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A174503 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A087799(n)) ), where A087799(n) = (5+sqrt(24))^n + (5-sqrt(24))^n.

%I A174503 #21 Mar 25 2024 06:38:48
%S A174503 1,8,1,96,1,968,1,9600,1,95048,1,940896,1,9313928,1,92198400,1,
%T A174503 912670088,1,9034502496,1,89432354888,1,885289046400,1,8763458109128,
%U A174503 1,86749292044896,1,858729462339848,1,8500545331353600,1
%N A174503 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A087799(n)) ), where A087799(n) = (5+sqrt(24))^n + (5-sqrt(24))^n.
%H A174503 Peter Bala, <a href="/A174500/a174500_2.pdf">Some simple continued fraction expansions for an infinite product, Part 1</a>
%H A174503 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,11,0,-11,0,1).
%F A174503 a(2n-2) = 1, a(2n-1) = A087799(n) - 2, for n>=1 [conjecture].
%F A174503 The above conjectures are correct. See the Bala link for details. - _Peter Bala_, Jan 08 2013
%F A174503 a(n) = 11*a(n-2)-11*a(n-4)+a(n-6). G.f.: -(x^4+8*x^3-10*x^2+8*x+1) / ((x-1)*(x+1)*(x^4-10*x^2+1)). [_Colin Barker_, Jan 20 2013]
%e A174503 Let L = Sum_{n>=1} 1/(n*A087799(n)) or, more explicitly,
%e A174503 L = 1/10 + 1/(2*98) + 1/(3*970) + 1/(4*9602) + 1/(5*95050) +...
%e A174503 so that L = 0.1054740177896236251618898675297390156061405857647...
%e A174503 then exp(L) = 1.1112372317482311056432125938345153306039099019639...
%e A174503 equals the continued fraction given by this sequence:
%e A174503 exp(L) = [1;8,1,96,1,968,1,9600,1,95048,1,940896,1,...]; i.e.,
%e A174503 exp(L) = 1 + 1/(8 + 1/(1 + 1/(96 + 1/(1 + 1/(968 + 1/(1 +...)))))).
%e A174503 Compare these partial quotients to A087799(n), n=1,2,3,...:
%e A174503 [10,98,970,9602,95050,940898,9313930,92198402,912670090,9034502498,...].
%o A174503 (PARI) {a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((5+sqrt(24))^m+(5-sqrt(24))^m))));contfrac(exp(L))[n]}
%Y A174503 Cf. A087799, A174500, A174502, A174504, A174509.
%K A174503 cofr,nonn,easy
%O A174503 0,2
%A A174503 _Paul D. Hanna_, Mar 21 2010