OFFSET
0,1
COMMENTS
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 = 4. For other cases see A221073 (m = 2), A221074 (m = 3) and A221076 (m = 5).
If we denote the present sequence by [2; 12, 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} [1-2*{(2-sqrt(3))^(2*k+1)}^(4*n+3)]/[1 - 2*{(2-sqrt(3))^(2*k+1)}^(4*n+1)]. An example is given below
LINKS
Index entries for linear recurrences with constant coefficients, signature (0,1,0,14,0,-14,0,-1,0,1).
FORMULA
a(2*n) = 1 for n >= 1. For n >= 1 we have
a(4*n - 3) = (2 + sqrt(3))^(2*n) + (2 - sqrt(3))^(2*n) - 2;
a(4*n - 1) = 1/2*{(2 + sqrt(3) )^(2*n + 1) + (2 - sqrt(3))^(2*n + 1)} - 2.
O.g.f.: 2 + x^2/(1 - x^2) + 12*x*(1 + x^2)^2/(1 - 15*x^4 + 15*x^8 - x^12) = 2 + 12*x + x^2 + 24*x^3 + x^4 + 192*x^5 + ....
O.g.f.: (x^10-2*x^8-14*x^6+28*x^4-12*x^3+x^2-12*x-2) / ((x-1)*(x+1)*(x^4-4*x^2+1)*(x^4+4*x^2+1)). - Colin Barker, Jan 10 2014
EXAMPLE
Product {n >= 0} {1 - 2*(2 - sqrt(3))^(4*n+3)}/{1 - 2*(2 - sqrt(3))^(4*n+1)} = 2.07715 13807 08976 70415 ...
= 2 + 1/(12 + 1/(1 + 1/(24 + 1/(1 + 1/(192 + 1/(1 + 1/(360 + ...))))))).
Since (2 - sqrt(3))^3 = 26 - 15*sqrt(3) we have the following simple continued fraction expansion:
product {n >= 0} {1 - 2*(26 - 15*sqrt(3))^(4*n+3)}/{1 - 2*(26 - 15*sqrt(3))^(4*n+1)} = 1.04000 05921 62729 43797 ... = 1 + 1/(24 + 1/(1 + 1/(2700 + 1/(1 + 1/(70224 + 1/(1 + 1/(7300800 + ...))))))).
CROSSREFS
KEYWORD
nonn,easy,cofr
AUTHOR
Peter Bala, Jan 06 2013
STATUS
approved