OFFSET
0,2
COMMENTS
Simple continued fraction expansion of product {k >= 0} (1 - 2*(N - sqrt(N^2-1))^(4*k+3))/(1 - 2*(N - sqrt(N^2-1))^(4*k+1)) at N = 4. For other cases see A221075 (N = 2), A221193 (N = 3) and A221195 (N = 5).
Denoting the present sequence by [1, c(1), 1, c(2), 1, c(3), 1, ...] then for n >= 0 the sequence [1, c(2*n+1), 1, c(2*(2*n+1)), 1, c(3*(2*n+1)), 1, ...] gives the simple continued fraction expansion of product {k >= 0} (1 - 2*((4 - sqrt(15))^(2*n+1))^(4*k+3))/(1 - 2*((4 - sqrt(15))^(2*n+1))^(4*k+1)).
LINKS
Index entries for linear recurrences with constant coefficients, signature (0,1,0,62,0,-62,0,-1,0,1).
FORMULA
a(4*n-1) = (4 + sqrt(15))^(2*n) + (4 - sqrt(15))^(2*n) - 2;
a(4*n+1) = 1/2*{(4 + sqrt(15))^(2*n+1) + (4 - sqrt(15))^(2*n+1)} - 2; a(2*n) = 1.
G.f.: -(x^4-2*x^3+12*x^2-2*x+1)*(x^4+4*x^3-4*x^2+4*x+1) / ((x-1)*(x+1)*(x^4-8*x^2+1)*(x^4+8*x^2+1)). [Colin Barker, Jan 14 2013]
EXAMPLE
product {k >= 0} (1 - 2*(4 - sqrt(15))^(4*k+3))/(1 - 2*(4 - sqrt(15))^(4*k+1)) = 1.33513 52548 90793 94897 ... = 1 + 1/(2 + 1/(1 + 1/(60 + 1/(1 + 1/(242 + ...))))).
MATHEMATICA
LinearRecurrence[{0, 1, 0, 62, 0, -62, 0, -1, 0, 1}, {1, 2, 1, 60, 1, 242, 1, 3840, 1, 15122}, 40] (* Harvey P. Dale, Aug 03 2023 *)
CROSSREFS
KEYWORD
nonn,easy,cofr
AUTHOR
Peter Bala, Jan 08 2013
EXTENSIONS
More terms from Colin Barker, Jan 14 2013
STATUS
approved