = 7, one string achieving a(7) = 3 is 0001011.

Michael S. Branicky, Python program

Cf. A162330, A183041, A285869.

# see links for faster version
from itertools import product
def css(k, n):
    cs = ((k>>i) | ((((1<1 else 0
print([a(n) for n in range(1, 16)]) # Michael S. Branicky, Jan 15 2024

a(31)-a(36) from Michael S. Branicky, Jan 15 2024

cp's OEIS Frontend

A285869 a(n) is the number of zeros of the Chebyshev S(n, x) polynomial in the open interval (-sqrt(2), +sqrt(2)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A083219 a(n) = n - 2*floor(n/4).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A183041 Least number of knight's moves from (0,0) to (n,1) on infinite chessboard.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Python

Formula

A226279 a(4n) = a(4n+2) = 2*n , a(4n+1) = a(4n+3) = 2*n-1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A293990 a(n) = (3*n + ((n-2) mod 4))/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

PARI

Formula

A330239 Minimum circular (strong) similarity of a length-n binary word.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Python

Extensions

A226279 a(4n) = a(4n+2) = 2n , a(4n+1) = a(4n+3) = 2n-1.