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

%I A330170 #29 Jul 29 2020 14:05:43
%S A330170 10,48,250,1392,8050,47448,282250,1686432,10097890,60526248,362976250,
%T A330170 2177317872,13062296530,78368963448,470199366250,2821153019712,
%U A330170 16926788715970,101560344351048,609360902796250,3656161927895952,21936961102828210
%N A330170 a(n) = 2^n + 3^n + 6^n - 1.
%C A330170 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].
%C A330170 Answer to the question: 1 is the only positive integer that is relatively prime to every term of the sequence.
%C A330170 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).
%H A330170 Colin Barker, <a href="/A330170/b330170.txt">Table of n, a(n) for n = 1..1000</a>
%H A330170 IMO, <a href="https://imomath.com/othercomp/I/Imo2005.pdf">IMO 2005 - Problem 4</a>
%H A330170 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (12,-47,72,-36).
%F A330170 a(n) = A000079(n) + A000244(n) + A000400(n) - 1.
%F A330170 From _Colin Barker_, Dec 04 2019: (Start)
%F A330170 G.f.: 2*x*(5 - 36*x + 72*x^2 - 36*x^3) / ((1 - x)*(1 - 2*x)*(1 - 3*x)*(1 - 6*x)).
%F A330170 a(n) = 12*a(n-1) - 47*a(n-2) + 72*a(n-3) - 36*a(n-4) for n>5.
%F A330170 (End)
%e A330170 a(9) = 2^9 + 3^9 + 6^9 - 1 = 10097890 = 11 * 917990.
%p A330170 A330170 := seq(2^n+3^n+6^n-1, n=1..50);
%t A330170 Table[2^n + 3^n + 6^n - 1, {n, 1, 21}] (* _Amiram Eldar_, Dec 04 2019 *)
%o A330170 (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
%Y A330170 Cf. A000079 (2^n), A000244 (3^n), A000400 (6^n), A318760 (a(p-2)/p).
%K A330170 nonn,easy
%O A330170 1,1
%A A330170 _Bernard Schott_, Dec 04 2019