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A168566 a(n) = (n-1)*(n+2)*(n^2 + n + 2)/4.

%I A168566 #41 Jun 01 2025 22:15:18
%S A168566 0,8,35,99,224,440,783,1295,2024,3024,4355,6083,8280,11024,14399,
%T A168566 18495,23408,29240,36099,44099,53360,64008,76175,89999,105624,123200,
%U A168566 142883,164835,189224,216224,246015,278783,314720,354024,396899,443555,494208
%N A168566 a(n) = (n-1)*(n+2)*(n^2 + n + 2)/4.
%C A168566 The products of two consecutive numbers in this sequence may be evaluated in terms of the Frobenius numbers for 5 consecutive integers, A138985(n) = F(n): for n>0, a(2n-1)*a(2n) = F(4n^2-2)^2 - (2n)^2; a(2n)*a(2n+1) = F(4n^2+4n)^2 - (2n+1)^2. - _Charlie Marion_, Jan 23 2012
%H A168566 Vincenzo Librandi, <a href="/A168566/b168566.txt">Table of n, a(n) for n = 1..10000</a>
%H A168566 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A168566 G.f.: x^2*(-8 + 5*x - 4*x^2 + x^3)/(x-1)^5. - _R. J. Mathar_, Jan 04 2011
%F A168566 a(n) = A000217(n)^2 - 1 = A000537(n)-1. - _Charlie Marion_, Sep 27 2011
%F A168566 From _G. C. Greubel_, Jul 26 2016: (Start)
%F A168566 a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
%F A168566 E.g.f.: (1/4)*(4 - (4 - 4*x + 14*x^2 + 8*x^3)*exp(x)). (End)
%F A168566 Sum_{n>=2} 1/a(n) = 49/36 - tanh(sqrt(7)*Pi/2)*Pi/sqrt(7). - _Amiram Eldar_, Mar 02 2023
%t A168566 s=0;lst={s};Do[s+=n^3;AppendTo[lst,s],{n,2,5!}];lst (* _Vladimir Joseph Stephan Orlovsky_, Apr 27 2010 *)
%t A168566 Table[(n-1)*(n+2)*(n^2 + n + 2)/4, {n,1,25}] (* _G. C. Greubel_, Jul 26 2016 *)
%t A168566 LinearRecurrence[{5,-10,10,-5,1},{0,8,35,99,224},40] (* _Harvey P. Dale_, Jan 21 2019 *)
%o A168566 (Magma) [(n-1)*(n+2)*(n^2+n+2)/4: n in [1..50]];
%o A168566 (PARI) a(n)=(n-1)*(n+2)*(n^2+n+2)/4 \\ _Charles R Greathouse IV_, Jul 26 2016
%Y A168566 Cf. A000578 (first differences), A000217, A000537, A138985.
%K A168566 nonn,easy
%O A168566 1,2
%A A168566 _Vincenzo Librandi_, Nov 30 2009