A248750 Decimal expansion of limit of the imaginary part of f(1+i,n), where f(x,0) = 1 and f(x,n) = x + 1/f(x,n-1).
7, 4, 2, 9, 3, 4, 1, 3, 5, 8, 7, 8, 3, 2, 2, 8, 3, 9, 0, 9, 1, 4, 3, 1, 9, 3, 7, 9, 4, 7, 2, 6, 6, 2, 8, 1, 0, 9, 6, 2, 4, 2, 9, 9, 2, 0, 0, 1, 1, 8, 6, 5, 0, 5, 4, 7, 5, 8, 6, 9, 2, 0, 6, 2, 1, 9, 0, 5, 7, 7, 6, 3, 9, 5, 6, 8, 7, 8, 5, 4, 9, 0, 5, 9, 2, 3
Offset: 0
Examples
0.742934135878322839091431937947266281096242992001186505475869206219... n f(x,n) Re(f(1+i,n)) Im(f(1+i,n)) 0 1 1 0 1 1 + x 2 1 2 (1 + x + x^2)/(1 + x) 7/5 4/5 3 (1 + 2*x + x^2 + x^3)/(1 + x + x^2) 20/13 9/13 Re(f(1+i,10)) = 815/533 = 1.529080... Im(f(1+i,10)) = 396/533 = 0.742964...
Programs
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Maple
evalf((1+sqrt(sqrt(5)-2))/2, 120); # Vaclav Kotesovec, Oct 19 2014
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Mathematica
$RecursionLimit = Infinity; $MaxExtraPrecision = Infinity; z = 20; (* For more accuracy, increase z *) f[x_, n_] := x + 1/f[x, n - 1]; f[x_, 1] = 1; t = Table[Factor[f[x, n]], {n, 1, z}]; u = t /. x -> I + 1; t = Table[Factor[f[x, n]], {n, 1, z}]; u = t /. x -> I + 1; r1 = N[Re[u][[z]], 130] r2 = N[Im[u][[z]], 130] d1 = RealDigits[r1] (*A248749*) d2 = RealDigits[r2] (*A248750*)
Formula
Equals (1+sqrt(sqrt(5)-2))/2. - Vaclav Kotesovec, Oct 19 2014
From Wolfdieter Lang, Mar 02 2018: (Start)
Equals (1 + (2 - phi)*sqrt(phi))/2, with phi = A001622.
Equals (1/10)*y*(1 - (1/50)*y^2) with y = -A300072. (End)
Comments