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%I #21 Feb 13 2024 03:23:11
%S 1,3,1,96,1,483,1,9600,1,47523,1,940896,1,4656963,1,92198400,1,
%T 456335043,1,9034502496,1,44716177443,1,885289046400,1,4381729054563,
%U 1,86749292044896,1,429364731169923,1,8500545331353600,1
%N Simple continued fraction expansion of product {k >= 0} (1 - 2*(N - sqrt(N^2-1))^(4*k+3))/(1 - 2*(N - sqrt(N^2-1))^(4*k+1)) at N = 5.
%C Simple continued fraction expansion of product {k >= 0} (1 - 2*(N - sqrt(N^2-1))^(4*k+3))/(1 - 2*(N - sqrt(N^2-1))^(4*k+1)) at N = 5. For other cases see A221075 (N = 2), A221193 (N = 3) and A221194 (N = 5).
%C Denoting the present sequence by [1, c(1), 1, c(2), 1, c(3), 1, ...] then for n >= 0 the sequence [1, c(2*n+1), 1, c(2*(2*n+1)), 1, c(3*(2*n+1)), 1, ...] gives the simple continued fraction expansion of product {k >= 0} (1 - 2*((5 - sqrt(24))^(2*n+1))^(4*k+3))/(1 - 2*((5 - sqrt(24))^(2*n+1))^(4*k+1)).
%H Peter Bala, <a href="/A174500/a174500_2.pdf">Some simple continued fraction expansions for an infinite product, Part 1</a>
%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (0,1,0,98,0,-98,0,-1,0,1).
%F a(4*n-1) = (5 + sqrt(24))^(2*n) + (5 - sqrt(24))^(2*n) - 2;
%F a(4*n+1) = 1/2*((5 + sqrt(24))^(2*n+1) + (5 - sqrt(24))^(2*n+1)) - 2; a(2*n) = 1.
%F G.f.: -(x^8+3*x^7+93*x^5-98*x^4+93*x^3+3*x+1) / ((x-1)*(x+1)*(x^4-10*x^2+1)*(x^4+10*x^2+1)). [_Colin Barker_, Jan 14 2013]
%e Product {k >= 0} (1 - 2*(5 - sqrt(24))^(4*k+3))/(1 - 2*(5 - sqrt(24))^(4*k+1)) = 1.25063 93996 76216 17350 ... = 1 + 1/(3 + 1/(1 + 1/(96 + 1/(1 + 1/(483 + ...))))).
%Y Cf. A221075 (N = 2), A221193 (N = 3), A221194 (N = 4).
%K nonn,easy,cofr
%O 0,2
%A _Peter Bala_, Jan 08 2013
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