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A116668 a(n) = (5*n^2 + n + 2)/2.

%I A116668 #37 Oct 25 2024 15:35:31
%S A116668 1,4,12,25,43,66,94,127,165,208,256,309,367,430,498,571,649,732,820,
%T A116668 913,1011,1114,1222,1335,1453,1576,1704,1837,1975,2118,2266,2419,2577,
%U A116668 2740,2908,3081,3259,3442,3630,3823,4021,4224,4432,4645,4863,5086,5314,5547
%N A116668 a(n) = (5*n^2 + n + 2)/2.
%C A116668 Binomial transform of (1, 3, 5, 0, 0, 0, ...).
%H A116668 G. C. Greubel, <a href="/A116668/b116668.txt">Table of n, a(n) for n = 0..5000</a>
%H A116668 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A116668 Product of Pascal's triangle as an infinite lower triangular matrix and the vector (1, 3, 5, 0, 0, 0, ...).
%F A116668 O.g.f.: (1+x+3*x^2)/(1-x)^3. - _R. J. Mathar_, Apr 02 2008
%F A116668 a(n) = 5*n + a(n-1) - 2 (with a(0)=1) - _Vincenzo Librandi_, Nov 13 2010
%F A116668 From _Elmo R. Oliveira_, Oct 25 2024: (Start)
%F A116668 E.g.f.: exp(x)*(1 + 3*x + 5*x^2/2).
%F A116668 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)
%e A116668 a(3) = 1*1 + 3*3 + 3*5 + 1*0 = 25.
%p A116668 a:=n->(5*n^2+n+2)/2: seq(a(n),n=0..50); # _Emeric Deutsch_, Feb 28 2006
%t A116668 s = 1; lst = {s}; Do[s += n + 2; AppendTo[lst, s], {n, 1, 235, 5}] lst (* _Zerinvary Lajos_, Jul 11 2009 *)
%t A116668 LinearRecurrence[{3,-3,1}, {1,4,12}, 50] (* _G. C. Greubel_, Jan 29 2018 *)
%o A116668 (PARI) a(n)=(5*n^2+n+2)/2 \\ _Charles R Greathouse IV_, Jun 17 2017
%o A116668 (Magma) [(5*n^2 + n+2)/2: n in [0..50]]; // _G. C. Greubel_, Jan 29 2018
%o A116668 (GAP) List([0..1000],n->(5*n^2+n+2)/2); # _Muniru A Asiru_, Jan 30 2018
%Y A116668 Cf. A116666.
%K A116668 nonn,easy
%O A116668 0,2
%A A116668 _Gary W. Adamson_, Feb 22 2006
%E A116668 More terms from _Emeric Deutsch_, Feb 28 2006