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Decimal expansion of 1/999.
18

%I #42 Dec 27 2023 11:53:41

%S 0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,

%T 0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,

%U 1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1

%N Decimal expansion of 1/999.

%C Expansion in any base b of 1/(b^3-1). E.g., 1/7 in base 2, 1/26 in base 3, 1/63 in base 4, etc. - _Franklin T. Adams-Watters_, Nov 07 2006

%C a(n) = A130196(n) - A131534(n). - _Reinhard Zumkeller_, Nov 12 2009

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 1).

%F From Mario Catalani (mario.catalani(AT)unito.it), Jan 07 2003: (Start)

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

%F a(n) = -(1/2)*((-1)^floor((2n-1)/3) + (-1)^floor((2n+1)/3)). (End)

%F From _Hieronymus Fischer_, May 29 2007: (Start)

%F a(n) = ((n+2) mod 3) mod 2.

%F a(n) = (1/2)*(1 - (-1)^(n + floor((n+2)/3))). (End)

%F a(n) = (1 + (-1)^Fibonacci(n+1))/2. - _Hieronymus Fischer_, Jun 14 2007

%F a(n) = (n^5 - n^2) mod 3. - _Gary Detlefs_, Mar 20 2010

%F a(n) = ((-1)^(a(n-1) + a(n-2)) + 1)/2 starting from n=3. - _Adriano Caroli_, Nov 21 2010

%F a(n) = 1 - Fibonacci(n+1) mod 2. - _Gary Detlefs_, Dec 26 2010

%F a(n) = floor((n+1)/3) - floor(n/3). - _Tani Akinari_, Oct 22 2012

%e 0.001001001001001001001...

%t Join[{0,0},RealDigits[1/999,10,120][[1]]] (* or *) PadRight[{},120,{0,0,1}] (* _Harvey P. Dale_, May 24 2012 *)

%o (PARI) a(n)=n%3==2 \\ _Jaume Oliver Lafont_, Mar 24 2009

%Y Essentially the same as A079978.

%Y Cf. A068601.

%Y Partial sums are given by A002264(n+1).

%K nonn,cons

%O 0,1

%A _N. J. A. Sloane_