OFFSET
0,1
COMMENTS
The discriminant D of the Cardano Tartaglia equation x^3 + p*x + q = 0 is D = -27*q^2 - 4*p^3. Let q = p = -n then D = -27*(-n)^2 - 4*(-n)^3 = n^2*(4*n - 27), D > 0 if n >= 7, « casus irreducibilis ». To start with (offset 0,1) n is substituted by (n + 7) with the result a(n) = 4*(n+7)^3 - 27*(n + 7)^2 equivalent to a(n) = (4*n +1)*(n+7)^2.
LINKS
Freimut Marschner, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
FORMULA
a(n) = 4*(n + 7)^3 - 27*(n + 7)^2.
G.f.: (108*x^3-257*x^2+124*x+49) / (x-1)^4. - Colin Barker, Jul 11 2014
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n >= 4, with inputs a(0), ..., a(3).
EXAMPLE
n = 0, a(0) = 4*7^3 - 27*7^2 = (4*7 - 27)*7^2 = 49.
MAPLE
MATHEMATICA
Table[4 (n + 7)^3 - 27 (n + 7)^2, {n, 0, 30}] (* Wesley Ivan Hurt, Jul 12 2014 *)
LinearRecurrence[{4, -6, 4, -1}, {49, 320, 729, 1300}, 40] (* Harvey P. Dale, Dec 26 2014 *)
PROG
(PARI) vector(100, n, (n+6)^2*(4*n-3)) \\ Colin Barker, Jul 11 2014
(PARI) Vec((108*x^3-257*x^2+124*x+49)/(x-1)^4 + O(x^100)) \\ Colin Barker, Jul 11 2014
(Magma) [4*(n + 7)^3 - 27*(n + 7)^2 : n in [0..30]]; // Wesley Ivan Hurt, Jul 12 2014
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Freimut Marschner, Jul 10 2014
EXTENSIONS
xref deleted.
Edited: recurrence according to index link added and checked. - Wolfdieter Lang, Jul 28 2014
STATUS
approved