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a(n) = 2*n*(n+3).
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%I #33 Dec 23 2022 07:40:09

%S 0,8,20,36,56,80,108,140,176,216,260,308,360,416,476,540,608,680,756,

%T 836,920,1008,1100,1196,1296,1400,1508,1620,1736,1856,1980,2108,2240,

%U 2376,2516,2660,2808,2960,3116,3276,3440,3608,3780,3956,4136,4320,4508,4700,4896

%N a(n) = 2*n*(n+3).

%C Numbers n such that 2*n+9 is a square. - _Vincenzo Librandi_, Nov 24 2010

%C a(n) appears also as the fourth member of the quartet [p0(n), p1(n), p2(n), a(n)] of the square of [n, n+1, n+2, n+3] in the Clifford algebra Cl_2 for n >= 0. p0(n) = -A147973(n+3), p1 = A046092(n) and p2(n) = A054000(n+1). See a comment on A147973, also with a reference. - _Wolfdieter Lang_, Oct 15 2014

%H Vincenzo Librandi, <a href="/A139570/b139570.txt">Table of n, a(n) for n = 0..1000</a>

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

%F a(n) = A028552(n)*2 = 2*n^2+6n = n(2n+6).

%F a(n) = a(n-1)+4*n+4 (with a(0)=0). - _Vincenzo Librandi_, Nov 24 2010

%F a(n) = A022998(n) * A022998(n+3). - _Paul Curtz_, Mar 27 2011

%F a(n) = 4 * A000096(n). - _Paul Curtz_, Mar 27 2011

%F G.f.: 4*x*(2 - x)/(1 - x)^3. - _Arkadiusz Wesolowski_, Dec 31 2011

%F From _Amiram Eldar_, Dec 23 2022: (Start)

%F Sum_{n>=1} 1/a(n) = 11/36.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = log(2)/3 - 5/36. (End)

%t CoefficientList[Series[4 x (2 - x)/(1 - x)^3, {x, 0, 40}], x] (* _Vincenzo Librandi_, May 23 2014 *)

%o (PARI) a(n)=2*n*(n+3) \\ _Charles R Greathouse IV_, Jun 17 2017

%Y Cf. A001105, A028552, A046092, A054000, A067728.

%K easy,nonn

%O 0,2

%A _Omar E. Pol_, May 19 2008