cp's OEIS Frontend

A174510 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A080040(n)) ), where A080040(n) = (1+sqrt(3))^n + (1-sqrt(3))^n.

%I A174510 #11 Mar 25 2024 09:21:55
%S A174510 1,1,3,1,9,13,1,37,51,1,141,193,1,529,723,1,1977,2701,1,7381,10083,1,
%T A174510 27549,37633,1,102817,140451,1,383721,524173,1,1432069,1956243,1,
%U A174510 5344557,7300801,1,19946161,27246963,1,74440089,101687053,1,277814197
%N A174510 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A080040(n)) ), where A080040(n) = (1+sqrt(3))^n + (1-sqrt(3))^n.
%H A174510 <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,5,0,0,-5,0,0,1).
%F A174510 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].
%F A174510 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]
%e A174510 Let L = Sum_{n>=1} 1/(n*A080040(n)) or, more explicitly,
%e A174510 L = 1/2 + 1/(2*8) + 1/(3*20) + 1/(4*56) + 1/(5*152) + 1/(6*416) +...
%e A174510 so that L = 0.5855329921665857283309456463364081071245363598803...
%e A174510 then exp(L) = 1.7959479567807442397990076546690432122217738278933...
%e A174510 equals the continued fraction given by this sequence:
%e A174510 exp(L) = [1;1,3,1,9,13,1,37,51,1,141,193,1,529,723,1,...]; i.e.,
%e A174510 exp(L) = 1 + 1/(1 + 1/(3 + 1/(1 + 1/(9 + 1/(13 + 1/(1 +...)))))).
%e A174510 Compare these partial quotients to A080040(n)/2^[n/2], n=1,2,3,...:
%e A174510 [2,4,10,14,38,52,142,194,530,724,1978,2702,7382,10084,27550,...],
%e A174510 where A080040 begins:
%e A174510 [2,8,20,56,152,416,1136,3104,8480,23168,63296,172928,472448,...].
%o A174510 (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]}
%Y A174510 Cf. A080040, A174500, A174504, A174509.
%K A174510 cofr,nonn,easy
%O A174510 0,3
%A A174510 _Paul D. Hanna_, Mar 21 2010