This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A221075 #19 Jun 13 2015 00:54:37
%S A221075 2,12,1,24,1,192,1,360,1,2700,1,5040,1,37632,1,70224,1,524172,1,
%T A221075 978120,1,7300800,1,13623480,1,101687052,1,189750624,1,1416317952,1,
%U A221075 2642885280,1,19726764300,1,36810643320,1
%N A221075 Simple continued fraction expansion of an infinite product.
%C A221075 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).
%C A221075 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
%H A221075 P. Bala, <a href="/A174500/a174500_2.pdf">Some simple continued fraction expansions for an infinite product, Part 1</a>
%H A221075 <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (0,1,0,14,0,-14,0,-1,0,1).
%F A221075 a(2*n) = 1 for n >= 1. For n >= 1 we have
%F A221075 a(4*n - 3) = (2 + sqrt(3))^(2*n) + (2 - sqrt(3))^(2*n) - 2;
%F A221075 a(4*n - 1) = 1/2*{(2 + sqrt(3) )^(2*n + 1) + (2 - sqrt(3))^(2*n + 1)} - 2.
%F A221075 a(4*n - 3) = 12*A098301(n) = 12*A001353(n)^2 = 4*A007654(n);
%F A221075 a(4*n - 1) = 24*A076139(n) = 12*A217855 = 8*A076140(n) = 6*A123480(n) = 3*A045899(n).
%F A221075 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 + ....
%F A221075 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
%e A221075 Product {n >= 0} {1 - 2*(2 - sqrt(3))^(4*n+3)}/{1 - 2*(2 - sqrt(3))^(4*n+1)} = 2.07715 13807 08976 70415 ...
%e A221075 = 2 + 1/(12 + 1/(1 + 1/(24 + 1/(1 + 1/(192 + 1/(1 + 1/(360 + ...))))))).
%e A221075 Since (2 - sqrt(3))^3 = 26 - 15*sqrt(3) we have the following simple continued fraction expansion:
%e A221075 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 + ...))))))).
%Y A221075 Cf. A001353, A007654, A045899, A076139, A076140, A098301, A123480, A174500, A217855, A221073 (m = 2), A221074 (m = 3), A221076 (m = 5).
%K A221075 nonn,easy,cofr
%O A221075 0,1
%A A221075 _Peter Bala_, Jan 06 2013