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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

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

Views

Author

Giovanni Artico, Oct 26 2013

Keywords

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

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)

A228937 Expansion of (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).

Original entry on oeis.org

1, 2, 30, 13, -2, -17, -90, -28, 2, 32, 150, 43, -2, -47, -210, -58, 2, 62, 270, 73, -2, -77, -330, -88, 2, 92, 390, 103, -2, -107, -450, -118, 2, 122, 510, 133, -2, -137, -570, -148, 2, 152, 630, 163, -2, -167, -690, -178, 2, 182
Offset: 0

Views

Author

Giovanni Artico, Oct 28 2013

Keywords

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.

Examples

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

Links

Crossrefs

Cf. A133593, A133570, A228935, A228936.

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(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 *)

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.
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