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a(n) = 5*n^2 + 10*n.
8

%I #34 Oct 01 2023 08:59:30

%S 15,40,75,120,175,240,315,400,495,600,715,840,975,1120,1275,1440,1615,

%T 1800,1995,2200,2415,2640,2875,3120,3375,3640,3915,4200,4495,4800,

%U 5115,5440,5775,6120,6475,6840,7215,7600,7995,8400,8815,9240,9675

%N a(n) = 5*n^2 + 10*n.

%C Positive numbers m such that 5*(5 + m) is a perfect square.

%H Vincenzo Librandi, <a href="/A067724/b067724.txt">Table of n, a(n) for n = 1..1000</a>

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

%F From _Vincenzo Librandi_, Jul 08 2012: (Start)

%F G.f.: 5*x*(3 - x)/(1 - x)^3.

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)

%F a(n) = A055998(3*n) + A055998(n). - _Bruno Berselli_, Sep 23 2016

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

%F Sum_{n>=1} 1/a(n) = 3/20.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 1/20. (End)

%F E.g.f.: 5*exp(x)*x*(3 + x). - _Stefano Spezia_, Oct 01 2023

%t Select[Range[10000], IntegerQ[ Sqrt[5 (5 + # )]] &]

%t CoefficientList[Series[5 (3 - x)/(1 - x)^3, {x, 0, 40}], x] (* _Vincenzo Librandi_, Jul 08 2012 *)

%t Table[5n^2+10n,{n,60}] (* or *) LinearRecurrence[{3,-3,1},{15,40,75},60] (* _Harvey P. Dale_, May 22 2018 *)

%o (PARI) a(n)=5*n*(n+2) \\ _Charles R Greathouse IV_, Dec 07 2011

%o (Magma) [5*n*(n+2): n in [1..50]]; // _Vincenzo Librandi_, Jul 08 2012

%Y Cf. numbers k such that k*(k + m) is a perfect square: A028560 (k=9), A067728 (k=8), A067727 (k=7), A067726 (k=6), A028347 (k=4), A067725 (k=3), A054000 (k=2), A067998 (k=1).

%Y Cf. A055998.

%K nonn,easy

%O 1,1

%A _Robert G. Wilson v_, Feb 05 2002