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a(n) = 4*n^2 + 28*n + 10.
1

%I #39 Mar 02 2023 02:54:30

%S 42,82,130,186,250,322,402,490,586,690,802,922,1050,1186,1330,1482,

%T 1642,1810,1986,2170,2362,2562,2770,2986,3210,3442,3682,3930,4186,

%U 4450,4722,5002,5290,5586,5890,6202,6522,6850,7186,7530,7882,8242,8610,8986,9370

%N a(n) = 4*n^2 + 28*n + 10.

%C Sequence gives values of x such that x^3 + 39x^2 = y^2 since a(n)^3 + 39*a(n)^2 = (8n^3 + 84n^2 + 216n + 70)^2.

%C a(n) = 2*(seventh diagonal to A153238).

%C About the first comment, naturally, the complete list of nonnegative values of x in x^3 + 39*x^2 = y^2 is given by x = m^2-39 with m>6. - _Bruno Berselli_, Jan 25 2012

%H Vincenzo Librandi, <a href="/A153644/b153644.txt">Table of n, a(n) for n = 1..10000</a>

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

%F From _Colin Barker_, Jan 24 2012: (Start)

%F a(1)=42, a(2)=82, a(3)=130, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

%F G.f.: 2*x*((3-x)*(7-5*x))/(1-x)^3. (End)

%F E.g.f.: 2*(-5 + (5 + 16*x + 2*x^2)*exp(x)). - _G. C. Greubel_, Aug 23 2016

%F Sum_{n>=1} 1/a(n) = 62/1995 + tan(sqrt(39)*Pi/2)*Pi/(4*sqrt(39)). - _Amiram Eldar_, Mar 02 2023

%t LinearRecurrence[{3,-3,1},{42,82,130}, 25] (* _G. C. Greubel_, Aug 23 2016 *)

%t Table[4n^2+28n+10,{n,70}] (* _Harvey P. Dale_, Jan 15 2023 *)

%o (PARI) a(n)=4*n*(n+7)+10 \\ _Charles R Greathouse IV_, Jan 24 2012

%o (Magma) [4*n^2 + 28*n + 10: n in [1..50]]; // _Vincenzo Librandi_, Jan 25 2012

%Y Cf. A153238, A067076.

%K nonn,easy,less

%O 1,1

%A _Vincenzo Librandi_, Dec 30 2008