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A228936 Expansion of (1+3*x-3*x^3-x^4)/(1+2*x^2+x^4). 2
1, 3, -2, -9, 2, 15, -2, -21, 2, 27, -2, -33, 2, 39, -2, -45, 2, 51, -2, -57, 2, 63, -2, -69, 2, 75, -2, -81, 2, 87, -2, -93, 2, 99, -2, -105, 2, 111, -2, -117, 2, 123, -2, -129, 2, 135, -2, -141, 2, 147 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

Optimal simple continued fraction (with signed denominators) of exp(1/3). See A228935.

The convergents are a subset of those of the standard regular continued fraction; the sequence of the signs of the difference between the convergents and exp(1/3) starts with: -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, ...

For every couple of successive equal signs in this sequence there is a convergent of the standard expansion not present in this one.

Repeating the expansion for other numbers of type 1/k a common pattern seems to emerge. Examples:

exp(1/4) gives 1, 4, -2, -12, 2, 20, -2, -28, 2, 36, -2, -44, 2, 52, ...

exp(1/5) gives 1, 5, -2, -15, 2, 25, -2, -35, 2, 45, -2, -55, 2, 65, ...

so it seems that in general the terms for exp(1/k) are generated by the formulas a(0)=1 , a(2n+1)=(-1)^n*k*(2n+1) for n>=0 , a(2n)=(-1)^n*2 for n>0 . These formulas give this expansion for exp(1/k):

exp(1/k) = 1+1/(k+1/(-2+1/(-3k+1/(2+1/(5k+1/(-2+1/(-7k+1/(2+....)))))))).

that can be rewritten in this equivalent form:

exp(1/k) = 1+1/(k-1/(2+1/(3k-1/(2+1/(5k-1/(2+1/(7k-1/(2+....)))))))).

This general expansion seems to be valid for any real value of k.

Closed form for the general case exp(1/k): b(n) = (1+(-1)^n-(1-(-1)^n)*k*n/2)*i^(n*(n+1)) for n>0 and with i=sqrt(-1). [Bruno Berselli, Nov 01 2013]

LINKS

Table of n, a(n) for n=0..49.

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

FORMULA

This sequence can be generated by these formulas:

a(0)=1; for n>=0, a(2n+1) = 3*(-1)^n*(2n+1), a(2n) = 2*(-1)^n for n>0.

Formulae for the general case exp(1/k):

b(0)=1; for n>=0, b(2n+1) = (-1)^n*k*(2n+1), b(2n) = 2*(-1)^n.

b(n) = 2*cos(n*Pi/2)+k*n*sin(n*Pi/2) for n>0.

exp(1/k) = 1+1/(k-1/(2+1/(3k-1/(2+1/(5k-1/(2+1/(7k-1/(2+....)))))))).

G.f. : (1-x)*(1+x)*(1+k*x+x^2)/(1+x^2)^2.

From Colin Barker, Oct 26 2013: (Start)

a(n) = (-i)^n+i^n+1/2*(((-i)^n-i^n)*n)*(3*i) for n>0, where i=sqrt(-1).

a(2n) = 2*(3*n*sin(Pi*n)+cos(Pi*n)) for n>0.

a(2n+1) = (6*n+3)*cos(Pi*n)-2*sin(Pi*n) for n>=0.

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

G.f.: -(x-1)*(x+1)*(x^2+3*x+1) / (x^2+1)^2. (End)

EXAMPLE

exp(1/3)=1+1/(3+1/(-2+1/(-9+1/(2+1/(15+1/(-2+1/(-21+1/(2+....)))))))) or

exp(1/3)=1+1/(3-1/(2+1/(9-1/(2+1/(15-1/(2+1/(21-1/(2+....))))))))

MAPLE

SCF := proc (n, q::posint)::list; local L, i, z; Digits := 10000; L := [round(n)]; z := n; for i from 2 to q do if z = op(-1, L) then break end if; z := 1/(z-op(-1, L)); L := [op(L), round(z)] end do; return L end proc

SCF(exp(1/3), 50)  # Giovanni Artico , Oct 26 2013

PROG

(PARI) Vec(-(x-1)*(x+1)*(x^2+3*x+1)/(x^2+1)^2+O(x^100)) \\ Colin Barker, Oct 26 2013

CROSSREFS

Cf. A133593, A133570, A228935.

Sequence in context: A143074 A173624 A182023 * A169862 A245884 A192492

Adjacent sequences:  A228933 A228934 A228935 * A228937 A228938 A228939

KEYWORD

sign,cofr,easy

AUTHOR

Giovanni Artico, Oct 26 2013

STATUS

approved

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Last modified March 19 18:49 EDT 2019. Contains 321330 sequences. (Running on oeis4.)