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

0,1

Comments

See A046854 for a triangle of coefficients of the numerators and denominators of f(x,n). Note that the limit of f(1,n) is the golden ratio.

Links

Table of n, a(n) for n=0..85.

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)

Example

0.742934135878322839091431937947266281096242992001186505475869206219...
n   f(n,x)                                 Re(f(n,1+i))  Im(f(n,1+i))
1   1                                      1             0
2   1 + x                                  2             1
3   (1 + x + x^2)/ (1 + x)                 7/5           4/5
4   (1 + 2x + x^2 + x^3)/(1 + x + x^2)     20/13         9/13
Re(f(11,1+i) = 815/533 = 1.529162...
Im(f(11,1+i) = 396/533 = 0.742964...

Maple

evalf((1+sqrt(sqrt(5)-2))/2, 120); # Vaclav Kotesovec, Oct 19 2014

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*)

Crossrefs

Cf. A248749, A046854, A248751, A248752, A001622. A300072.

Sequence in context: A245055 A335020 A225410 * A071875 A200687 A200121

Adjacent sequences: A248747 A248748 A248749 * A248751 A248752 A248753

Keyword

nonn,easy,cons

Author

Clark Kimberling, Oct 13 2014

Status

approved