A132596 X-values of solutions to the equation X*(X + 1) - 6*Y^2 = 0.

0, 2, 24, 242, 2400, 23762, 235224, 2328482, 23049600, 228167522, 2258625624, 22358088722, 221322261600, 2190864527282, 21687323011224, 214682365584962, 2125136332838400, 21036680962799042, 208241673295152024

Offset: 0

Author

Mohamed Bouhamida, Nov 14 2007

Keywords

• nonn
• easy

Comments

Or, 3*A000217(X) is a square, (3*A004189(n))^2. - Zak Seidov, Apr 08 2009

"You can find an infinite number of [different] triangular numbers such that when multiplied together form a square number. For example, for every triangular number, T_n, there are an infinite number of other triangular numbers, T_m, such that T_n*T_m is a square. For example, T_2 * T_24 = 30^2." [Pickover] - Robert G. Wilson v, Apr 01 2010

References

• Clifford A. Pickover, The Loom of God, Tapestries of Mathematics and Mysticism, Sterling, NY, 2009, page 33.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Crossrefs

Cf. A007654, A001079, A000217, A098297, A108741 (Y^2), A004189 (Y).

Programs

Mathematica

LinearRecurrence[{11, -11, 1}, {0, 2, 24}, 19] (* Jean-François Alcover, Feb 26 2019 *)

Formula

a(n) = 10*a(n-1) - a(n-2) + 4, a(0)=0, a(1)=2.

a(n) = (A001079(n) - 1)/2. - Max Alekseyev, Nov 13 2009

From R. J. Mathar, Apr 20 2010: (Start)

a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3) = 2*A098297(n).

G.f.: -2*x*(1+x) / ( (x-1)*(x^2-10*x+1) ). (End)

a(n) = 2*A098297(n) = (1/2)*(T(2*n,sqrt(3)) - 1), where T(n,x) is the n-th Chebyshev polynomial of the first kind. - Peter Bala, Dec 31 2012