A171088 To find 3 consecutive integers in the sequence, you have to take 4 consecutive terms, no more and no less.

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

Offset

0,2

Links

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

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

Formula

a(3n) = n-1; a(3n+1) = n+1; a(3n+2) = n+3. Or, a(n) = 1 + a(n-3). - Alex Meiburg, Aug 21 2010

a(n) = a(n-1) + a(n-3) - a(n-4) = (n + 3 + 5*A049347(n-1))/3. - R. J. Mathar, Aug 21 2010

G.f.: (1+2*x+x^3-3*x^2)/((1+x+x^2)*(x-1)^2). - R. J. Mathar, Aug 21 2010

a(n) = 1 + n/3 - (5*I/9)*sqrt(3)*(-1/2+(I/2)*sqrt(3))^n + (5*I/9)*sqrt(3)*(-1/2-(I/2)*sqrt(3))^n for each n >= 0. - Alexander R. Povolotsky, Aug 21 2010

a(n) = (3*n+9+10*sqrt(3)*sin(2*n*Pi/3))/9. - Wesley Ivan Hurt, Oct 01 2017

Example

Taking the first 3 terms doesn't give 3 consecutive integers as they are 0, 1 and ... 3. But if you take the fourth term (2), you'll have in hand 0,1,2 [and even another triple of consecutive integers, which is (1,2,3)].

Mathematica

f[n_] := Switch[ Mod[n, 3], 0, n/3 + 1, 1, (n + 8)/3, 2, (n - 2)/3]; Array[f, 78, 0] (* Robert G. Wilson v, Sep 10 2010 *)
CoefficientList[Series[(1 + 2*x + x^3 - 3*x^2)/((1 + x + x^2)*(x - 1)^2), {x, 0, 50}], x] (* G. C. Greubel, Feb 25 2017 *)

Prog

(PARI) x='x+O('x^50); Vec((1+2*x+x^3-3*x^2)/((1+x+x^2)*(x-1)^2)) \\ _G. C. Greubel, Feb 25 2017

Crossrefs

Sequence in context: A136163 A178313 A190013 * A058624 A145856 A092154

Adjacent sequences: A171085 A171086 A171087 * A171089 A171090 A171091

Keyword

nonn,easy

Author

N. J. A. Sloane, Sep 08 2010, based on a posting to the Sequence Fans Mailing List by Eric Angelini, Aug 21 2010

Extensions

a(28) onwards from Robert G. Wilson v, Sep 10 2010

Status

approved