OFFSET
0,2
COMMENTS
Optimal simple continued fraction (with signed denominators) of exp(2/5)
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(2/5) starts with:
-1, 1, -1, -1, 1, -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 2/k a common pattern seems to emerge. Examples:
exp(2/7) gives 1, 3, 42, 18, -2, -24, -126, -39, 2, 45, 210, 60, -2,...
exp(2/9) gives 1, 4, 54, 23, -2, -31, -162, -50, 2, 58, 270, 77, -2,...
so it seems that in general the terms for exp(2/k) are generated by the following formulas:
b(0)=1, b(1)=k/2-1/2, b(2)=6*k, b(3)=5*k/2+1/2, b(4)=-2, b(5)=7*k/2+1/2, b(6)=-18*k, b(7)=-11*k/2-1/2, b(8)=2; b(n) = -2*b(n-4) -b(n-8) for n>8, recurrence which corresponds to the g.f. 1/2*(1-x)*(1+x)*(2*(1+x^6)+(k-1)*(x+x^5)+(12*k+2)*(x^2+x^4)+6*k*x^3)/(1+x^4)^2; also:
b(0)=1 , b(4m+1)=(-1)^m*((k-1)/2+3*k*m), b(4m+3)=(-1)^m*((5*k+1)/2+3*k*m), b(4m+2)=(-1)^m*(6*k+12*k*m), b(4m+4)=(-1)^(m+1)*2 for n>=0.
These formulas give this expansion for exp(2/k):
exp(2/k)=1+1/((k-1)/2+1/(6k+1/((5k+1)/2+1/(-2+1/(-(7k-1)/2+1/...)))))
that can be rewritten in this equivalent form:
exp(2/k)=1+1/(k/2-1/2+1/(6k+1/(5k/2+1/2-1/(2+1/(7x/2-1/2+1/...))))).
This general expansion seems to be valid for any real value of k.
LINKS
Index entries for linear recurrences with constant coefficients, signature (0,0,0,-2,0,0,0,-1).
FORMULA
G.f.: (1+2*x+30*x^2+13*x^3-13*x^5-30*x^6-2*x^7-x^8)/(1+2*x^4+x^8).
a(0)=1, a(1)=2, a(2)=30, a(3)=13, a(4)=-2, a(5)=-17, a(6)=-90, a(7)=-28, a(8)=2; for n>8, a(n) = -2*a(n-4) -a(n-8).
a(0)=1 , a(4m+1) = (-1)^m*(2+15*m), a(4m+3) = (-1)^m*(13+15*m), a(4m+2) = (-1)^m*(30+60*m), a(4m+4) = 2*(-1)^(m+1) for m>=0.
EXAMPLE
exp(2/5)=1+1/(2+1/(30+1/(13+1/(-2+1/(-17+1/(-90+1/(-28+1/(2+...)))))))),
or equivalently:
exp(2/5)=1+1/(2+1/(30+1/(13-1/(2+1/(17+1/(90+1/(28-1/(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(2/5), 50)
MATHEMATICA
Join[{1}, LinearRecurrence[{0, 0, 0, -2, 0, 0, 0, -1}, {2, 30, 13, -2, -17, -90, -28, 2}, 50]] (* Bruno Berselli, Nov 06 2013 *)
CROSSREFS
KEYWORD
cofr,sign,easy
AUTHOR
Giovanni Artico, Oct 28 2013
STATUS
approved