A221366 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) when x = (1/2)*(7 - 3*sqrt(5)).
1, 5, 1, 45, 1, 320, 1, 2205, 1, 15125, 1, 103680, 1, 710645, 1, 4870845, 1, 33385280, 1, 228826125, 1, 1568397605, 1, 10749957120, 1, 73681302245, 1, 505019158605, 1, 3461452808000, 1, 23725150497405, 1
Offset: 0
Examples
F(1/2*(7 - sqrt(45))) = 1.16725 98258 10214 95210 ... = 1 + 1/(5 + 1/(1 + 1/(45 + 1/(1 + 1/(320 + 1/(1 + 1/(2205 + ...))))))). F((1/2*(7 - sqrt(45)))^2) = 1.02173 93445 69104 86504 ... = 1 + 1/(45 + 1/(1 + 1/(2205 + 1/(1 + 1/(103680 + 1/(1 + 1/(4870845 + ...))))))). F((1/2*(7 - sqrt(45)))^3) = 1.00311 52648 91110 10148 ... = 1 + 1/(320 + 1/(1 + 1/(103680 + 1/(1 + 1/(33385280 + 1/(1 + 1/(10749957120 + ...))))))).
Links
- Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
- Index entries for linear recurrences with constant coefficients, signature (0,8,0,-8,0,1).
Crossrefs
Programs
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Mathematica
LinearRecurrence[{0,8,0,-8,0,1},{1,5,1,45,1,320},40] (* or *) Riffle[ LinearRecurrence[{8,-8,1},{5,45,320},20],1,{1,-1,2}] (* Harvey P. Dale, Jan 04 2018 *)
Formula
a(2*n-1) = (1/2*(7 + sqrt(45)))^n + (1/2*(7 - sqrt(45)))^n - 2 = A081070(n); a(2*n) = 1.
a(4*n+1) = 5*(A033890(n))^2.
a(n) = 8*a(n-2)-8*a(n-4)+a(n-6). G.f.: -(x^4+5*x^3-7*x^2+5*x+1) / ((x-1)*(x+1)*(x^2-3*x+1)*(x^2+3*x+1)). - Colin Barker, Jan 20 2013
Comments