OFFSET
0,2
COMMENTS
"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
MATHEMATICA
LinearRecurrence[{11, -11, 1}, {0, 2, 24}, 19] (* Jean-François Alcover, Feb 26 2019 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Mohamed Bouhamida, Nov 14 2007
STATUS
approved