A131534 -id:A131534 - cp's OEIS frontend

A146325 Period 3: repeat [1, 4, 1].

1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 1

Offset: 1

Artur Jasinski, Oct 30 2008

Continued fraction of (1 + sqrt(26))/5 = A188659.
Digital roots of the centered triangular numbers A005448. - Ant King, May 08 2012
Also the digital roots of centered 12-gonal numbers A003154. - Peter M. Chema, Dec 20 2023

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

Cf. A003154, A005448, A021337, A131534 (square roots), A188659.

&cat [[1,4,1]^^40]; // Bruno Berselli, Jun 27 2016

seq(op([1, 4, 1]), n=1..50); # Wesley Ivan Hurt, Jul 01 2016

Table[Round[N[4 (Cos[(2 n - 1) ArcTan[Sqrt[3]]])^2, 100]], {n, 1, 100}]
PadLeft[{},111,{1,4,1}] (* Harvey P. Dale, Sep 18 2011 *)

a(n)=1+3*(n%3==2) \\ Jaume Oliver Lafont, Mar 24 2009

a(n) = 4*(cos((2*n - 1)*Pi/3))^2 = 4 - 4*(sin((2*n - 1)*Pi/3))^2.
a(n+3) = a(n).
a(n) = 2 - cos(2*Pi*n/3) + sqrt(3)*sin(2*Pi*n/3).
O.g.f.: x*(1+4*x+x^2)/(1-x^3). [Richard Choulet, Nov 03 2008]
a(n) = 6 - a(n-1) - a(n-2) for n>2. - Ant King, Jun 12 2012
a(n) = (n mod 3)^(n mod 3). - Bruno Berselli, Jun 27 2016
a(n) = 1 + A021337(n) for n>0. - Wesley Ivan Hurt, Jul 01 2016

A177702 Period 3: repeat [1, 1, 2].

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2

Offset: 0

Klaus Brockhaus, May 11 2010

Continued fraction expansion of (2+sqrt(10))/3.
Decimal expansion of 112/999.
a(n) = A131534(n+2) = |A132419(n)| = |A132367(n)| = |A131556(n+2)|= |A122876(n)|.

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

Cf. A022003, A131534, A177703.

Magma
```
&cat[ [1, 1, 2]: k in [1..35] ];
```

seq(op([1, 1, 2]), n=1..50); # Wesley Ivan Hurt, Jul 01 2016

PadRight[{},120,{1,1,2}] (* or *) LinearRecurrence[{0,0,1},{1,1,2},120] (* Harvey P. Dale, Dec 19 2014 *)

a(n)=max(n%3,1) \\ Charles R Greathouse IV, Jul 17 2016

a(n) = a(n-3) for n > 2, with a(0) = 1, a(1) = 1, a(2) = 2.
G.f.: (1+x+2*x^2)/(1-x^3).
a(n) = 4/3 - cos(2*Pi*n/3)/3 - sin(2*Pi*n/3)/sqrt(3). - R. J. Mathar, Oct 08 2011
a(n) = 1 + A022003(n). - Wesley Ivan Hurt, Jul 01 2016

A195547 Denominators a(n) of Pythagorean approximations b(n)/a(n) to 1/2.

Original entry on oeis.org

1, 4, 12, 15, 80, 208, 273, 1428, 3740, 4895, 25632, 67104, 87841, 459940, 1204140, 1576239, 8253296, 21607408, 28284465, 148099380, 387729212, 507544127, 2657535552, 6957518400, 9107509825, 47687540548, 124847601996, 163427632719, 855718194320, 2240299317520

Offset: 1

Clark Kimberling, Sep 20 2011

See A195500 for a discussion and references.

a(n) is the numerator of the harmonic mean of F(n) and F(n+1), where F = A000045 (Fibonacci numbers). Example: 2*F(9)*F(10)/(F(9)+F(10)) = 2*34*55/(34+55) = 3740/89, therefore a(9) = 3740. - Francesco Daddi, Jul 04 2018

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A195500, A195548, A195549, A131534.

