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A100040 a(n) = 2*n^2 + n - 5.

%I A100040 #39 Sep 08 2022 08:45:15
%S A100040 -5,-2,5,16,31,50,73,100,131,166,205,248,295,346,401,460,523,590,661,
%T A100040 736,815,898,985,1076,1171,1270,1373,1480,1591,1706,1825,1948,2075,
%U A100040 2206,2341,2480,2623,2770,2921,3076,3235,3398,3565,3736,3911,4090,4273
%N A100040 a(n) = 2*n^2 + n - 5.
%C A100040 a(n) is the result of taking five consecutive numbers starting at n-2, then adding the products of the first and the last and of the second with the fourth and finally adding the middle term. That is, a(n) = (n^2-4) + (n^2-1) + n. - _J. M. Bergot_, Mar 06 2018
%H A100040 G. C. Greubel, <a href="/A100040/b100040.txt">Table of n, a(n) for n = 0..5000</a>
%H A100040 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A100040 A100035(a(n)) = 4 for n>3;
%F A100040 a(n) = A014105(n) - 5 = A084849(n) - 6 = A100041(n) + 2.
%F A100040 a(n) = 2*a(n-1)-a(n-2)+4; a(0)=-5, a(1)=-2. - _Vincenzo Librandi_, Dec 26 2010
%F A100040 G.f.: (-5 + 13*x - 4*x^2)/(1 - x)^3. - _Arkadiusz Wesolowski_, Dec 25 2011
%F A100040 E.g.f.: (2*x^2 + 3*x - 5)*exp(x). - _G. C. Greubel_, Jul 15 2017
%p A100040 [seq(2*n^2+n-5,n=0..50)]; # _Muniru A Asiru_, Mar 20 2018
%t A100040 Table[2*n^2 + n - 5, {n, 0, 50}] (* _G. C. Greubel_, Jul 15 2017 *)
%t A100040 LinearRecurrence[{3,-3,1},{-5,-2,5},50] (* _Harvey P. Dale_, Sep 21 2017 *)
%o A100040 (Magma) [ 2*n^2+n-5: n in [0..50] ];
%o A100040 (PARI) a(n)=2*n^2+n-5 \\ _Charles R Greathouse IV_, Sep 24 2015
%o A100040 (GAP) List([0..50],n->2*n^2+n-5); # _Muniru A Asiru_, Mar 20 2018
%Y A100040 Cf. A100036, A100037, A100038, A100039.
%K A100040 sign,easy
%O A100040 0,1
%A A100040 _Reinhard Zumkeller_, Oct 31 2004