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a(n) = 4*(4 + 9*n^2 + 15*n).
2

%I #29 Dec 12 2023 08:30:20

%S 16,112,280,520,832,1216,1672,2200,2800,3472,4216,5032,5920,6880,7912,

%T 9016,10192,11440,12760,14152,15616,17152,18760,20440,22192,24016,

%U 25912,27880,29920,32032,34216,36472,38800,41200,43672,46216,48832,51520,54280,57112

%N a(n) = 4*(4 + 9*n^2 + 15*n).

%C A decimation: A061039(6n+5).

%C a(n) mod 9 = period 3: repeat 7,4,1 = A070403(n+1).

%H Vincenzo Librandi, <a href="/A144449/b144449.txt">Table of n, a(n) for n = 0..10000</a>

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

%F a(n) = a(n-1) + 24*(3*n+1) = a(n-1) + 72*n + 24, a(0)=16.

%F A061039(6n+2) = A061039(6n-4) + 24*(3n+1) = a(6n-4) + 72*n + 24, a(2)=16.

%F From _G. C. Greubel_, Mar 06 2022: (Start)

%F G.f.: 8*(2 + 8*x - x^2)/(1-x)^3.

%F E.g.f.: 4*(4 + 24*x + 9*x^2)*exp(x). (End)

%F From _Amiram Eldar_, Mar 11 2022: (Start)

%F Sum_{n>=0} 1/a(n) = 1/12.

%F Sum_{n>=0} (-1)^n/a(n) = Pi/(18*sqrt(3)) + log(2)/18 - 1/12. (End)

%t Table[36n^2+60n+16,{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{16,112,280},40] (* _Harvey P. Dale_, Apr 04 2020 *)

%o (Magma) [36*n^2 + 60*n + 16: n in [0..40]]; // _Vincenzo Librandi_, Aug 07 2011

%o (PARI) a(n)=36*n^2+60*n+16 \\ _Charles R Greathouse IV_, Jun 17 2017

%o (Sage) [(6*n+5)^2 - 9 for n in (0..40)] # _G. C. Greubel_, Mar 06 2022

%Y Cf. A061039, A070403.

%Y Subsequence of A008590.

%K nonn,easy

%O 0,1

%A _Paul Curtz_, Oct 06 2008

%E Edited by _Charles R Greathouse IV_, Jul 25 2010