r = 1/2; z = 30;
p[{f_, n_}] := (#1[[2]]/#1[[
      1]] &)[({2 #1[[1]] #1[[2]], #1[[1]]^2 - #1[[
         2]]^2} &)[({Numerator[#1], Denominator[#1]} &)[
     Array[FromContinuedFraction[
        ContinuedFraction[(#1 + Sqrt[1 + #1^2] &)[f], #1]] &, {n}]]]];
{a, b} = ({Denominator[#1], Numerator[#1]} &)[
  p[{r, z}]]  (* A195547, A195548 *)
Sqrt[a^2 + b^2] (* A195549 *)
(* Peter J. C. Moses, Sep 02 2011 *)
Table[Numerator[2 Fibonacci[n] Fibonacci[n+1] / ( Fibonacci[n] + Fibonacci[n+1])], {n, 1, 40}] (* Vincenzo Librandi, Jul 21 2018 *)

a(n) = 2*F(n)*F(n+1)/(2-((n+2)^2 mod 3)), where F(n)=Fibonacci(n). - Gary Detlefs, Oct 15 2011

Empirical G.f.: x*(1+4*x+12*x^2-2*x^3+12*x^4+4*x^5+x^6)/(1-17*x^3-17*x^6+x^9). - Colin Barker, Apr 15 2012

A195549 Hypotenuses of primitive Pythagorean triples in A195547 and A195548.

Original entry on oeis.org

1, 5, 13, 17, 89, 233, 305, 1597, 4181, 5473, 28657, 75025, 98209, 514229, 1346269, 1762289, 9227465, 24157817, 31622993, 165580141, 433494437, 567451585, 2971215073, 7778742049, 10182505537, 53316291173, 139583862445, 182717648081

Offset: 1

Clark Kimberling, Sep 20 2011

nonn

See A195500 for discussion and list of related sequences; see A195547 for Mathematica program.

Cf. A000045, A195500, A195547, A195548, A131534.

with(combinat):f:= n-> fibonacci(n):seq((f(n)^2+f(n+1)^2)/(2-((n+2)^2 mod 3)), n=1..25);

a(n) = (F(n)^2 + F(n+1)^2)/(2 - ((n+2)^2 mod 3)), F(n) = Fibonacci(n). - Gary Detlefs, Oct 14 2011
Conjectures from Colin Barker, Apr 08 2012: (Start)
a(n) = 18*a(n-3) - a(n-6).
G.f.: x*(1+5*x+13*x^2-x^3-x^4-x^5)/((1-3*x+x^2)*(1+3*x+8*x^2+3*x^3+x^4)). (End)
Conjecture: a(n) is the denominator of the reduced fraction (F(2*n+1)-2)/F(2*n+1), F(n) = Fibonacci(n). - Sébastien Labbé, May 06 2022

A157334 First differences of A156799.

Original entry on oeis.org

1, 2, 1, 5, 1, 2, 1, 6, 1, 2, 1, 5, 1, 2, 1, 15, 1, 2, 1, 5, 1, 2, 1, 6, 1, 2, 1, 5, 1, 2, 1, 19, 1, 2, 1, 5, 1, 2, 1, 6, 1, 2, 1, 5, 1, 2, 1, 15, 1, 2, 1, 5, 1, 2, 1, 6, 1, 2, 1, 5, 1, 2, 1, 43, 1, 2, 1, 5, 1, 2, 1, 6, 1, 2, 1, 5, 1, 2, 1, 15, 1, 2, 1, 5, 1, 2, 1, 6, 1, 2, 1, 5, 1, 2, 1, 19

Offset: 0

Paul Curtz, Feb 27 2009

nonn

For 1,2,1 see A131534. First of a family of four: second is 3,1,1,4,3,1,1,2,3=A157076, third is 9,10,9,14,4,1,10 (from 3's in A090822), fourth is 220,221,220,224, (from A157107). Many consequences.

A195548 Numerators b(n) of Pythagorean approximations b(n)/a(n) to 1/2.

Original entry on oeis.org

0, 3, 5, 8, 39, 105, 136, 715, 1869, 2448, 12815, 33553, 43920, 229971, 602069, 788120, 4126647, 10803705, 14142232, 74049691, 193864605, 253772064, 1328767775, 3478759201, 4553754912, 23843770275, 62423800997, 81713816360, 427859097159, 1120149658761

Offset: 1

Clark Kimberling, Sep 20 2011

nonn

See A195500 for discussion and list of related sequences; see A195547 for Mathematica program.

