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A116149 a(n) = sum of n consecutive cubes after n^3.

%I A116149 #28 Mar 10 2025 13:23:24
%S A116149 8,91,405,1196,2800,5643,10241,17200,27216,41075,59653,83916,114920,
%T A116149 153811,201825,260288,330616,414315,512981,628300,762048,916091,
%U A116149 1092385,1292976,1520000,1775683,2062341,2382380,2738296,3132675,3568193
%N A116149 a(n) = sum of n consecutive cubes after n^3.
%H A116149 G. C. Greubel, <a href="/A116149/b116149.txt">Table of n, a(n) for n = 1..1000</a>
%H A116149 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A116149 a(n) = n^2*(1 + 3*n)*(3 + 5*n)/4.
%F A116149 G.f.: x*(8 +51*x + 30*x^2 + x^3)/(1-x)^5. - _Colin Barker_, Dec 17 2012
%F A116149 a(n) = A000217(2*n)^2 - A000217(n)^2. - _Bruno Berselli_, Aug 31 2017
%F A116149 From _G. C. Greubel_, May 10 2019: (Start)
%F A116149 a(n) = Sum_{k=(n+1)..2*n} k^3.
%F A116149 E.g.f.: x*(32 + 150*x + 104*x^2 + 15*x^3)*exp(x)/4. (End)
%e A116149 a(1) = sum of 1 cube  after 1^3 = 2^3 = 8,
%e A116149 a(2) = sum of 2 cubes after 2^3 = 3^3+4^3 = 91,
%e A116149 a(3) = sum of 3 cubes after 3^3 = 4^3+5^3+6^3 = 405,
%e A116149 a(4) = sum of 4 cubes after 4^3 = 5^3+6^3+7^3+8^3 = 1196.
%t A116149 With[{cbs=Range[100]^3},Table[Total[Take[cbs,{n+1,2n}]],{n,35}]]  (* _Harvey P. Dale_, Feb 13 2011 *)
%o A116149 (PARI) {a(n) = n^2*(1+3*n)*(3+5*n)/4}; \\ _G. C. Greubel_, May 10 2019
%o A116149 (Magma) [n^2*(1+3*n)*(3+5*n)/4: n in [1..40]]; // _G. C. Greubel_, May 10 2019
%o A116149 (Sage) [n^2*(1+3*n)*(3+5*n)/4 for n in (1..40)] # _G. C. Greubel_, May 10 2019
%o A116149 (GAP) List([1..40], n-> n^2*(1+3*n)*(3+5*n)/4); # _G. C. Greubel_, May 10 2019
%Y A116149 Cf. A000217, A240137.
%K A116149 nonn,easy
%O A116149 1,1
%A A116149 _Zak Seidov_, Apr 14 2007