A058212 a(n) = 1 + floor(n*(n-3)/6).

1, 0, 0, 1, 1, 2, 4, 5, 7, 10, 12, 15, 19, 22, 26, 31, 35, 40, 46, 51, 57, 64, 70, 77, 85, 92, 100, 109, 117, 126, 136, 145, 155, 166, 176, 187, 199, 210, 222, 235, 247, 260, 274, 287, 301, 316, 330, 345, 361, 376, 392, 409, 425, 442, 460, 477, 495, 514, 532, 551

Offset

0,6

Comments

For n >= 3, number of solutions to x+y+z == 0 (mod n) with 0 <= x < y < z < n. E.g., for n=3 there is a unique solution, x=0, y=1, z=2.

Links

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

S. A. Burr, B. Grünbaum, and N. J. A. Sloane, The Orchard Problem, Geometriae Dedicata, 2 (1974), 397-424.

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

Formula

From Paul Barry, Mar 18 2004: (Start)

G.f.: (1 - 2x + x^2 + x^4)/((1 - x)^2(1 - x^3)).

a(n) = 4*cos(2*Pi*n/3)/9 + (3*n^2 - 9*n + 10)/18. (End)

E.g.f.: (exp(x)*(10 - 6*x + 3*x^2) + 8*exp(-x/2)*cos(sqrt(3)*x/2))/18. - Stefano Spezia, May 03 2023

Sum_{n>=3} 1/a(n) = 6 - (2*Pi/sqrt(3))*(1 - tanh(sqrt(5/3)*Pi/2)/sqrt(5)). - Amiram Eldar, May 06 2023

Mathematica

Table[Floor[(n(n-3))/6]+1, {n, 0, 70}] (* or *) LinearRecurrence[{2, -1, 1, -2, 1}, {1, 0, 0, 1, 1}, 70] (* Harvey P. Dale, Jun 21 2021 *)

Prog

(Sage) [ceil(binomial(n, 2)/3) for n in range(-1, 55)] # Zerinvary Lajos, Dec 03 2009
(Haskell)
a058212 n = 1 + n * (n - 3) `div` 6  -- Reinhard Zumkeller, May 08
(PARI) a(n)=n*(n-3)\6 + 1 \\ Charles R Greathouse IV, Jun 11 2015

Crossrefs

Cf. A003035.

Apart from initial term, same as A007997.

The third diagonal of A061857.

Sequence in context: A186386 A247589 A361070 * A007997 A123120 A194462

Adjacent sequences: A058209 A058210 A058211 * A058213 A058214 A058215

Keyword

nonn,easy

Author

N. J. A. Sloane, Nov 30 2000

Extensions

Zerinvary Lajos, Dec 07 2009

Status

approved