Cf. A195500, A195547, A195549, A131534.

a(n) = (F(n+1)^2-F(n)^2)/(2-((n+2)^2 mod 3)), where F(n)=Fibonacci(n). [Gary Detlefs, Oct 15 2011]

Empirical G.f.: x*(3+5*x+8*x^2-12*x^3+20*x^4+x^6-x^7)/(1-17*x^3-17*x^6+x^9). [Colin Barker, Apr 15 2012]

A210209 GCD of all sums of n consecutive Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 4, 1, 3, 2, 11, 1, 8, 1, 29, 2, 21, 1, 76, 1, 55, 2, 199, 1, 144, 1, 521, 2, 377, 1, 1364, 1, 987, 2, 3571, 1, 2584, 1, 9349, 2, 6765, 1, 24476, 1, 17711, 2, 64079, 1, 46368, 1, 167761, 2, 121393, 1, 439204, 1, 317811, 2, 1149851, 1, 832040

Offset: 0

Alonso del Arte, Mar 18 2012

Early on in the Posamentier & Lehmann (2007) book, the fact that the sum of any ten consecutive Fibonacci numbers is a multiple of 11 is presented as an interesting property of the Fibonacci numbers. Much later in the book a proof of this fact is given, using arithmetic modulo 11. An alternative proof could demonstrate that 11*F(n + 6) = Sum_{i=n..n+9} F(i).

			a(3) = 2 because all sums of three consecutive Fibonacci numbers are divisible by 2 (F(n) + F(n-1) + F(n-2) = 2F(n)), but since the GCD of 3 + 5 + 8 = 16 and 5 + 8 + 13 = 26 is 2, no number larger than 2 divides all sums of three consecutive Fibonacci numbers.
a(4) = 1 because the GCD of 1 + 1 + 2 + 3 = 7 and 1 + 2 + 3 + 5 = 11 is 1, so the sums of four consecutive Fibonacci numbers have no factors in common.

Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers, Prometheus Books, New York (2007) p. 33.

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Dan Guyer and aBa Mbirika, GCD of sums of k consecutive Fibonacci, Lucas, and generalized Fibonacci numbers, Journal of Integer Sequences, 24 No.9, Article 21.9.8 (2021), 25pp; arXiv preprint, arXiv:2104.12262 [math.NT], 2021.
I. D. Ruggles, Elementary problem B-1, Fibonacci Quarterly, Vol. 1, No. 1, February 1963, p. 73; Solution by Marjorie R. Bicknell, published in Vol. 1, No. 3, October 1963, pp. 76-77.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,1,0,-1,0,-3,0,0,0,1).

Cf. A000045, A000071, sum of the first n Fibonacci numbers, A001175 (Pisano periods). Cf. also A229339.
Bisections give: A005013 (even part), A131534 (odd part).
Sums of m consecutive Fibonacci numbers: A055389 (m = 3, ignoring the initial 1); A000032 (m = 4, these are the Lucas numbers); A013655 (m = 5); A022087 (m = 6); A022096 (m = 7); A022379 (m = 8).

a:= n-> (Matrix(7, (i, j)-> `if`(i=j-1, 1, `if`(i=7, [1, 0, -3, -1, 1, 3, 0][j], 0)))^iquo(n, 2, 'r'). `if`(r=0, <<0, 1, 1, 4, 3, 11, 8>>, <<1, 2, 1, 1, 2, 1, 1>>))[1, 1]: seq(a(n), n=0..80);  # Alois P. Heinz, Mar 18 2012

