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.
%I A228936 #35 Feb 10 2024 12:20:34 %S A228936 1,3,-2,-9,2,15,-2,-21,2,27,-2,-33,2,39,-2,-45,2,51,-2,-57,2,63,-2, %T A228936 -69,2,75,-2,-81,2,87,-2,-93,2,99,-2,-105,2,111,-2,-117,2,123,-2,-129, %U A228936 2,135,-2,-141,2,147 %N A228936 Expansion of (1 + 3*x - 3*x^3 - x^4)/(1 + 2*x^2 + x^4). %C A228936 Optimal simple continued fraction (with signed denominators) of exp(1/3). See A228935. %C A228936 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, ... %C A228936 For every couple of successive equal signs in this sequence there is a convergent of the standard expansion not present in this one. %C A228936 Repeating the expansion for other numbers of type 1/k a common pattern seems to emerge. Examples: %C A228936 exp(1/4) gives 1, 4, -2, -12, 2, 20, -2, -28, 2, 36, -2, -44, 2, 52, ... %C A228936 exp(1/5) gives 1, 5, -2, -15, 2, 25, -2, -35, 2, 45, -2, -55, 2, 65, ... %C A228936 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): %C A228936 exp(1/k) = 1+1/(k+1/(-2+1/(-3k+1/(2+1/(5k+1/(-2+1/(-7k+1/(2+...)))))))). %C A228936 which can be rewritten in this equivalent form: %C A228936 exp(1/k) = 1+1/(k-1/(2+1/(3k-1/(2+1/(5k-1/(2+1/(7k-1/(2+...)))))))). %C A228936 This general expansion seems to be valid for any real value of k. %C A228936 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] %H A228936 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,-2,0,-1). %F A228936 This sequence can be generated by these formulas: %F A228936 a(0)=1; for n >= 0, a(2n+1) = 3*(-1)^n*(2n+1), a(2n) = 2*(-1)^n for n > 0. %F A228936 Formulae for the general case exp(1/k): %F A228936 b(0)=1; for n >= 0, b(2n+1) = (-1)^n*k*(2n+1), b(2n) = 2*(-1)^n. %F A228936 b(n) = 2*cos(n*Pi/2) + k*n*sin(n*Pi/2) for n > 0. %F A228936 exp(1/k) = 1+1/(k-1/(2+1/(3k-1/(2+1/(5k-1/(2+1/(7k-1/(2+...)))))))). %F A228936 G.f.: (1-x)*(1+x)*(1+k*x+x^2)/(1+x^2)^2. %F A228936 From _Colin Barker_, Oct 26 2013: (Start) %F A228936 a(n) = (-i)^n + i^n + (1/2)*(((-i)^n-i^n)*n)*(3*i) for n > 0, where i=sqrt(-1). %F A228936 a(2n) = 2*(3*n*sin(Pi*n) + cos(Pi*n)) for n > 0. %F A228936 a(2n+1) = (6*n+3)*cos(Pi*n) - 2*sin(Pi*n) for n >= 0. %F A228936 a(n) = -2*a(n-2) - a(n-4) for n > 4. %F A228936 G.f.: -(x-1)*(x+1)*(x^2+3*x+1) / (x^2+1)^2. (End) %e A228936 exp(1/3) = 1+1/(3+1/(-2+1/(-9+1/(2+1/(15+1/(-2+1/(-21+1/(2+...)))))))) or %e A228936 exp(1/3) = 1+1/(3-1/(2+1/(9-1/(2+1/(15-1/(2+1/(21-1/(2+...)))))))) %p A228936 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 %p A228936 SCF(exp(1/3), 50) # _Giovanni Artico_, Oct 26 2013 %o A228936 (PARI) Vec(-(x-1)*(x+1)*(x^2+3*x+1)/(x^2+1)^2+O(x^100)) \\ _Colin Barker_, Oct 26 2013 %Y A228936 Cf. A133593, A133570, A228935. %K A228936 sign,cofr,easy %O A228936 0,2 %A A228936 _Giovanni Artico_, Oct 26 2013