A221194 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.

1, 2, 1, 60, 1, 242, 1, 3840, 1, 15122, 1, 238140, 1, 937442, 1, 14760960, 1, 58106402, 1, 914941500, 1, 3601659602, 1, 56711612160, 1, 223244789042, 1, 3515205012540, 1, 13837575261122, 1, 217885999165440, 1, 857706421400642, 1, 13505416743244860, 1

Offset: 0

Author

Peter Bala, Jan 08 2013

Keywords

• nonn
• easy
• cofr

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)).

Examples

			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 + ...))))).
		

Links

• Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
• Index entries for linear recurrences with constant coefficients, signature (0,1,0,62,0,-62,0,-1,0,1).

Crossrefs

Cf. A221075 (N = 2), A221193 (N = 3) and A221195 (N = 5).

Programs

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 *)

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]

Extensions

More terms from Colin Barker, Jan 14 2013