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%I #15 Mar 25 2024 09:21:41
%S 1,7,65,1,535,4353,1,35367,287297,1,2333751,18957313,1,153992263,
%T 1250895425,1,10161155671,82540140801,1,670482282087,5446398397505,1,
%U 44241669462135,359379754094593,1,2919279702218887,23713617371845697,1
%N Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A086594(n)) ), where A086594(n) = (4+sqrt(17))^n + (4-sqrt(17))^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 (0,0,67,0,0,-67,0,0,1).
%F a(3n-3) = 1, a(3n-2) = A086594(2n-1) - 1, a(3n-1) = A086594(2n) - 1, for n>=1 [conjecture].
%F a(n) = 67*a(n-3)-67*a(n-6)+a(n-9). G.f.: -(x^2-x+1)*(x^6-8*x^5-8*x^4-2*x^3+72*x^2+8*x+1) / ((x-1)*(x^2+x+1)*(x^6-66*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*A086594(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(17) - 4. 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 = 8. See the Bala link for details.
%F The theory also provides the simple continued fraction expansion of the numbers F({sqrt(17) - 4}^(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(17) - 4}^(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*A086594(n)) or, more explicitly,
%e L = 1/8 + 1/(2*66) + 1/(3*536) + 1/(4*4354) + 1/(5*35368) +...
%e so that L = 0.1332613701545977545822925541573311424901819508933...
%e then exp(L) = 1.1425485874089841897117810754210805471767735522069...
%e equals the continued fraction given by this sequence:
%e exp(L) = [1;7,65,1,535,4353,1,35367,287297,1,2333751,...]; i.e.,
%e exp(L) = 1 + 1/(7 + 1/(65 + 1/(1 + 1/(535 + 1/(4353 + 1/(1 +...)))))).
%e Compare these partial quotients to A086594(n), n=1,2,3,...:
%e [8,66,536,4354,35368,287298,2333752,18957314,153992264,...].
%o (PARI) {a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((4+sqrt(17))^m+(4-sqrt(17))^m))));contfrac(exp(L))[n]}
%Y Cf. A086594, A174502, A174507, A174509.
%K cofr,nonn,easy
%O 0,2
%A _Paul D. Hanna_, Mar 21 2010
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