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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.

%I A221074 #21 Feb 13 2024 10:20:57
%S A221074 2,8,1,16,1,96,1,176,1,968,1,1760,1,9600,1,17440,1,95048,1,172656,1,
%T A221074 940896,1,1709136,1,9313928,1,16918720,1,92198400,1,167478080,1,
%U A221074 912670088,1,1657862096,1
%N 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.
%C 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. For other cases see A221073 (m = 2), A221075 (m = 4) and A221076 (m = 5).
%H A221074 Peter Bala, <a href="/A174500/a174500_2.pdf">Some simple continued fraction expansions for an infinite product, Part 1</a>
%H A221074 <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (0,1,0,10,0,-10,0,-1,0,1).
%F A221074 a(2*n) = 1 for n >= 1. For n >= 1 we have
%F A221074 a(4*n - 3) = (sqrt(3) + sqrt(2))^(2*n) + (sqrt(3) - sqrt(2))^(2*n) - 2;
%F A221074 a(4*n - 1) = 1/sqrt(3)*{(sqrt(3) + sqrt(2))^(2*n + 1) + (sqrt(3) - sqrt(2))^(2*n + 1)} - 2.
%F A221074 a(4*n - 3) = 8*A098297(n) = 4*A132596(n); a(4*n - 1) = 4*A105038(n).
%F A221074 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 + ....
%F A221074 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.
%F A221074 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
%e A221074 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 ...
%e A221074 = 2 + 1/(8 + 1/(1 + 1/(16 + 1/(1 + 1/(96 + 1/(1 + 1/(176 + ...))))))).
%e A221074 Since (sqrt(3) - sqrt(2))^3 = 9*sqrt(3) - 11*sqrt(2) we have the following simple continued fraction expansion:
%e A221074 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 + ...))))))).
%Y A221074 Cf. A098297, A105438, A132596, A174500, A221073 (m = 2), A221075 (m = 4), A221076 (m = 5).
%K A221074 nonn,easy,cofr
%O A221074 0,1
%A A221074 _Peter Bala_, Jan 06 2013