A166012 a(n) = 2*(A000045(n)-(n mod 2)) + 1 + (n mod 2).

1, 2, 3, 4, 7, 10, 17, 26, 43, 68, 111, 178, 289, 466, 755, 1220, 1975, 3194, 5169, 8362, 13531, 21892, 35423, 57314, 92737, 150050, 242787, 392836, 635623, 1028458, 1664081, 2692538, 4356619, 7049156, 11405775, 18454930, 29860705, 48315634, 78176339

Offset

0,2

Comments

This is an auxiliary sequence for computing A138606.

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (1,2,-1,-1).

Formula

a(2n) = 2*A000045(2n) + 1, a(2n+1) = 2*A000045(2n+1).

Without reference to A000045: a(n)=2*Floor(a(n-1)/2)+a(n-2). - Clark Kimberling, Nov 07 2009

If n mod 2 = 0 then a(n) = a(n-1) + a(n-2), else a(n) = a(n-1) + a(n-2) - 1.

a(n) = 2*Fibonacci(n) + (1+(-1)^n)/2.

a(n) = 2*Fibonacci(n) + [(n+1)mod 2]. - Gary Detlefs, Dec 29 2010

G.f.: (1 + x - x^2 - 2*x^3)/((1 - x^2)*(1 - x - x^2)). - Ilya Gutkovskiy, Apr 22 2016

From Colin Barker, Apr 22 2016: (Start)

a(n) = a(n-1)+2*a(n-2)-a(n-3)-a(n-4) for n>3.

a(n) = (1/2+(-1)^n/2-(2*((1/2*(1-sqrt(5)))^n-(1/2*(1+sqrt(5)))^n))/sqrt(5)).

(End)

Mathematica

Table[2*Fibonacci[n] + (1 + (-1)^n)/2, {n, 0, 100}] (* G. C. Greubel, Apr 21 2016 *)
LinearRecurrence[{1, 2, -1, -1}, {1, 2, 3, 4}, 40] (* Harvey P. Dale, May 01 2018 *)

Prog

(MIT Scheme:) (define (A166012 n) (+ 1 (modulo n 2) (* 2 (- (A000045 n) (modulo n 2)))))
(PARI) Vec((1+x-x^2-2*x^3)/((1-x)*(1+x)*(1-x-x^2)) + O(x^50)) \\ Colin Barker, Apr 22 2016

Crossrefs

Sequence in context: A119016 A082958 A218495 * A060166 A053634 A094863

Adjacent sequences: A166009 A166010 A166011 * A166013 A166014 A166015

Keyword

nonn,easy

Author

Antti Karttunen, Oct 05 2009

Status

approved