editing
approved
editing
approved
More terms by _from _Peter Luschny_, Feb 27 2019
proposed
editing
editing
proposed
1, 1, 3, 6, 4, 13, 53, 111, 231, 160, 1000, 13, 4329, 693, 2083, 39014, 81188, 84477, 351597, 243893, 1522639, 3168640, 6594000, 21441, 1359821, 59426081, 123666803, 19796382, 535556412, 61916837, 2319302053, 4826511631, 10044062391, 20901884640, 14499073000
A091808 := n -> numer(Im(numtheory[cfrac]([I, [I, I]$n-1]))):
seq(A091808(n), n=1..35); # Peter Luschny, Feb 27 2019
GenerateA091808[1] := I; GenerateA091808[n_] := I + I/(GenerateA091808[n-1]); GenerateNumeratorsA091808[n_] := Table[Numerator[Im[GenerateA091808[x]]], {x, 1, n}]; (* GenerateNumeratorsA091808[20] would give the first 20 terms. *)
More terms by Peter Luschny, Feb 27 2019
approved
editing
editing
approved
The sequence of complex numbers (which this sequence is part of) converges to (i+sqrt(-1+4i))/2, found by simply solving the equation A = i + (i/A) for A using the quadratic formula. When plotted in the complex plane, these numbers form a counter-clockwise counterclockwise spiral that quickly converges to a point.
a(6) = 13 since the sixth convergent is (3/5) + (13/10)i and hence the numerator of the imaginary part is 13.
approved
editing
Cf., A123457.
cofr,frac,nonn,new
Given the infinite continued fraction i+(i/(i+(i/(i+...)))), where i is the square root of (-1), this is the numerator of the imaginary part of the convergents.
1, 1, 3, 6, 4, 13, 53, 111, 231, 160, 1000, 13, 4329, 693, 2083, 39014, 81188, 84477, 351597
1,3
The sequence of complex numbers (which this sequence is part of) converges to (i+sqrt(-1+4i))/2, found by simply solving the equation A=i+(i/A) for A using the quadratic formula. When plotted in the complex plane, these numbers form a counter-clockwise spiral that quickly converges to a point.
a(6)=13 since the sixth convergent is (3/5)+(13/10)i and hence the numerator of the imaginary part is 13.
GenerateA091808[1] := I; GenerateA091808[n_] := I + I/(GenerateA091808[n-1]); GenerateNumeratorsA091808[n_] := Table[Numerator[Im[GenerateA091808[x]]], {x, 1, n}]; GenerateNumeratorsA091808[20] would give the first 20 terms.
cofr,frac,nonn
Ryan Witko (witko(AT)nyu.edu), Mar 06 2004
approved