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%I A294070 #86 Sep 08 2022 08:46:20
%S A294070 4,6,15,40,96,204,391,690,1140,1786,2679,3876,5440,7440,9951,13054,
%T A294070 16836,21390,26815,33216,40704,49396,59415,70890,83956,98754,115431,
%U A294070 134140,155040,178296,204079,232566,263940,298390,336111,377304,422176,470940,523815
%N A294070 a(n) = (1/4)*(n^2 - 2*n)^2 + (9/4)*(n^2 - 2*n) + 6.
%H A294070 Colin Barker, <a href="/A294070/b294070.txt">Table of n, a(n) for n = 1..1000</a>
%H A294070 SESC NSU Correspondence School, <a href="http://kvant.mccme.ru/pdf/2018/2018-07.pdf#page=44">First assignments for 2018/2019</a> (in Russian), Kvant, 2018, No. 7, p. 42, Mathematics section, 6th grade, exercise no. 2. "Calculate and show in a reduced fraction form the following sum: 1/(2*3) + 2/(3*5) + 3/(5*8) + 4/(8*12) + 5/(12*17)."
%H A294070 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A294070 a(n) = A152948(n) * A152948(n+1).
%F A294070 From _Muniru A Asiru_, Aug 16 2018: (Start)
%F A294070 a(n) = (n^2 - 3*n + 6)*(n^2 - n + 4)/4.
%F A294070 a(n) = A152948(n)*A027689(n-1)/2. (End)
%F A294070 a(n) = A266883(A061925(n-1)). - _Bruno Berselli_, Aug 30 2018
%F A294070 From _Colin Barker_, Nov 26 2018: (Start)
%F A294070 G.f.: x*(4 - 14*x + 25*x^2 - 15*x^3 + 6*x^4)/(1 - x)^5.
%F A294070 a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 5.
%F A294070 a(n) = (24 - 18*n + 13*n^2 - 4*n^3 + n^4)/4. (End)
%F A294070 E.g.f.: (1/4)*exp(x)*(16 + 8*x + 14*x^2 + 6*x^3 + x^4). - _Stefano Spezia_, Nov 30 2018
%e A294070 2*2, 2*3, 3*5, 5*8, 8*12, 12*17, 17*23, 23*30, 30*38, ...
%p A294070 b:=n->(n^2-3*n+6)/2: seq(b(n)*b(n+1),n=1..40); # _Muniru A Asiru_, Aug 16 2018
%t A294070 Times@@@Partition[Array[(#^2 -3# +6)/2 &, 40], 2, 1] (* _Michael De Vlieger_, Sep 24 2018 *)
%t A294070 LinearRecurrence[{5,-10,10,-5,1}, {4,6,15,40,96}, 40] (* _G. C. Greubel_, Feb 10 2019 *)
%o A294070 (GAP) List([1..40],n->(n^2-3*n+6)*(n^2-n+4)/4); # _Muniru A Asiru_, Aug 16 2018
%o A294070 (Magma) [(n^2-3*n+6)*(n^2-n+4)/4: n in [1..40]]; // _Vincenzo Librandi_, Aug 30 2018
%o A294070 (PARI) Vec(x*(4 - 14*x + 25*x^2 - 15*x^3 + 6*x^4)/(1-x)^5 + O(x^40)) \\ _Colin Barker_, Nov 26 2018
%o A294070 (Sage) [(n^2-3*n+6)*(n^2-n+4)/4 for n in (1..40)] # _G. C. Greubel_, Feb 10 2019
%Y A294070 Cf. A027689, A152948.
%K A294070 nonn,easy
%O A294070 1,1
%A A294070 _Jan Lakota Nono_, Aug 14 2018

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