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a(n) = n*(27*n + 1)/2.
3

%I #35 Aug 23 2017 23:39:58

%S 0,14,55,123,218,340,489,665,868,1098,1355,1639,1950,2288,2653,3045,

%T 3464,3910,4383,4883,5410,5964,6545,7153,7788,8450,9139,9855,10598,

%U 11368,12165,12989,13840,14718,15623

%N a(n) = n*(27*n + 1)/2.

%H G. C. Greubel, <a href="/A022285/b022285.txt">Table of n, a(n) for n = 0..5000</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) + 27*n - 13 for n>0, a(0)=0. - _Vincenzo Librandi_, Aug 04 2010

%F a(0)=0, a(1)=14, a(2)=55; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). -_Harvey P. Dale_, Sep 20 2011

%F G.f.: x*(13*x + 14)/(1-x)^3. - _Harvey P. Dale_, Sep 20 2011

%F a(n) = 12/(n+2)!*Sum_{j=0..n} (-1)^(n-j)*binomial(n,j)*(j+n)^(n+2). - _Vladimir Kruchinin_, Jun 04 2013

%F a(n) = A000217(14*n) - A000217(13*n). - _Bruno Berselli_, Oct 13 2016

%F E.g.f.: (x/2)*(27*x + 28)*exp(x). - _G. C. Greubel_, Aug 23 2017

%p A022285:=n->n*(27*n+1)/2; seq(A022285(k), k=0..100); # _Wesley Ivan Hurt_, Nov 04 2013

%t Table[n (27 n + 1)/2, {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 14, 55}, 40] (* _Harvey P. Dale_, Sep 20 2011 *)

%o (PARI) a(n)=n*(27*n+1)/2 \\ _Charles R Greathouse IV_, Jun 17 2017

%Y Cf. similar sequences listed in A022289.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_