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Nonnegative numbers of form n*(n^2+-1)/2.
4

%I #32 Mar 02 2024 16:36:21

%S 0,1,3,5,12,15,30,34,60,65,105,111,168,175,252,260,360,369,495,505,

%T 660,671,858,870,1092,1105,1365,1379,1680,1695,2040,2056,2448,2465,

%U 2907,2925,3420,3439,3990,4010,4620,4641,5313,5335,6072,6095,6900,6924,7800

%N Nonnegative numbers of form n*(n^2+-1)/2.

%C Taking alternate terms gives A027480 and A006003. - _Jeremy Gardiner_, Apr 10 2005

%H Colin Barker, <a href="/A057587/b057587.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-3,-3,3,1,-1).

%F a(n) = (2*n^3+9*n^2+15*n+5+(3*n^2+n-5)*(-1)^n)/32. - _Luce ETIENNE_, Nov 18 2014

%F From _Wesley Ivan Hurt_, Mar 27 2015: (Start)

%F G.f.: x*(1 + 2 x - x^2 + x^3)/((x - 1)^4*(x + 1)^3).

%F a(n) = a(n-1)+3*a(n-2)-3*a(n-3)-3*a(n-4)+3*a(n-5)+a(n-6)-a(n-7). (End)

%F a(2*n+1) + a(2*n) = (a(2*n+1) - a(2*n))^3. - _Greg Dresden_, Feb 10 2022

%F a(2*n+1)^2 - a(2*n)^2 = (n+1)^4. - _Adam Michael Bere_, Feb 15 2024

%p A057587:=n->(2*n^3+9*n^2+15*n+5+(3*n^2+n-5)*(-1)^n)/32: seq(A057587(n), n=0..50); # _Wesley Ivan Hurt_, Mar 27 2015

%t CoefficientList[Series[x*(1 + 2 x - x^2 + x^3)/((x - 1)^4*(x + 1)^3), {x, 0, 50}], x] (* _Wesley Ivan Hurt_, Mar 27 2015 *)

%t Table[(2 n^3 + 9 n^2 + 15 n + 5 + (3 n^2 + n - 5) (-1)^n) / 32, {n, 0, 50}] (* _Vincenzo Librandi_, Mar 28 2015 *)

%t LinearRecurrence[{1,3,-3,-3,3,1,-1},{0,1,3,5,12,15,30},50] (* _Harvey P. Dale_, Nov 29 2022 *)

%o (PARI) concat(0, Vec(x*(x^3-x^2+2*x+1)/((x-1)^4*(x+1)^3) + O(x^100))) \\ _Colin Barker_, Nov 18 2014

%o (Magma) [(2*n^3+9*n^2+15*n+5+(3*n^2+n-5)*(-1)^n)/32 : n in [0..50]]; // _Wesley Ivan Hurt_, Mar 27 2015

%Y Cf. A027480, A006003.

%K nonn,easy

%O 0,3

%A _N. J. A. Sloane_, Oct 05 2000