A330170 a(n) = 2^n + 3^n + 6^n - 1.

10, 48, 250, 1392, 8050, 47448, 282250, 1686432, 10097890, 60526248, 362976250, 2177317872, 13062296530, 78368963448, 470199366250, 2821153019712, 16926788715970, 101560344351048, 609360902796250, 3656161927895952, 21936961102828210

Offset: 1

Author

Bernard Schott, Dec 04 2019

Keywords

• nonn
• easy

Comments

This sequence is the subject of the 4th problem, proposed by Poland, of the 46th International Mathematical Olympiad in 2005 at Mérida (Mexico) [see the link IMO].

Answer to the question: 1 is the only positive integer that is relatively prime to every term of the sequence.

Proof: p=2 divides a(1) = 10, p=3 divides a(2) = 48, and if p prime >= 5, then p divides a(p-2). So, for every prime p, there exists n >= 1 such that p divides a(n).

Examples

			a(9) = 2^9 + 3^9 + 6^9 - 1 = 10097890 = 11 * 917990.
		

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• IMO, IMO 2005 - Problem 4
• Index entries for linear recurrences with constant coefficients, signature (12,-47,72,-36).

Crossrefs

Cf. A000079 (2^n), A000244 (3^n), A000400 (6^n), A318760 (a(p-2)/p).

Programs

Maple

A330170 := seq(2^n+3^n+6^n-1, n=1..50);

Mathematica

Table[2^n + 3^n + 6^n - 1, {n, 1, 21}] (* Amiram Eldar, Dec 04 2019 *)

PARI

Vec(2*x*(5 - 36*x + 72*x^2 - 36*x^3) / ((1 - x)*(1 - 2*x)*(1 - 3*x)*(1 - 6*x)) + O(x^40)) \\ Colin Barker, Dec 04 2019

Formula

a(n) = A000079(n) + A000244(n) + A000400(n) - 1.

From Colin Barker, Dec 04 2019: (Start)

G.f.: 2*x*(5 - 36*x + 72*x^2 - 36*x^3) / ((1 - x)*(1 - 2*x)*(1 - 3*x)*(1 - 6*x)).

a(n) = 12*a(n-1) - 47*a(n-2) + 72*a(n-3) - 36*a(n-4) for n>5.

(End)