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%I #17 Mar 25 2024 09:21:45
%S 1,9,101,1,1029,10401,1,105049,1060901,1,10714069,108201601,1,
%T 1092730089,11035502501,1,111447755109,1125513053601,1,11366578291129,
%U 114791295964901,1,1159279537940149,11707586675366401,1
%N Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A086927(n)) ), where A086927(n) = (5+sqrt(26))^n + (5-sqrt(26))^n.
%H Peter Bala, <a href="/A174500/a174500_2.pdf">Some simple continued fraction expansions for an infinite product, Part 1</a>.
%H Peter Bala, <a href="/A174504/a174504.pdf">Some simple continued fraction expansions for an infinite product, Part 2</a>.
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,103,0,0,-103,0,0,1).
%F a(3n-3) = 1, a(3n-2) = A086927(2n-1) - 1, a(3n-1) = A086927(2n) - 1, for n>=1 [conjecture].
%F a(n) = 103*a(n-3)-103*a(n-6)+a(n-9). G.f.: -(x^2-x+1)*(x^6-10*x^5-10*x^4-2*x^3+110*x^2+10*x+1) / ((x-1)*(x^2+x+1)*(x^6-102*x^3+1)). [_Colin Barker_, Jan 20 2013]
%F From _Peter Bala_, Jan 25 2013: (Start)
%F The above conjectures are correct. The real number exp( Sum {n>=1} 1/(n*A086927(n)) ) is equal to the infinite product F(x) := product {n >= 0} (1 + x^(4*n+3))/(1 - x^(4*n+1)) evaluated at x = sqrt(26) - 5. Ramanujan has given a continued fraction expansion for the product F(x). Using this we can find the simple continued fraction expansion of the numbers F(1/2*(sqrt(N^2 + 4) - N)), N a positive integer. The present case is when N = 10. See the Bala link for details.
%F The theory also provides the simple continued fraction expansion of the numbers F({sqrt(26) - 5}^(2*k+1)), k = 1, 2, 3, ...: if [1; c(1), c(2), 1, c(3), c(4), 1, ...] denotes the present sequence then the simple continued fraction expansion of F({sqrt(26) - 5}^(2*k+1)) is given by [1; c(2*k+1), c(2*(2*k+1)), 1, c(3*(2*k+1)), c(4*(2*k+1)), 1, ...].
%F (End)
%e Let L = Sum_{n>=1} 1/(n*A086927(n)) or, more explicitly,
%e L = 1/10 + 1/(2*102) + 1/(3*1030) + 1/(4*10402) + 1/(5*105050) +...
%e so that L = 0.1052516947742519131304505213983109248819463097531...
%e then exp(L) = 1.1109902055968924364755807035083159869000358017128...
%e equals the continued fraction given by this sequence:
%e exp(L) = [1;9,101,1,1029,10401,1,105049,1060901,1,...]; i.e.,
%e exp(L) = 1 + 1/(9 + 1/(101 + 1/(1 + 1/(1029 + 1/(10401 +1/(1+...)))))).
%e Compare these partial quotients to A086927(n), n=1,2,3,...:
%e [10,102,1030,10402,105050,1060902,10714070,108201602,...].
%t LinearRecurrence[{0,0,103,0,0,-103,0,0,1},{1,9,101,1,1029,10401,1,105049,1060901},30] (* _Harvey P. Dale_, Dec 24 2014 *)
%o (PARI) {a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((5+sqrt(26))^m+(5-sqrt(26))^m))));contfrac(exp(L))[n]}
%Y Cf. A086927, A174503, A174508, A174510.
%K cofr,nonn,easy
%O 0,2
%A _Paul D. Hanna_, Mar 21 2010
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