A221367 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) when x = (1/2)*(9 - sqrt(77)).

1, 7, 1, 77, 1, 700, 1, 6237, 1, 55447, 1, 492800, 1, 4379767, 1, 38925117, 1, 345946300, 1, 3074591597, 1, 27325378087, 1, 242853811200, 1, 2158358922727, 1, 19182376493357, 1, 170483029517500, 1, 1515164889164157, 1, 13466000972959927, 1

Offset: 0

Author

Peter Bala, Jan 15 2013

Keywords

• nonn
• easy
• cofr

Comments

The function F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) is analytic for |x| < 1. When x is a quadratic irrational of the form x = 1/2*(N - sqrt(N^2 - 4)), N an integer greater than 2, the real number F(x) has a predictable simple continued fraction expansion. The first examples of these expansions, for N = 2, 4, 6 and 8, are due to Hanna. See A174500 through A175503. The present sequence is the case N = 9. See also A221364 (N = 3), A221365 (N = 5) and A221366 (N = 7).

If we denote the present sequence by [1, c(1), 1, c(2), 1, c(3), ...] then for k = 1, 2, ..., the simple continued fraction expansion of F((1/2*(9 - sqrt(77)))^k) is given by the sequence [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...]. Examples are given below.

Examples

			F(1/2*(9 - sqrt(77))) = 1.12519 81018 34502 81936 ... = 1 + 1/(7 + 1/(1 + 1/(77 + 1/(1 + 1/(700 + 1/(1 + 1/(6237 + ...))))))).
F((1/2*(9 - sqrt(77)))^2) = 1.01282 05391 65421 74656 ... = 1 + 1/(77 + 1/(1 + 1/(6237 + 1/(1 + 1/(492800 + 1/(1 + 1/(38925117 + ...))))))).
F((1/2*(9 - sqrt(77)))^3) = 1.00142 65335 27667 24640 ... = 1 + 1/(700 + 1/(1 + 1/(492800 + 1/(1 + 1/(345946300 + 1/(1 + 1/(242853811200 + ...))))))).
		

Links

• Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
• Index entries for linear recurrences with constant coefficients, signature (0,10,0,-10,0,1).

Crossrefs

Cf. A018193, A057081, A174500 (N = 4), A221364 (N = 3), A221365 (N = 5), A221366 (N = 7).

Formula

a(2*n-1) = (1/2*(9 + sqrt(77)))^n + (1/2*(9 - sqrt(77)))^n - 2; a(2*n) = 1.

a(4*n-1) = 77*(A018913(n))^2; a(4*n+1) = 7*(A057081(n))^2.

a(n) = 10*a(n-2)-10*a(n-4)+a(n-6). G.f.: -(x^4+7*x^3-9*x^2+7*x+1) / ((x-1)*(x+1)*(x^4-9*x^2+1)). - Colin Barker, Jan 20 2013