A174501 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A003499(n)) ), where A003499(n) = (3+sqrt(8))^n + (3-sqrt(8))^n.
1, 4, 1, 32, 1, 196, 1, 1152, 1, 6724, 1, 39200, 1, 228484, 1, 1331712, 1, 7761796, 1, 45239072, 1, 263672644, 1, 1536796800, 1, 8957108164, 1, 52205852192, 1, 304278004996, 1, 1773462177792, 1, 10336495061764, 1, 60245508192800, 1, 351136554095044, 1
Offset: 1
Examples
Let L = Sum_{n>=1} 1/(n*A003499(n)) or, more explicitly,
L = 1/6 + 1/(2*34) + 1/(3*198) + 1/(4*1154) + 1/(5*6726) +...
so that L = 0.1833074113563494600094468694966574405706183998044...
then exp(L) = 1.2011836088120841844713993433258934531421726294252...
equals the continued fraction given by this sequence:
exp(L) = [1;4,1,32,1,196,1,1152,1,6724,1,39200,1,...]; i.e.,
exp(L) = 1 + 1/(4 + 1/(1 + 1/(32 + 1/(1 + 1/(196 + 1/(1 +...)))))).
Compare these partial quotients to A003499(n), n=1,2,3,...:
[6,34,198,1154,6726,39202,228486,1331714,7761798,45239074,...].
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
- Index entries for linear recurrences with constant coefficients, signature (0,7,0,-7,0,1).
Programs
-
Mathematica
LinearRecurrence[{0,7,0,-7,0,1},{1,4,1,32,1,196},50] (* Harvey P. Dale, Jul 14 2021 *) -
PARI
{a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((3+sqrt(8))^m+(3-sqrt(8))^m))));contfrac(exp(L))[n]} -
PARI
Vec(-x*(x^4+4*x^3-6*x^2+4*x+1)/((x-1)*(x+1)*(x^2-2*x-1)*(x^2+2*x-1)) + O(x^50)) \\ Colin Barker, May 11 2016
Formula
a(2n-1) = 1, a(2n) = A003499(n) - 2, for n>=1 [conjecture].
The above conjectures are correct. See the Bala link for details. - Peter Bala, Jan 08 2013
a(n) = 7*a(n-2)-7*a(n-4)+a(n-6). G.f.: -x*(x^4+4*x^3-6*x^2+4*x+1) / ((x-1)*(x+1)*(x^2-2*x-1)*(x^2+2*x-1)). - Colin Barker, Jan 20 2013
a(n) = (((-1-sqrt(2))^n+(1-sqrt(2))^n+(sqrt(2)-1)^n+(1+sqrt(2))^n-4))/2 for n even. - Colin Barker, May 11 2016