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Revision History for A135300

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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.
(history; published version)
#23 by Michael De Vlieger at Sat Apr 22 13:10:19 EDT 2023
STATUS

reviewed

approved

#22 by Michel Marcus at Sat Apr 22 10:53:30 EDT 2023
STATUS

proposed

reviewed

#21 by Stefano Spezia at Sat Apr 22 08:44:59 EDT 2023
STATUS

editing

proposed

#20 by Stefano Spezia at Sat Apr 22 06:47:39 EDT 2023
FORMULA

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

CROSSREFS
STATUS

approved

editing

#19 by N. J. A. Sloane at Sat Feb 15 10:52:27 EST 2020
AUTHOR

_Mohamed Bouhamida (bhmd95(AT)yahoo.fr), _, Dec 04 2007

Discussion
Sat Feb 15
10:52
OEIS Server: https://oeis.org/edit/global/2849
#18 by N. J. A. Sloane at Mon Dec 24 21:40:21 EST 2018
STATUS

proposed

approved

#17 by Jon E. Schoenfield at Mon Dec 24 16:38:41 EST 2018
STATUS

editing

proposed

#16 by Jon E. Schoenfield at Mon Dec 24 16:38:35 EST 2018
NAME

Sequence allows us to find Positive X positive -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.

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 it , 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

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

STATUS

approved

editing

#15 by Charles R Greathouse IV at Sun Oct 09 23:30:17 EDT 2016
STATUS

editing

approved

#14 by Charles R Greathouse IV at Sun Oct 09 23:30:05 EDT 2016
PROG

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

STATUS

proposed

editing