A179805 a(0) = 1, a(1) = 3, a(2) = 6 and a(n) = 2*a(n-1) - a(n-2) for n > 3.

1, 3, 6, 15, 24, 33, 42, 51, 60, 69, 78, 87, 96, 105, 114, 123, 132, 141, 150, 159, 168, 177, 186, 195, 204, 213, 222, 231, 240, 249, 258, 267, 276, 285, 294, 303, 312, 321, 330, 339, 348, 357, 366, 375, 384, 393, 402

Offset: 0

Author

Gary W. Adamson, Jul 27 2010

Keywords

• nonn
• easy

Comments

Apart from the second term, the same as A122709. - R. J. Mathar, Jul 30 2010

For n > 1, a(n) is the maximum value of the sum of the vertices in a normal magic triangle of order n (see formula 10 in Trotter). - Stefano Spezia, Mar 03 2021

Examples

			a(4) = 24 = 9 + a(3) = 9 + 15.
a(4) = 24 = 2*a(3) - a(2) = 2*15 - 6.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000
• Terrel Trotter, Normal Magic Triangles of Order n, Journal of Recreational Mathematics Vol. 5, No. 1, 1972, pp. 28-32.
• Index entries for linear recurrences with constant coefficients, signature (2,-1).

Crossrefs

Cf. A122709.

Programs

Mathematica

LinearRecurrence[{2,-1},{1,3,6,15},50] (* Harvey P. Dale, Sep 25 2018 *)

Formula

(1 + 3*x + 6*x^2 + 15*x^3 + ...) = (1 + 3*x^2 + 3*x^3 + 3*x^4 + ...) * (1 + 3*x + 3*x^2 + 3*x^3 + 3*x^4 + ...).

a(0) = 1, a(1) = 3, a(2) = 6 and a(n) = 2*a(n-1) - a(n-2) for n > 3.

a(n) = a(n-1) + 9 for n > 2.

For n > 1, a(n) == 6 (mod 9).

From Colin Barker, Oct 28 2012: (Start)

a(n) = 9*n - 12 for n > 1.

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

E.g.f.: 13 + 6*x + 3*exp(x)*(3x - 4). - Stefano Spezia, Mar 03 2021