A133570 "Exact" continued fraction of e.

3, -4, 2, 5, -2, -7, 2, 9, -2, -11, 2, 13, -2, -15, 2, 17, -2, -19, 2, 21, -2, -23, 2, 25, -2, -27, 2, 29, -2, -31, 2, 33, -2, -35, 2, 37, -2, -39, 2, 41, -2, -43, 2, 45, -2, -47, 2, 49, -2, -51, 2, 53, -2, -55, 2, 57, -2, -59, 2, 61, -2, -63, 2, 65, -2, -67, 2, 69, -2, -71, 2, 73, -2, -75, 2, 77, -2, -79, 2, 81, -2, -83, 2, 85, -2, -87, 2

Offset

0,1

Comments

See comments in A133593. Just as for the usual continued fraction for e, the exact continued fraction also has a simple pattern.

Links

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (0,-2,0,-1).

Formula

x(0) = e, a(n) = floor( |x(n)| + 0.5 ) * Sign(x(n)), x(n+1) = 1 / (x(n)-a(n)).

From Colin Barker, Sep 13 2013 and Jan 08 2016: (Start)

a(n) = 1/2*((2-i*2)*((-i)^n-i*i^n)+((-i)^n-i^n)*n)*(-1)*i for n>1.

a(n) = -2*a(n-2)-a(n-4) for n>5.

G.f.: -(x^5-5*x^4+3*x^3-8*x^2+4*x-3)/(x^2+1)^2.

(End)

Mathematica

$MaxExtraPrecision = Infinity; x[0] = E; a[n_] := a[n] = Round[Abs[x[n]]]*Sign[x[n]]; x[n_] := 1/(x[n - 1] - a[n - 1]); Table[a[n], {n, 0, 120}] (* Clark Kimberling, Sep 04 2013 *)
Join[{3, -4}, LinearRecurrence[{0, -2, 0, -1}, {2, 5, -2, -7}, 100]] (* Vincenzo Librandi, Jan 09 2016 *)

Prog

(PARI) Vec(-(x^5-5*x^4+3*x^3-8*x^2+4*x-3)/(x^2+1)^2 + O(x^100)) \\ Colin Barker, Sep 13 2013

Crossrefs

Cf. A133593, A003417.

Sequence in context: A152833 A139525 A246832 * A117041 A209688 A143939

Adjacent sequences: A133567 A133568 A133569 * A133571 A133572 A133573

Keyword

cofr,sign,easy

Author

Serhat Sevki Dincer (jfcgauss(AT)gmail.com), Dec 30 2007

Status

approved