OFFSET
0,1
COMMENTS
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).
LINKS
Freimut Marschner, Table of n, a(n) for n = 0..9999
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
FORMULA
a(n) = 27*(n - 6)^2 + 4*(n - 6)^3 = ((n - 6)^2)*(4*n + 3).
G.f.: (49*x^3+124*x^2-257*x+108) / (x-1)^4. - Colin Barker, Jul 11 2014
EXAMPLE
n = 0, a(0) = 27*(-6)^2 + 4*(-6)^3 = 27*36 + 4*(-216) = 108;
n = 7, a(7) = 27*(1)^2 + 4*(1)^3 = 27 + 4 = 31.
MAPLE
MATHEMATICA
Table[(4 n + 3)(n - 6)^2, {n, 0, 50}] (* Wesley Ivan Hurt, Jul 18 2014 *)
LinearRecurrence[{4, -6, 4, -1}, {108, 175, 176, 135}, 50] (* Harvey P. Dale, Oct 13 2019 *)
PROG
(PARI) vector(100, n, (n-7)^2*(4*n-1)) \\ Colin Barker, Jul 11 2014
(PARI) Vec((49*x^3+124*x^2-257*x+108)/(x-1)^4 + O(x^100)) \\ Colin Barker, Jul 11 2014
(Magma) [(4*n+3)*(n-6)^2 : n in [0..50]]; // Wesley Ivan Hurt, Jul 18 2014
CROSSREFS
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).
KEYWORD
nonn,easy
AUTHOR
Freimut Marschner, Jul 10 2014
STATUS
approved