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%I #22 Mar 25 2024 06:38:44
%S 1,6,1,60,1,486,1,3840,1,30246,1,238140,1,1874886,1,14760960,1,
%T 116212806,1,914941500,1,7203319206,1,56711612160,1,446489578086,1,
%U 3515205012540,1,27675150522246,1,217885999165440,1,1715412842801286,1
%N Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A086903(n)) ), where A086903(n) = (4+sqrt(15))^n + (4-sqrt(15))^n.
%H Peter Bala, <a href="/A174500/a174500_2.pdf">Some simple continued fraction expansions for an infinite product, Part 1</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,9,0,-9,0,1).
%F a(2n-2) = 1, a(2n-1) = A086903(n) - 2, for n>=1 [conjecture].
%F The above conjectures are correct. See the Bala link for details. - _Peter Bala_, Jan 08 2013
%F a(n) = 9*a(n-2)-9*a(n-4)+a(n-6). G.f.: -(x^4+6*x^3-8*x^2+6*x+1) / ((x-1)*(x+1)*(x^4-8*x^2+1)). [_Colin Barker_, Jan 20 2013]
%e Let L = Sum_{n>=1} 1/(n*A086903(n)) or, more explicitly,
%e L = 1/8 + 1/(2*62) + 1/(3*488) + 1/(4*3842) + 1/(5*30248) +...
%e so that L = 0.1338200441271648228100625767561479963630539052200...
%e then exp(L) = 1.1431870779045667085973926071888878662387686835715...
%e equals the continued fraction given by this sequence:
%e exp(L) = [1;6,1,60,1,486,1,3840,1,30246,1,238140,1,...]; i.e.,
%e exp(L) = 1 + 1/(6 + 1/(1 + 1/(60 + 1/(1 + 1/(486 + 1/(1 +...)))))).
%e Compare these partial quotients to A086903(n), n=1,2,3,...:
%e [8,62,488,3842,30248,238142,1874888,14760962,116212808,914941502,...].
%t LinearRecurrence[{0,9,0,-9,0,1},{1,6,1,60,1,486},50] (* _Harvey P. Dale_, Jun 09 2013 *)
%o (PARI) {a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((4+sqrt(15))^m+(4-sqrt(15))^m))));contfrac(exp(L))[n]}
%Y Cf. A086903, A174500, A174501, A174503, A174508.
%K cofr,nonn,easy
%O 0,2
%A _Paul D. Hanna_, Mar 21 2010