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%I #19 Dec 11 2022 02:38:41
%S 43,28,15,4,-5,-12,-17,-20,-21,-20,-17,-12,-5,4,15,28,43,60,79,100,
%T 123,148,175,204,235,268,303,340,379,420,463,508,555,604,655,708,763,
%U 820,879,940,1003,1068,1135,1204,1275,1348,1423,1500,1579,1660,1743,1828,1915,2004,2095,2188
%N Q(n,5), where Q(m,k) is defined in A127080 and A127137.
%C Numbers m such that m + 21 is a square. The product of two consecutive terms belongs to the sequence, see formula. - _Klaus Purath_, Oct 30 2022
%D V. van der Noort and N. J. A. Sloane, Paper in preparation, 2007.
%H Colin Barker, <a href="/A127147/b127147.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = n^2 - 16*n + 43.
%F From _Colin Barker_, Nov 12 2014: (Start)
%F a(n) = 3*a(n-1) -3*a(n-2) +a(n-3).
%F G.f.: (43 -101*x + 60*x^2)/(1-x)^3. (End)
%F E.g.f.: (43 - 15*x + x^2)*exp(x). - _G. C. Greubel_, Aug 12 2019
%F From _Klaus Purath_, Oct 30 2022: (Start)
%F According to the formula a(n) = n^2 - 16*n + 43 when expanded to negative indices, a(n)*a(n+1) = a(a(n)+n) = (a(n)+n)*(a(n+1)-(n+1)) + 43.
%F a(n) = 2*a(n-1) - a(n-2) + 2. (End)
%p seq((n-8)^2 -21, n=0..60); # _G. C. Greubel_, Aug 12 2019
%t CoefficientList[Series[(60x^2 -101x +43)/(1-x)^3, {x,0,60}], x] (* _Vincenzo Librandi_, Nov 12 2014 *)
%t (Range[0,60] -8)^2 -21 (* _G. C. Greubel_, Aug 12 2019 *)
%o (PARI) Vec(-(60*x^2-101*x+43)/(x-1)^3 + O(x^60)) \\ _Colin Barker_, Nov 12 2014
%o (Magma) [n^2-16*n+43: n in [0..60]]; // _Vincenzo Librandi_, Nov 12 2014
%o (Sage) [(n-8)^2 -21 for n in (0..60)] # _G. C. Greubel_, Aug 12 2019
%o (GAP) List([0..60], n-> (n-8)^2 -21); # _G. C. Greubel_, Aug 12 2019
%Y A row of A127080.
%K sign,easy
%O 0,1
%A _N. J. A. Sloane_, Mar 24 2007
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