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 A221193 Simple continued fraction expansion of an infinite product. 2
 1, 1, 1, 32, 1, 97, 1, 1152, 1, 3361, 1, 39200, 1, 114241, 1, 1331712, 1, 3880897, 1, 45239072, 1, 131836321, 1, 1536796800, 1, 4478554081, 1, 52205852192, 1, 152139002497, 1, 1773462177792, 1, 5168247530881, 1, 60245508192800, 1, 175568277047521, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS 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 = 3. For other cases see A221075 (N = 2), A221194 (N = 4) and A221195 (N = 5). 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*((3 - 2*sqrt(2))^(2*n+1))^(4*k+3))/(1 - 2*((3 - 2*sqrt(2))^(2*n+1))^(4*k+1)). LINKS Index entries for linear recurrences with constant coefficients, signature (0,1,0,34,0,-34,0,-1,0,1). FORMULA a(4*n-1) = (3 + 2*sqrt(2))^(2*n) + (3 - 2*sqrt(2))^(2*n) - 2; a(4*n+1) = 1/2*((3 + 2*sqrt(2))^(2*n+1) + (3 - 2*sqrt(2))^(2*n+1)) - 2; a(2*n) = 1. G.f.: -(x^8+x^7+31*x^5-34*x^4+31*x^3+x+1) / ((x-1)*(x+1)*(x^2-2*x-1)*(x^2+2*x-1)*(x^4+6*x^2+1)). [Colin Barker, Jan 14 2013] EXAMPLE Product {k >= 0} (1 - 2*(3 - 2*sqrt(2))^(4*k+3))/(1 - 2*(3 - 2*sqrt(2))^(4*k+1)) = 1.50746 49374 34879 05211 ... = 1 + 1/(1 + 1/(1 + 1/(32 + 1/(1 + 1/(97 + ...))))). CROSSREFS A221075 (N = 2), A221194 (N = 4), A221195 (N = 5). Sequence in context: A303323 A301910 A302154 * A103325 A221760 A037932 Adjacent sequences:  A221190 A221191 A221192 * A221194 A221195 A221196 KEYWORD nonn,easy,cofr AUTHOR Peter Bala, Jan 08 2013 EXTENSIONS More terms from Colin Barker, Jan 14 2013 STATUS approved

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Last modified May 26 00:32 EDT 2020. Contains 334613 sequences. (Running on oeis4.)