A221074 Simple continued fraction expansion of product {n >= 0} {1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+3)}/{1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+1)} at m = 3.
2, 8, 1, 16, 1, 96, 1, 176, 1, 968, 1, 1760, 1, 9600, 1, 17440, 1, 95048, 1, 172656, 1, 940896, 1, 1709136, 1, 9313928, 1, 16918720, 1, 92198400, 1, 167478080, 1, 912670088, 1, 1657862096, 1
Offset: 0
Examples
Product {n >= 0} {1 - sqrt(3)*(sqrt(3) - sqrt(2))^(4*n+3)}/{1 - sqrt(3)*(sqrt(3) - sqrt(2))^(4*n+1)} = 2.11180 16361 44098 52896 ...
= 2 + 1/(8 + 1/(1 + 1/(16 + 1/(1 + 1/(96 + 1/(1 + 1/(176 + ...))))))).
Since (sqrt(3) - sqrt(2))^3 = 9*sqrt(3) - 11*sqrt(2) we have the following simple continued fraction expansion:
product {n >= 0} {1 - sqrt(3)*(9*sqrt(3) - 11*sqrt(2))^(4*n+3)}/{1 - sqrt(3)*(9*sqrt(3) - 11*sqrt(2))^(4*n+1)} = 1 + 1/(16 + 1/(1 + 1/(968 + 1/(1 + 1/(17440 + 1/(1 + 1/(940896 + ...))))))).
Links
- Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
- Index entries for linear recurrences with constant coefficients, signature (0,1,0,10,0,-10,0,-1,0,1).
Crossrefs
Formula
a(2*n) = 1 for n >= 1. For n >= 1 we have
a(4*n - 3) = (sqrt(3) + sqrt(2))^(2*n) + (sqrt(3) - sqrt(2))^(2*n) - 2;
a(4*n - 1) = 1/sqrt(3)*{(sqrt(3) + sqrt(2))^(2*n + 1) + (sqrt(3) - sqrt(2))^(2*n + 1)} - 2.
O.g.f.: 2 + x^2/(1 - x^2) + 8*x*(1 + x^2)^2/(1 - 11*x^4 + 11*x^8 - x^12) = 2 + 8*x + x^2 + 16*x^3 + x^4 + 96*x^5 + ....
If we denote the present sequence by [2; 8, 1, c(3), 1, c(4), 1, ...] then for k >= 1 the sequence [1; c(2*k+1), 1, c(2*(2*k+1)), 1, c(3*(2*k+1)), 1, ...] gives the simple continued fraction expansion of product {n >= 0} [1-sqrt(3)*{(sqrt(3)-sqrt(2))^(2*k+1)}^(4*n+3)]/[1 - sqrt(3)*{(sqrt(3)-sqrt(2))^(2*k+1)}^(4*n+1)]. An example is given below.
O.g.f.: (x^10-2*x^8-10*x^6+20*x^4-8*x^3+x^2-8*x-2) / ((x-1)*(x+1)*(x^8-10*x^4+1)). - Colin Barker, Jan 10 2014
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