Table[GCD[Fibonacci[n + 1] - 1, Fibonacci[n]], {n, 1, 50}] (* Horst H. Manninger, Dec 19 2021 *)

a(n)=([0,1,0,0,0,0,0,0,0,0,0,0,0,0; 0,0,1,0,0,0,0,0,0,0,0,0,0,0; 0,0,0,1,0,0,0,0,0,0,0,0,0,0; 0,0,0,0,1,0,0,0,0,0,0,0,0,0; 0,0,0,0,0,1,0,0,0,0,0,0,0,0; 0,0,0,0,0,0,1,0,0,0,0,0,0,0; 0,0,0,0,0,0,0,1,0,0,0,0,0,0; 0,0,0,0,0,0,0,0,1,0,0,0,0,0; 0,0,0,0,0,0,0,0,0,1,0,0,0,0; 0,0,0,0,0,0,0,0,0,0,1,0,0,0; 0,0,0,0,0,0,0,0,0,0,0,1,0,0; 0,0,0,0,0,0,0,0,0,0,0,0,1,0; 0,0,0,0,0,0,0,0,0,0,0,0,0,1; 1,0,0,0,-3,0,-1,0,1,0,3,0,0,0]^n*[0;1;1;2;1;1;4;1;3;2;11;1;8;1])[1,1] \\ Charles R Greathouse IV, Jun 20 2017

G.f.: -x*(x^12-x^11+2*x^10-x^9-2*x^8-x^7-6*x^6+x^5-2*x^4+x^3+2*x^2+x+1) / (x^14-3*x^10-x^8+x^6+3*x^4-1) = -1/(x^4+x^2-1) + (x^2+1)/(x^4-x^2-1) + (x+2)/(6*(x^2+x+1)) + (x-2)/(6*(x^2-x+1)) - 2/(3*(x+1)) - 2/(3*(x-1)). - Alois P. Heinz, Mar 18 2012
a(n) = gcd(Fibonacci(n+1)-1, Fibonacci(n)). - Horst H. Manninger, Dec 19 2021
From Aba Mbirika, Jan 21 2022: (Start)
a(n) = gcd(F(n+1)-1, F(n+2)-1).
a(n) = Lcm_{A001175(m) divides n} m.
Proofs of these formulas are given in Theorems 15 and 25 of the Guyer-Mbirika paper. (End)

More terms from Alois P. Heinz, Mar 18 2012

A364447 Repeat [1,2,1,3].

Original entry on oeis.org

1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1

Offset: 1

Rok Cestnik, Jul 25 2023

Continued fraction of sqrt(5) - 3/2 = 0.7360679... (without integer part); and (4*sqrt(5) + 6)/11 = 1.3585701... (with integer part).

Lexicographically earliest sequence in which n is banned for n terms after each appearance (see A364448 for n^2 and A364449 for n^3).

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

Cf. A131743 (repeat [0,1,0,2]).
Cf. A000034, A010882, A131534.
Cf. A364448, A364449.

PadRight[{}, 100, {1, 2, 1, 3}] (* Paolo Xausa, Jan 23 2025 *)

def A364447(n): return (3,1,2,1)[n&3] # Chai Wah Wu, Jul 29 2023

a(n) = A131743(n-1) + 1.

A245477 Period 6: repeat [1, 1, 1, 1, 1, 2].

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1

Offset: 0

Hailey R. Olafson, Jul 23 2014

First differences of A047368. The first differences of this sequence are in A131533. - Wesley Ivan Hurt, Jul 24 2014

Binomial Transform of a(n) gives: 1, 2, 4, 8, 16, 33, 70, 149, 312, 638, 1276, 2511, ... - Wesley Ivan Hurt, Aug 13 2014

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

Cf. A024643, A047368, A097325, A130782, A131534, A172051, A177702, A177704, A177706.

[Floor((n+1)*7/6) - Floor((n)*7/6) : n in [0..100]]; // Wesley Ivan Hurt, Aug 06 2014

A:= n -> piecewise(n mod 6 = 5, 2, 1);
seq(A(n), n=0..100); # Robert Israel, Jul 23 2014

Table[2 - Sign[Mod[n + 1, 6]], {n, 0, 100}] (* Wesley Ivan Hurt, Jul 24 2014 *)
PadRight[{},120,{1,1,1,1,1,2}] (* Harvey P. Dale, Jun 02 2016 *)

a(n) = 7*(n+1)\6 - 7*n\6; \\ Michel Marcus, Jul 23 2014

[floor((n+1)*7/6) - floor((n)*7/6) for n in [0..200]]

