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A255499 a(n) = (n^4 + 2*n^3 - n^2)/2.

%I A255499 #44 Sep 08 2022 08:46:11
%S A255499 0,1,14,63,184,425,846,1519,2528,3969,5950,8591,12024,16393,21854,
%T A255499 28575,36736,46529,58158,71839,87800,106281,127534,151823,179424,
%U A255499 210625,245726,285039,328888,377609,431550,491071,556544,628353,706894,792575,885816,987049,1096718,1215279,1343200
%N A255499 a(n) = (n^4 + 2*n^3 - n^2)/2.
%H A255499 G. C. Greubel, <a href="/A255499/b255499.txt">Table of n, a(n) for n = 0..1000</a>
%H A255499 L. Kaylor and D. Offner, <a href="http://dx.doi.org/10.2140/involve.2014.7.627">Counting matrices over a finite field with all eigenvalues in the field</a>, Involve, a Journal of Mathematics, Vol. 7 (2014), No. 5, 627-645, DOI: 10.2140/involve.2014.7.627; see Eq. (1).
%H A255499 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A255499 G.f.: x*(1+9*x+3*x^2-x^3)/(1-x)^5. - _Vincenzo Librandi_, Sep 05 2015
%F A255499 a(n) = 5*a(n-1) -10*a(n-2) +10*a(n-3) -5*a(n-4) +a(n-5). - _Vincenzo Librandi_, Sep 05 2015
%F A255499 a(n) = Sum_{k=n..n+n^2-1} k (the sum of the first n^2 integers starting with n). - _Matthew Niemiro_, Jun 26 2020
%F A255499 E.g.f.: (x/2)*(2 +12*x +8*x^2 +x^3)*exp(x). - _G. C. Greubel_, Sep 24 2021
%t A255499 Table[n^4/2 + n^3 - n^2/2, {n, 0, 60}] (* or *) CoefficientList[Series[x (1 + 9 x + 3 x^2 - x^3)/(1 - x)^5, {x, 0, 45}], x] (* _Vincenzo Librandi_, Sep 05 2015 *)
%t A255499 LinearRecurrence[{5,-10,10,-5,1},{0,1,14,63,184},50] (* _Harvey P. Dale_, Nov 11 2017 *)
%o A255499 (PARI) a(n) = n^4/2+n^3-n^2/2; \\ _Michel Marcus_, Sep 05 2015
%o A255499 (Magma) [n^4/2+n^3-n^2/2: n in [0..40]] // _Vincenzo Librandi_, Sep 05 2015
%o A255499 (Sage) [n^2*(n^2 +2*n -1)/2 for n in (0..40)] # _G. C. Greubel_, Sep 24 2021
%Y A255499 Cf. A229738.
%K A255499 nonn,easy
%O A255499 0,3
%A A255499 _N. J. A. Sloane_, Mar 13 2015