OFFSET
1,2
COMMENTS
The problem is to find p and r such that 6*(2*p-1)^2 = 5*(2*r+1)^2 + 1 equivalent to 3*p^2 - 3*p + 1 = (5*r^2 + 5*r + 2)/2. The Diophantine equation (6*X)^2 = 30*Y^2 + 6 is such that
X is given by 1, 21, 461, 10121, ... with a(n+2) = 22*a(n+1) - a(n) and also a(n+1) = 11*a(n) + sqrt(120*a(n)^2 - 20);
Y is given by 1, 23, 805, 11087, ... with a(n+2) = 22*a(n+1) - a(n) and also a(n+1) = 11*a(n) + sqrt(120*a(n)^2+24);
r is given by 0, 11, 252, 5543, 121704, ... with a(n+2) = 22*a(n+1) - a(n) + 10 and also a(n+1) = 11*a(n) + 5 + sqrt(120*a(n)^2 + 120*a(n) + 36);
p is given by 1, 11, 231, 5061, ... with a(n+2) = 22*a(n+1) - a(n) - 10 and also a(n+1) = 11*a(n) - 5 + sqrt(120*a(n)^2 - 120*a(n) + 25).
LINKS
Colin Barker, Table of n, a(n) for n = 1..373
Index entries for linear recurrences with constant coefficients, signature (483,-483,1).
FORMULA
a(n+2) = 482*a(n+1) - a(n) - 150.
a(n+1) = 241*a(n) - 75 + 11*sqrt(480*a(n)^2 - 300*a(n) + 45).
G.f.: z*(1-152*z+z^2)/((1-z)*(1-482*z+z^2)).
MATHEMATICA
LinearRecurrence[{483, -483, 1}, {1, 331, 159391}, 20] (* Paolo Xausa, Jan 07 2024 *)
PROG
(PARI) Vec(-x*(x^2-152*x+1)/((x-1)*(x^2-482*x+1)) + O(x^100)) \\ Colin Barker, Feb 07 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Richard Choulet, Sep 21 2007
EXTENSIONS
More terms from Paolo P. Lava, Sep 26 2008
STATUS
approved