a(n) = floor((n+1)*7/6) - floor((n)*7/6).
G.f.: 1/(1-x) + x^5/(1-x^6). - Robert Israel, Jul 23 2014
From Wesley Ivan Hurt, Jul 24 2014, Aug 06-29 2014: (Start)
  a(n) = 2 - sign((n+1) mod 6).
  a(n) = 3 - 2^sign((n+1) mod 6).
  a(n) = A172051(n) + 1.
  a(2n) = 1, a(2n+1) = A177702(n).
  Sum_{i=0..n-2} a(i) = A047368(n), n>0.
  a(n) = 1 + mod(n, 1 + mod(n-1, 3)).
  a(n) = 1 + binomial(mod(5n + 10, 6), 5). (End)
From Wesley Ivan Hurt, Jun 23 2016: (Start)
a(n) = a(n-6) for n>5.
a(n) = (7 - cos(n*Pi) + cos(n*Pi/3) - cos(2*n*Pi/3) - sqrt(3)*sin(n*Pi/3) - sqrt(3)*sin(2*n*Pi/3))/6. (End)

A106740 Triangle read by rows: greatest common divisors of pairs of Fibonacci numbers greater than 1: T(n, k) = gcd(Fibonacci(n), Fibonacci(k)).

Original entry on oeis.org

2, 1, 3, 1, 1, 5, 2, 1, 1, 8, 1, 1, 1, 1, 13, 1, 3, 1, 1, 1, 21, 2, 1, 1, 2, 1, 1, 34, 1, 1, 5, 1, 1, 1, 1, 55, 1, 1, 1, 1, 1, 1, 1, 1, 89, 2, 3, 1, 8, 1, 3, 2, 1, 1, 144, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 233, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 377, 2, 1, 5, 2, 1, 1, 2, 5, 1, 2, 1, 1, 610

Offset: 3

Reinhard Zumkeller, May 15 2005

			Triangle begins as:
  2;
  1, 3;
  1, 1, 5;
  2, 1, 1, 8;
  1, 1, 1, 1, 13;
  1, 3, 1, 1,  1, 21;
  2, 1, 1, 2,  1,  1, 34;
  1, 1, 5, 1,  1,  1,  1, 55;
  1, 1, 1, 1,  1,  1,  1,  1, 89;
  2, 3, 1, 8,  1,  3,  2,  1,  1, 144;
  1, 1, 1, 1,  1,  1,  1,  1,  1,   1, 233;
  1, 1, 1, 1, 13,  1,  1,  1,  1,   1,   1, 377;
  2, 1, 5, 2,  1,  1,  2,  5,  1,   2,   1,   1, 610;

G. C. Greubel, Rows n = 3..52 of the triangle, flattened

Cf. A000012, A000045, A010125, A060904, A061347, A093148, A131534.

T[n_, k_]:= GCD[Fibonacci[n], Fibonacci[k]];
Table[T[n, k], {n,3,18}, {k,3,n}]//Flatten (* G. C. Greubel, Sep 11 2021 *)

def T(n,k): return gcd(fibonacci(n), fibonacci(k))
flatten([[T(n,k) for k in (3..n)] for n in (3..18)]) # G. C. Greubel, Sep 11 2021

T(n, k) = gcd(A000045(n), A000045(k)) for n >= 3 and 3 <= k <= n.
T(n, 3) = abs(A061347(n)).
T(n, 4) = A093148(n-1).
T(n, n) = A000045(n).
From G. C. Greubel, Sep 11 2021: (Start)
T(n, 3) = A131534(n-2).
T(n, 5) = A060904(n).
T(n, 6) = A010125(n).
T(n, n-1) = T(n, n-2) = A000012(n).
T(n, n-3) = A093148(n-5).
T(n, n-4) = A093148(n-5).
T(n, n-5) = A060904(n-5).
T(n, n-6) = A010125(n-6). (End)

cp's OEIS Frontend

A146325 Period 3: repeat [1, 4, 1].

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A177702 Period 3: repeat [1, 1, 2].

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A195547 Denominators a(n) of Pythagorean approximations b(n)/a(n) to 1/2.

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A195549 Hypotenuses of primitive Pythagorean triples in A195547 and A195548.

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A157334 First differences of A156799.

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A195548 Numerators b(n) of Pythagorean approximations b(n)/a(n) to 1/2.

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A210209 GCD of all sums of n consecutive Fibonacci numbers.

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Maple

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A364447 Repeat [1,2,1,3].

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