A100063 A Chebyshev transform of Jacobsthal numbers.

1, 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

Offset: 0

Author

Paul Barry, Nov 02 2004

Keywords

• easy
• nonn
• mult

Comments

A Chebyshev transform of A001045(n+1): if A(x) is the g.f. of a sequence, map it to ((1-x^2)/(1+x^2))*A(x/(1+x^2)).

Also decimal expansion of 1111/9990. - Elmo R. Oliveira, Feb 18 2024

Also partial quotients of the continued fraction expansion of sqrt(5/2). - Hugo Pfoertner, Jan 10 2025

Examples

			G.f. = 1 + x + x^2 + 2*x^3 + x^4 + x^5 + 2*x^6 + x^7 + x^8 + 2*x^9 + ... - _Michael Somos_, Feb 20 2024
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Andrei Asinowski, Cyril Banderier, and Valerie Roitner, Generating functions for lattice paths with several forbidden patterns, (2019).
• Index entries for linear recurrences with constant coefficients, signature (0,0,1).

Crossrefs

Cf. A000012, A001045, A100051, A061347, A057079, A154272.

Cf. A000032, A000045.

Programs

Mathematica

PadRight[{1},120,{2,1,1}] (* or *) LinearRecurrence[{0,0,1},{1,1,1,2},120] (* Harvey P. Dale, Jul 08 2015 *)
a[ n_] := If[n<1, Boole[n==0], {2, 1, 1}[[1+Mod[n, 3]]]]; (* Michael Somos, Feb 20 2024 *)

PARI

my(x='x+O('x^50)); Vec((1+x)(1+x^2)/(1-x^3)) \\ G. C. Greubel, May 03 2017

PARI

{a(n) = if(n<1, n==0, [2, 1, 1][n%3+1])}; /* Michael Somos, Feb 20 2024 */

PARI

contfrac(sqrt(5/2),,80) \\ Hugo Pfoertner, Jan 10 2025

Formula

G.f.: (1+x)(1+x^2)/(1-x^3).

a(n) = n*Sum_{k=0..floor(n/2)} binomial(n-k, k)(-1)^k*A001045(n-2k+1)/(n-k).

Multiplicative with a(3^e) = 2, a(p^e) = 1 otherwise. - David W. Wilson, Jun 11 2005

Dirichlet g.f.: zeta(s)*(1+1/3^s). Dirichlet convolution of A154272 and A000012. - R. J. Mathar, Feb 07 2011

a(n) = 2 if n == 0 (mod 3) and n > 0, and a(n) = 1 otherwise. - Amiram Eldar, Nov 01 2022

a(n) = gcd(Fibonacci(n), Lucas(n)) = gcd(A000045(n), A000032(n)), for n >= 1. - Amiram Eldar, Jul 10 2023