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A091808 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. 7

%I #8 Feb 27 2019 09:13:10

%S 1,1,3,6,4,13,53,111,231,160,1000,13,4329,693,2083,39014,81188,84477,

%T 351597,243893,1522639,3168640,6594000,21441,1359821,59426081,

%U 123666803,19796382,535556412,61916837,2319302053,4826511631,10044062391,20901884640,14499073000

%N 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.

%C 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 counterclockwise spiral that quickly converges to a point.

%e a(6) = 13 since the sixth convergent is (3/5) + (13/10)i and hence the numerator of the imaginary part is 13.

%p A091808 := n -> numer(Im(numtheory[cfrac]([I, [I, I]$n-1]))):

%p seq(A091808(n), n=1..35); # _Peter Luschny_, Feb 27 2019

%t 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. *)

%Y Cf. A091806, A091807, A091808, A091809.

%K cofr,frac,nonn

%O 1,3

%A Ryan Witko (witko(AT)nyu.edu), Mar 06 2004

%E More terms from _Peter Luschny_, Feb 27 2019

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Last modified August 14 03:23 EDT 2024. Contains 375146 sequences. (Running on oeis4.)