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%I #33 Sep 08 2022 08:45:59
%S 0,4,116,3144,84904,2292428,61895580,1671180688,45121878608,
%T 1218290722452,32893849506244,888133936668632,23979616290053112,
%U 647449639831434076,17481140275448720108,471990787437115442976,12743751260802116960416,344081284041657157931300
%N a(n) = (3^(3*n + 3)- 26*n - 27)/169.
%C Second differences are four times the entries of A009971. - _R. J. Mathar_, Oct 25 2011
%H Vincenzo Librandi, <a href="/A198080/b198080.txt">Table of n, a(n) for n = 0..700</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (29,-55,27).
%F a(n) = (3^(3*n + 3) - 26*n - 27)/169.
%F G.f.: -4*x / ( (27*x-1)*(x-1)^2 ). - _R. J. Mathar_, Oct 25 2011
%e a(1) = (3^(3 + 3) - 26 - 27)/169 = 676/169 = 4.
%p for n from 0 to 30 do:x:=(3^(3*n+3) - 26*n - 27)/169 : printf(`%d, `, x):od:
%t LinearRecurrence[{29,-55,27},{0,4,116},50] (* _Vincenzo Librandi_, Nov 25 2011 *)
%o (Magma) I:=[0, 4, 116]; [n le 3 select I[n] else 29*Self(n-1)-55*Self(n-2)+27*Self(n-3): n in [1..20]]; // _Vincenzo Librandi_, Nov 25 2011
%o (PARI) a(n)=(3^(3*n+3)-26*n-27)/169 \\ _Charles R Greathouse IV_, Jul 06 2017
%K nonn,easy
%O 0,2
%A _Michel Lagneau_, Oct 24 2011
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