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a(n) = n*(23*n + 1)/2.
4

%I #24 Aug 23 2017 23:39:27

%S 0,12,47,105,186,290,417,567,740,936,1155,1397,1662,1950,2261,2595,

%T 2952,3332,3735,4161,4610,5082,5577,6095,6636,7200,7787,8397,9030,

%U 9686,10365,11067,11792,12540,13311

%N a(n) = n*(23*n + 1)/2.

%H G. C. Greubel, <a href="/A022281/b022281.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) = 23*n + a(n-1) - 11 for n>0, a(0)=0. - _Vincenzo Librandi_, Aug 04 2010

%F G.f.: x*(12 + 11*x)/(1 - x)^3 . - _R. J. Mathar_, Aug 04 2016

%F a(n) = A000217(12*n) - A000217(11*n). - _Bruno Berselli_, Oct 13 2016

%F E.g.f.: (x/2)*(23*x + 24)*exp(x). - _G. C. Greubel_, Aug 23 2017

%t Table[n (23 n + 1)/2, {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 12, 47}, 40] (* _Harvey P. Dale_, Aug 16 2016 *)

%o (PARI) a(n)=n*(23*n+1)/2 \\ _Charles R Greathouse IV_, Jun 16 2017

%Y Cf. similar sequences listed in A022289.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_