A171088 - OEIS

%I #24 Jan 19 2019 04:15:43

%S 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,

%T 13,10,12,14,11,13,15,12,14,16,13,15,17,14,16,18,15,17,19,16,18,20,17,

%U 19,21,18,20,22,19,21,23,20,22,24,21,23,25,22,24,26,23,25,27,24,26,28,25

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

%H G. C. Greubel, <a href="/A171088/b171088.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,1,-1).

%F 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

%F a(n) = a(n-1) + a(n-3) - a(n-4) = (n + 3 + 5*A049347(n-1))/3. - _R. J. Mathar_, Aug 21 2010

%F G.f.: (1+2*x+x^3-3*x^2)/((1+x+x^2)*(x-1)^2). - _R. J. Mathar_, Aug 21 2010

%F 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

%F a(n) = (3*n+9+10*sqrt(3)*sin(2*n*Pi/3))/9. - _Wesley Ivan Hurt_, Oct 01 2017

%e 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)].

%t 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 *)

%t 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 *)

%o (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

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_, Sep 08 2010, based on a posting to the Sequence Fans Mailing List by _Eric Angelini_, Aug 21 2010

%E a(28) onwards from _Robert G. Wilson v_, Sep 10 2010