login
a(n) = n(2n+17).
5

%I #22 Jul 07 2024 19:50:06

%S 0,19,42,69,100,135,174,217,264,315,370,429,492,559,630,705,784,867,

%T 954,1045,1140,1239,1342,1449,1560,1675,1794,1917,2044,2175,2310,2449,

%U 2592,2739,2890,3045,3204,3367,3534,3705,3880,4059

%N a(n) = n(2n+17).

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

%F a(n) = 2*n^2 + 17*n.

%F a(n) = a(n-1) + 4*n + 15; a(0) = 0. - _Vincenzo Librandi_, Nov 24 2010

%t a[n_]:=Sum[4*i+15, {i, 1, n}]; (* _Vladimir Joseph Stephan Orlovsky_, Dec 04 2008 *)

%t Table[n(2n+17),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,19,42},50] (* _Harvey P. Dale_, Jul 07 2024 *)

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

%Y Cf. A014105, A014106, A033537, A130861, A139576, A139577, A139578, A139579, A139581.

%K easy,nonn

%O 0,2

%A _Omar E. Pol_, May 19 2008