%I #38 Sep 08 2022 08:46:08
%S 108,175,176,135,76,23,0,31,140,351,688,1175,1836,2695,3776,5103,6700,
%T 8591,10800,13351,16268,19575,23296,27455,32076,37183,42800,48951,
%U 55660,62951,70848,79375,88556,98415,108976,120263,132300,145111,158720,173151,188428
%N a(n) = 27*(n - 6)^2 + 4*(n - 6)^3 = ((n - 6)^2)*(4*n + 3).
%C The discriminant D of the Cardano Tartaglia cubic equation x^3 + p*x +q = 0 is D = -27*q^2 - 4*p^3, D < 0. Let D = (-1)*D then D = 27*q^2 + 4*p^3, D > 0, q = p > -6. So D = 27*(n - 6)^2 + 4*(n - 6)^3, D > 0, q = p = n, n >= 0 for all real solutions of D as positive integers. The case D < 0 is treated in A245033. Remark: ((n - 6)^2)*(4*n + 3) = ((6 - n)^2)*(4*n + 3).
%H Freimut Marschner, <a href="/A245032/b245032.txt">Table of n, a(n) for n = 0..9999</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F a(n) = 27*(n - 6)^2 + 4*(n - 6)^3 = ((n - 6)^2)*(4*n + 3).
%F G.f.: (49*x^3+124*x^2-257*x+108) / (x-1)^4. - _Colin Barker_, Jul 11 2014
%F a(n) = (4n + 3) * (n - 6)^2 = A004767(n) * A000290(n-6). - _Wesley Ivan Hurt_, Jul 18 2014
%e n = 0, a(0) = 27*(-6)^2 + 4*(-6)^3 = 27*36 + 4*(-216) = 108;
%e n = 7, a(7) = 27*(1)^2 + 4*(1)^3 = 27 + 4 = 31.
%p A245032:=n->(4*n+3)*(n-6)^2: seq(A245032(n), n=0..50); # _Wesley Ivan Hurt_, Jul 18 2014
%t Table[(4 n + 3)(n - 6)^2, {n, 0, 50}] (* _Wesley Ivan Hurt_, Jul 18 2014 *)
%t LinearRecurrence[{4,-6,4,-1},{108,175,176,135},50] (* _Harvey P. Dale_, Oct 13 2019 *)
%o (PARI) vector(100, n, (n-7)^2*(4*n-1)) \\ _Colin Barker_, Jul 11 2014
%o (PARI) Vec((49*x^3+124*x^2-257*x+108)/(x-1)^4 + O(x^100)) \\ _Colin Barker_, Jul 11 2014
%o (Magma) [(4*n+3)*(n-6)^2 : n in [0..50]]; // _Wesley Ivan Hurt_, Jul 18 2014
%Y Cf. A028347 (Discriminant of quadratic equation : a(n) = n^2 - 4*n, n > 2), A245033 (a(n) = 4*(n + 7)^3 - 27*(n + 7)^2 = (4*n +1)*(n+7)^2), A245035 (a(n) = (-1)*( -27*(prime(n) - 7)^2 - 4*(prime(n) - 7)^3 ) = (prime(n) - 7)^2 * (4* prime(n) - 1) with n >= 1), A245036 (a(n) = 4*prime(n)^3 - 27*prime(n)^2 = (prime(n)^2)*[4*prime(n) - 27], n >= 4), A004767 (4*n+3).
%K nonn,easy
%O 0,1
%A _Freimut Marschner_, Jul 10 2014