A228936 Expansion of (1 + 3*x - 3*x^3 - x^4)/(1 + 2*x^2 + x^4).

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

Offset: 0

Author

Giovanni Artico, Oct 26 2013

Keywords

• sign
• cofr
• easy

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+...)))))))).

which 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]

Examples

			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+...))))))))
		

Links

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

Crossrefs

Cf. A133593, A133570, A228935.

Programs

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

PARI

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

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)