A022003 Decimal expansion of 1/999.

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, 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, 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

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• cons
• easy

Comments

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

a(n) = A130196(n) - A131534(n). - Reinhard Zumkeller, Nov 12 2009

Examples

			0.001001001001001001001...
		

Links

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

Crossrefs

Essentially the same as A079978.

Cf. A068601.

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

Programs

Mathematica

Join[{0,0},RealDigits[1/999,10,120][[1]]] (* or *) PadRight[{},120,{0,0,1}] (* Harvey P. Dale, May 24 2012 *)

PARI

a(n)=n%3==2 \\ Jaume Oliver Lafont, Mar 24 2009

Formula

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

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

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

From Hieronymus Fischer, May 29 2007: (Start)

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

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

a(n) = (1 + (-1)^Fibonacci(n+1))/2. - Hieronymus Fischer, Jun 14 2007

a(n) = (n^5 - n^2) mod 3. - Gary Detlefs, Mar 20 2010

a(n) = ((-1)^(a(n-1) + a(n-2)) + 1)/2 starting from n=3. - Adriano Caroli, Nov 21 2010

a(n) = 1 - Fibonacci(n+1) mod 2. - Gary Detlefs, Dec 26 2010

a(n) = floor((n+1)/3) - floor(n/3). - Tani Akinari, Oct 22 2012