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A245033
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4*(n + 7)^3 - 27*(n + 7)^2 = (4*n +1)*(n+7)^2.
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2
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49, 320, 729, 1300, 2057, 3024, 4225, 5684, 7425, 9472, 11849, 14580, 17689, 21200, 25137, 29524, 34385, 39744, 45625, 52052, 59049, 66640, 74849, 83700, 93217, 103424, 114345, 126004, 138425, 151632, 165649, 180500, 196209, 212800, 230297, 248724
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OFFSET
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0,1
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COMMENTS
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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.
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LINKS
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FORMULA
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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).
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EXAMPLE
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n = 0, a(0) = 4*7^3 - 27*7^2 = (4*7 - 27)*7^2 = 49.
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MAPLE
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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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).
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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xref deleted.
Edited: recurrence according to index link added and checked. - Wolfdieter Lang, Jul 28 2014
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STATUS
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approved
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