A245036 a(n) = 4*prime(n)^3 - 27*prime(n)^2 = (prime(n)^2)*[4*prime(n) - 27], n >= 4.

49, 2057, 4225, 11849, 17689, 34385, 74849, 93217, 165649, 230297, 268105, 355649, 519665, 727529, 807457, 1081849, 1295537, 1412185, 1803649, 2101145, 2606009, 3396649, 3845777, 4084465, 4591049, 4859329, 5426825, 7758049, 8529017, 9778649, 10220809, 12632369, 13156177, 14814049

Offset

4,1

Comments

The discriminant D of the solution of the Cardano-Tartaglia equation x^3 + p*x + q = 0 is written D = -27*q^2 - 4*p^3. Let q = p = -prime(n) then D = -27*prime(n)^2 - 4*[-prime(n)]^3 = 4*[prime(n)]^3 - 27*prime(n)^2 = (prime(n)^2)*[4*prime(n) - 27], D > 0, n >= 4. a(n) = -D, offset 4,1. Remark: a(1) = -76, a(2) = -135, a(3) = -175.

Because in the preceding comment q can also be taken as +prime(n) this shows that the equation x^3 - prime(n)*x - prime(n) = 0 as well as x^3 - prime(n)*x + prime(p) = 0 has for each n >= 4 three distinct real solutions. - Wolfdieter Lang, Jul 29 2014

Links

Freimut Marschner, Table of n, a(n) for n = 4..6320

Formula

a(n) = 4*[prime(n)]^3 - 27*prime(n)^2 = (prime(n)^2)*[4*prime(n) - 27], n >= 4.

Example

n = 6, prime(6) = 13, 13^2*(4*13 - 27) = 4225.

Mathematica

4#^3-27#^2&/@Prime[Range[4, 40]] (* Harvey P. Dale, Feb 08 2021 *)

Crossrefs

Cf. A000040 (prime(n)), A001248 (prime(n)^2), A030078 (prime(n)^3).

Sequence in context: A367378 A065785 A163927 * A061615 A049682 A162914

Adjacent sequences: A245033 A245034 A245035 * A245037 A245038 A245039

Keyword

nonn

Author

Freimut Marschner, Jul 10 2014

Extensions

Name and Data have been changed to be in accordance to the Cardano-Tartaglia discriminant.

Status

approved