A135300 Positive X-values of solutions to the equation 1!*X^4 - 2!*(X + 1)^3 + 3!*(X + 2)^2 - (4^2)*(X + 3) + 5^2 = Y^3.

1, 7, 26, 63, 124, 215, 342, 511, 728, 999, 1330, 1727, 2196, 2743, 3374, 4095, 4912, 5831, 6858, 7999, 9260, 10647, 12166, 13823, 15624, 17575, 19682, 21951, 24388, 26999, 29790, 32767, 35936, 39303, 42874

Offset: 1

Author

Mohamed Bouhamida, Dec 04 2007

Keywords

• nonn
• easy

Comments

To prove that X = 1 or X = n^3 - 1: Y^3 = 1!*X^4 - 2!*(X + 1)^3 + 3!*(X + 2)^2 - (4^2)*(X + 3) + 5^2 = X^4 - 2*(X + 1)^3 + 6*(X + 2)^2 - 16(X + 3) + 25 = X^4 - 2*X^3 + 2X - 1 = (X + 1)(X^3 - 3*X^2 + 3X - 1) = (X + 1)*(X - 1)^3, which means that X = 1 or (X + 1) must be a cube, so (X, Y) = (1, 0) or (X, Y) = (n^3 - 1, n(n^3 - 2)) with n >= 2.

Apart from the first term, the same as A068601. - R. J. Mathar, Apr 29 2008

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A001093, A068601.

Programs

Mathematica

Join[{1},LinearRecurrence[{4,-6,4,-1},{7,26,63,124},40]] (* Harvey P. Dale, Jul 12 2015 *)

PARI

a(n)=if(n>1, n^3-1, 1) \\ Charles R Greathouse IV, Oct 09 2016

Formula

a(1) = 1 and a(n) = n^3 - 1 with n >= 2.

G.f.: x*(1 + 3*x + 4*x^2 - 3*x^3 + x^4)/(1-x)^4. - Colin Barker, Oct 25 2012

E.g.f.: (1 + x)*(1 - exp(x)*(1 - 2*x - x^2)). - Stefano Spezia, Apr 22 2023