|
|
A330170
|
|
a(n) = 2^n + 3^n + 6^n - 1.
|
|
1
|
|
|
10, 48, 250, 1392, 8050, 47448, 282250, 1686432, 10097890, 60526248, 362976250, 2177317872, 13062296530, 78368963448, 470199366250, 2821153019712, 16926788715970, 101560344351048, 609360902796250, 3656161927895952, 21936961102828210
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
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).
|
|
LINKS
|
|
|
FORMULA
|
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)
|
|
EXAMPLE
|
a(9) = 2^9 + 3^9 + 6^9 - 1 = 10097890 = 11 * 917990.
|
|
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 *)
|
|
PROG
|
(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
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|