A086302 a(n) = 4*n^4 + 24*n^3 + 48*n^2 + 36*n + 8.

8, 120, 528, 1520, 3480, 6888, 12320, 20448, 32040, 47960, 69168, 96720, 131768, 175560, 229440, 294848, 373320, 466488, 576080, 703920, 851928, 1022120, 1216608, 1437600, 1687400, 1968408, 2283120, 2634128, 3024120, 3455880, 3932288, 4456320, 5031048

Offset: 0

Author

Neven Juric (neven.juric(AT)apis-it.hr), Aug 29 2003

Keywords

• nonn
• easy

Comments

Suppose one wishes to find sets of four positive integers (a,b,c,d) such that ab+1, ac+1, ad+1, bc+1, bd+1, cd+1 are perfect squares. Then one may take a = 1, b = x^2 + 2x, c = x^2 + 4x + 3, d = 4x^4 + 24x^3 + 48x^2 + 36x + 8.

Examples

			(a,b,c,d) = (1,3,8,120), (1,8,15,528), (1,15,24,1520), (1,24,35,3480), ...
		

Links

• Paolo Xausa, Table of n, a(n) for n = 0..10000
• Philip Gibbs, Diophantine quadruples and Cayley's hyperdeterminant, arXiv:math/0107203 [math.NT], 2001.
• Eric Weisstein's World of Mathematics, Diophantus Property.
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A057769, A062392.

Programs

Mathematica

LinearRecurrence[{5, -10, 10, -5, 1}, {8, 120, 528, 1520, 3480}, 50] (* or *)
A086302[n_] := 4 (n + 1) (n + 2) (n^2 + 3 n + 1);
Array[A086302, 50, 0] (* Paolo Xausa, Jan 16 2024 *)

Formula

a(n) = A057769(n+1) + 1. - N. J. A. Sloane, Jun 12 2004

G.f.: 8*(1 + 10*x + x^2)/(1 - x)^5. - Colin Barker, Mar 26 2012

a(n) = 4 * (n+1) * (n+2) * (n^2 + 3*n + 1). - Bruno Berselli, Mar 26 2012

a(n) = 8*A062392(n+1). - Bruce J. Nicholson, Jun 05 2017

Sum_{n>=0} 1/a(n) = tan(sqrt(5)*Pi/2)*Pi/(4*sqrt(5)). - Amiram Eldar, Jan 22 2024

E.g.f.: 4*exp(x)*(2 + 28*x + 37*x^2 + 12*x^3 + x^4). - Stefano Spezia, Apr 27 2025