A132596 - OEIS

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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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

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).

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