OFFSET
1,1
COMMENTS
Positive numbers k such that 6*(6 + k) is a perfect square.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Leo Tavares, Illustration: Star Cut Hexagons
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
G.f.: 6*x*(3 - x)/(1 - x)^3. - Vincenzo Librandi, Jul 08 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Jul 08 2012
E.g.f.: 6*x*(3 + x)*exp(x). - G. C. Greubel, Sep 01 2019
From Amiram Eldar, Feb 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 1/8.
Sum_{n>=1} (-1)^(n+1)/a(n) = 1/24. (End)
a(n) = 6*A005563(n). - Hugo Pfoertner, May 24 2023
MAPLE
seq(6*n*(n+2), n=1..45); # G. C. Greubel, Sep 01 2019
MATHEMATICA
Select[ Range[15000], IntegerQ[ Sqrt[ 6(6 + # )]] & ]
CoefficientList[Series[6*(3-x)/(1-x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Jul 08 2012 *)
6*(Range[2, 45]^2 -1) (* G. C. Greubel, Sep 01 2019 *)
LinearRecurrence[{3, -3, 1}, {18, 48, 90}, 60] (* Harvey P. Dale, May 10 2022 *)
PROG
(PARI) a(n)=6*n*(n+2) \\ Charles R Greathouse IV, Dec 07 2011
(Magma) [6*n*(n+2): n in [1..50]]; // Vincenzo Librandi, Jul 08 2012
(Sage) [6*n*(n+2) for n in (1..45)] # G. C. Greubel, Sep 01 2019
(GAP) List([1..45], n-> 6*n*(n+2)); # G. C. Greubel, Sep 01 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Robert G. Wilson v, Feb 05 2002
STATUS
approved