A174510 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A080040(n)) ), where A080040(n) = (1+sqrt(3))^n + (1-sqrt(3))^n.
1, 1, 3, 1, 9, 13, 1, 37, 51, 1, 141, 193, 1, 529, 723, 1, 1977, 2701, 1, 7381, 10083, 1, 27549, 37633, 1, 102817, 140451, 1, 383721, 524173, 1, 1432069, 1956243, 1, 5344557, 7300801, 1, 19946161, 27246963, 1, 74440089, 101687053, 1, 277814197
Offset: 0
Examples
Let L = Sum_{n>=1} 1/(n*A080040(n)) or, more explicitly,
L = 1/2 + 1/(2*8) + 1/(3*20) + 1/(4*56) + 1/(5*152) + 1/(6*416) +...
so that L = 0.5855329921665857283309456463364081071245363598803...
then exp(L) = 1.7959479567807442397990076546690432122217738278933...
equals the continued fraction given by this sequence:
exp(L) = [1;1,3,1,9,13,1,37,51,1,141,193,1,529,723,1,...]; i.e.,
exp(L) = 1 + 1/(1 + 1/(3 + 1/(1 + 1/(9 + 1/(13 + 1/(1 +...)))))).
Compare these partial quotients to A080040(n)/2^[n/2], n=1,2,3,...:
[2,4,10,14,38,52,142,194,530,724,1978,2702,7382,10084,27550,...],
where A080040 begins:
[2,8,20,56,152,416,1136,3104,8480,23168,63296,172928,472448,...].
Links
- Index entries for linear recurrences with constant coefficients, signature (0,0,5,0,0,-5,0,0,1).
Programs
-
PARI
{a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((1+sqrt(3))^m+(1-sqrt(3))^m))));contfrac(exp(L))[n]}
Formula
a(3n-3) = 1, a(3n-2) = A080040(2n-1)/2^(n-1) - 1, a(3n-1) = A080040(2n)/2^n - 1, for n>=1 [conjecture].
a(n) = 5*a(n-3)-5*a(n-6)+a(n-9). G.f.: -(x^2-x+1)*(x^6-2*x^5-2*x^4-2*x^3+4*x^2+2*x+1) / ((x-1)*(x^2+x+1)*(x^6-4*x^3+1)). [Colin Barker, Jan 20 2013]