A221195 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 = 5.

1, 3, 1, 96, 1, 483, 1, 9600, 1, 47523, 1, 940896, 1, 4656963, 1, 92198400, 1, 456335043, 1, 9034502496, 1, 44716177443, 1, 885289046400, 1, 4381729054563, 1, 86749292044896, 1, 429364731169923, 1, 8500545331353600, 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 = 5. For other cases see A221075 (N = 2), A221193 (N = 3) and A221194 (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*((5 - sqrt(24))^(2*n+1))^(4*k+3))/(1 - 2*((5 - sqrt(24))^(2*n+1))^(4*k+1)).

Examples

			Product {k >= 0} (1 - 2*(5 - sqrt(24))^(4*k+3))/(1 - 2*(5 - sqrt(24))^(4*k+1)) = 1.25063 93996 76216 17350 ... = 1 + 1/(3 + 1/(1 + 1/(96 + 1/(1 + 1/(483 + ...))))).
		

Links

• Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
• Index entries for linear recurrences with constant coefficients, signature (0,1,0,98,0,-98,0,-1,0,1).

Crossrefs

Cf. A221075 (N = 2), A221193 (N = 3), A221194 (N = 4).

Formula

a(4*n-1) = (5 + sqrt(24))^(2*n) + (5 - sqrt(24))^(2*n) - 2;

a(4*n+1) = 1/2*((5 + sqrt(24))^(2*n+1) + (5 - sqrt(24))^(2*n+1)) - 2; a(2*n) = 1.

G.f.: -(x^8+3*x^7+93*x^5-98*x^4+93*x^3+3*x+1) / ((x-1)*(x+1)*(x^4-10*x^2+1)*(x^4+10*x^2+1)). [Colin Barker, Jan 14 2013]