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A174504 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A002203(n)) ), where A002203(n) = (1+sqrt(2))^n + (1-sqrt(2))^n.

%I A174504 #21 Jul 06 2017 02:58:51
%S A174504 1,1,5,1,13,33,1,81,197,1,477,1153,1,2785,6725,1,16237,39201,1,94641,
%T A174504 228485,1,551613,1331713,1,3215041,7761797,1,18738637,45239073,1,
%U A174504 109216785,263672645,1,636562077,1536796801,1,3710155681,8957108165,1
%N A174504 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A002203(n)) ), where A002203(n) = (1+sqrt(2))^n + (1-sqrt(2))^n.
%H A174504 G. C. Greubel, <a href="/A174504/b174504.txt">Table of n, a(n) for n = 0..1000</a>
%H A174504 P. Bala, <a href="/A174500/a174500_2.pdf">Some simple continued fraction expansions for an infinite product, Part 1</a>
%H A174504 P. Bala, <a href="/A174504/a174504.pdf">Some simple continued fraction expansions for an infinite product, Part 2</a>
%H A174504 <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,7,0,0,-7,0,0,1).
%F A174504 a(3n-3) = 1, a(3n-2) = A002203(2n-1) - 1, a(3n-1) = A002203(2n) - 1, for n>=1 [conjecture].
%F A174504 From _Colin Barker_, Jan 20 2013: (Start)
%F A174504 a(n) = 7*a(n-3) - 7*a(n-6) + a(n-9).
%F A174504 G.f.: -(x^2-x+1)*(x^6-2*x^5-2*x^4-2*x^3+6*x^2+2*x+1) / ((x-1)*(x^2+x+1)*(x^6-6*x^3+1)). (End)
%F A174504 From _Peter Bala_, Jan 25 2013: (Start)
%F A174504 The above conjectures are correct. The real number exp( Sum {n>=1} 1/(n*A002203(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(2) - 1. 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 = 2. See the Bala link for details.
%F A174504 The theory also provides the simple continued fraction expansion of the numbers F({sqrt(2) - 1}^(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(2) - 1}^(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 A174504 (End)
%e A174504 Let L = Sum_{n>=1} 1/(n*A002203(n)) or, more explicitly,
%e A174504 L = 1/2 + 1/(2*6) + 1/(3*14) + 1/(4*34) + 1/(5*82) + 1/(6*198) +...
%e A174504 so that L = 0.6182475539420223547415636201969551910173293917288...
%e A174504 then exp(L) = 1.8556732235071087707741415885016794127474675663938...
%e A174504 equals the continued fraction given by this sequence:
%e A174504 exp(L) = [1;1,5,1,13,33,1,81,197,1,477,1153,1,2785,6725,1,...]; i.e.,
%e A174504 exp(L) = 1 + 1/(1 + 1/(5 + 1/(1 + 1/(13 + 1/(33 + 1/(1 +...)))))).
%e A174504 Compare these partial quotients to A002203(n), n=1,2,3,...:
%e A174504 [2,6,14,34,82,198,478,1154,2786,6726,16238,39202,94642,228486,...].
%t A174504 LinearRecurrence[{0,0,7,0,0,-7,0,0,1},{1,1,5,1,13,33,1,81,197},40] (* _Harvey P. Dale_, Sep 15 2016 *)
%o A174504 (PARI) {a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((1+sqrt(2))^m+(1-sqrt(2))^m))));contfrac(exp(L))[n]}
%Y A174504 Cf. A002203 (companion Pell numbers), A174500, A174503, A174505.
%K A174504 cofr,nonn,easy
%O A174504 0,3
%A A174504 _Paul D. Hanna_, Mar 21 2010