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%I #23 Sep 08 2022 08:46:08
%S 49,320,729,1300,2057,3024,4225,5684,7425,9472,11849,14580,17689,
%T 21200,25137,29524,34385,39744,45625,52052,59049,66640,74849,83700,
%U 93217,103424,114345,126004,138425,151632,165649,180500,196209,212800,230297,248724
%N 4*(n + 7)^3 - 27*(n + 7)^2 = (4*n +1)*(n+7)^2.
%C 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.
%H Freimut Marschner, <a href="/A245033/b245033.txt">Table of n, a(n) for n = 0..10000</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) = 4*(n + 7)^3 - 27*(n + 7)^2.
%F G.f.: (108*x^3-257*x^2+124*x+49) / (x-1)^4. - _Colin Barker_, Jul 11 2014
%F 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).
%e n = 0, a(0) = 4*7^3 - 27*7^2 = (4*7 - 27)*7^2 = 49.
%p A245033:=n->4*(n + 7)^3 - 27*(n + 7)^2: seq(A245033(n), n=0..30); # _Wesley Ivan Hurt_, Jul 12 2014
%t Table[4 (n + 7)^3 - 27 (n + 7)^2, {n, 0, 30}] (* _Wesley Ivan Hurt_, Jul 12 2014 *)
%t LinearRecurrence[{4,-6,4,-1},{49,320,729,1300},40] (* _Harvey P. Dale_, Dec 26 2014 *)
%o (PARI) vector(100, n, (n+6)^2*(4*n-3)) \\ _Colin Barker_, Jul 11 2014
%o (PARI) Vec((108*x^3-257*x^2+124*x+49)/(x-1)^4 + O(x^100)) \\ _Colin Barker_, Jul 11 2014
%o (Magma) [4*(n + 7)^3 - 27*(n + 7)^2 : n in [0..30]]; // _Wesley Ivan Hurt_, Jul 12 2014
%Y A000290 (squares: a(n) = n^2), A000578 (cubes: a(n) = n^3), A028347 (Discriminant of quadratic equation : a(n) = n^2 - 4*n, n > 2).
%K nonn,easy
%O 0,1
%A _Freimut Marschner_, Jul 10 2014
%E xref deleted.
%E Edited: recurrence according to index link added and checked. - _Wolfdieter Lang_, Jul 28 2014
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