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Zero together with the partial sums of A064412.
1

%I #27 Jun 19 2022 11:03:54

%S 0,1,6,20,52,112,215,375,613,948,1407,2013,2799,3793,5034,6554,8398,

%T 10603,13220,16290,19870,24006,28761,34185,40347,47302,55125,63875,

%U 73633,84463,96452,109668,124204,140133,157554,176544,197208,219628,243915,270155,298465,328936,361691,396825,434467,474717,517710,563550,612378,664303,719472

%N Zero together with the partial sums of A064412.

%H Colin Barker, <a href="/A275112/b275112.txt">Table of n, a(n) for n = 0..1000</a>

%H Luce ETIENNE, <a href="/A275112/a275112.pdf">Illustration of initial terms of this sequence and A064412</a>

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (3,-2,-2,4,-4,2,2,-3,1).

%F a(n) = (28*n^4+36*n^3+18*n^2+12*n+(1-(-1)^n))/16 for n even.

%F a(n) = (28*n^4+92*n^3+114*n^2+68*n+17-(-1)^n)/16 for n odd.

%F a(n) = (14*n^4+36*n^3+36*n^2+42*n+11+3*(2*n-1)*(-1)^n-8*(-1)^(((2*n-1+(-1)^n))/4))/128.

%F G.f.: x*(1+x+x^2)*(1+2*x+x^2+3*x^3) / ((1-x)^5*(1+x)^2*(1+x^2)). - _Colin Barker_, Jul 18 2016

%t {0}~Join~Accumulate@ CoefficientList[Series[(1 + x + x^2) (1 + 2 x + x^2 + 3 x^3)/((1 - x)^2 (1 - x^2) (1 - x^4)), {x, 0, 49}], x] (* _Michael De Vlieger_, Jul 18 2016, after _Wesley Ivan Hurt_ at A064412, or *)

%t Table[(14 n^4 + 36 n^3 + 36 n^2 + 42 n + 11 + 3 (2 n - 1) (-1)^n - 8 (-1)^(((2 n - 1 + (-1)^n))/4))/128, {n, 50}] (* _Michael De Vlieger_, Jul 18 2016 *)

%t LinearRecurrence[{3,-2,-2,4,-4,2,2,-3,1},{0,1,6,20,52,112,215,375,613},60] (* _Harvey P. Dale_, Jun 19 2022 *)

%o (PARI) concat(0, Vec(x*(1+x+x^2)*(1+2*x+x^2+3*x^3)/((1-x)^5*(1+x)^2*(1+x^2)) + O(x^50))) \\ _Colin Barker_, Jul 18 2016

%Y Cf. A064412, A213389, A006003.

%K nonn,easy

%O 0,3

%A _Luce ETIENNE_, Jul 17 2016