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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.
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%I #23 Apr 22 2023 13:10:19

%S 1,7,26,63,124,215,342,511,728,999,1330,1727,2196,2743,3374,4095,4912,

%T 5831,6858,7999,9260,10647,12166,13823,15624,17575,19682,21951,24388,

%U 26999,29790,32767,35936,39303,42874

%N 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.

%C 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.

%C Apart from the first term, the same as A068601. - _R. J. Mathar_, Apr 29 2008

%H G. C. Greubel, <a href="/A135300/b135300.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

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

%F G.f.: x*(1 + 3*x + 4*x^2 - 3*x^3 + x^4)/(1-x)^4. - _Colin Barker_, Oct 25 2012

%F E.g.f.: (1 + x)*(1 - exp(x)*(1 - 2*x - x^2)). - _Stefano Spezia_, Apr 22 2023

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

%o (PARI) a(n)=if(n>1, n^3-1, 1) \\ _Charles R Greathouse IV_, Oct 09 2016

%Y Cf. A001093, A068601.

%K nonn,easy

%O 1,2

%A _Mohamed Bouhamida_, Dec 04 2007