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a(n) = 6*n^2 + 12*n.
8

%I #50 May 25 2023 08:06:13

%S 18,48,90,144,210,288,378,480,594,720,858,1008,1170,1344,1530,1728,

%T 1938,2160,2394,2640,2898,3168,3450,3744,4050,4368,4698,5040,5394,

%U 5760,6138,6528,6930,7344,7770,8208,8658,9120,9594,10080,10578,11088,11610

%N a(n) = 6*n^2 + 12*n.

%C Positive numbers k such that 6*(6 + k) is a perfect square.

%H Vincenzo Librandi, <a href="/A067726/b067726.txt">Table of n, a(n) for n = 1..1000</a>

%H Leo Tavares, <a href="/A067726/a067726.jpg">Illustration: Star Cut Hexagons</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F G.f.: 6*x*(3 - x)/(1 - x)^3. - _Vincenzo Librandi_, Jul 08 2012

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - _Vincenzo Librandi_, Jul 08 2012

%F E.g.f.: 6*x*(3 + x)*exp(x). - _G. C. Greubel_, Sep 01 2019

%F From _Amiram Eldar_, Feb 25 2022: (Start)

%F Sum_{n>=1} 1/a(n) = 1/8.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 1/24. (End)

%F a(n) = A003215(2*n) - A003154(n). - _Leo Tavares_, May 20 2023

%F a(n) = 6*A005563(n). - _Hugo Pfoertner_, May 24 2023

%p seq(6*n*(n+2), n=1..45); # _G. C. Greubel_, Sep 01 2019

%t Select[ Range[15000], IntegerQ[ Sqrt[ 6(6 + # )]] & ]

%t CoefficientList[Series[6*(3-x)/(1-x)^3,{x,0,50}],x] (* _Vincenzo Librandi_, Jul 08 2012 *)

%t 6*(Range[2, 45]^2 -1) (* _G. C. Greubel_, Sep 01 2019 *)

%t LinearRecurrence[{3,-3,1},{18,48,90},60] (* _Harvey P. Dale_, May 10 2022 *)

%o (PARI) a(n)=6*n*(n+2) \\ _Charles R Greathouse IV_, Dec 07 2011

%o (Magma) [6*n*(n+2): n in [1..50]]; // _Vincenzo Librandi_, Jul 08 2012

%o (Sage) [6*n*(n+2) for n in (1..45)] # _G. C. Greubel_, Sep 01 2019

%o (GAP) List([1..45], n-> 6*n*(n+2)); # _G. C. Greubel_, Sep 01 2019

%Y Cf. numbers k such that k*(k + m) is a perfect square: A028560 (k=9), A067728 (k=8), A067727 (k=7), A067724 (k=5), A028347 (k=4), A067725 (k=3), A054000 (k=2), A005563 (k=1).

%Y Cf. A003154, A003215.

%K nonn,easy

%O 1,1

%A _Robert G. Wilson v_, Feb 05 2002