%I #33 Mar 02 2022 08:58:23
%S 1,2,1,12,1,50,1,192,1,722,1,2700,1,10082,1,37632,1,140450,1,524172,1,
%T 1956242,1,7300800,1,27246962,1,101687052,1,379501250,1,1416317952,1,
%U 5285770562,1,19726764300,1,73621286642,1,274758382272,1
%N Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A003500(n)) ), where A003500(n) = (2+sqrt(3))^n + (2-sqrt(3))^n.
%H P. 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 (0,5,0,-5,0,1).
%F a(2n-1) = 1, a(2n) = A003500(n) - 2, for n>=1 [conjecture].
%F Contribution from _Peter Bala_, Jan 04 2013: (Start)
%F The above conjectures are correct. The real number exp( sum {n>=1} 1/(n*A003500(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 = 2 - sqrt(3). Ramanujan has given a continued fraction expansion for F(x). Using this we can find the simple continued fraction expansion of the numbers F(1/2*(N - sqrt(N^2 - 4))), N an integer greater than 3. The present case is when N = 4. See the Bala link for details.
%F The theory also provides the simple continued fraction expansion of the numbers F({2 - sqrt(3)}^k), k = 1,2,3,...: if [1; c(1), 1, c(2), 1, c(3), 1, ...] denotes the present sequence then the simple continued fraction expansion of F({2 - sqrt(3)}^k) is given by [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...].
%F (End)
%F a(n) = 5*a(n-2)-5*a(n-4)+a(n-6). G.f.: -x*(x^4+2*x^3-4*x^2+2*x+1) / ((x-1)*(x+1)*(x^4-4*x^2+1)). [_Colin Barker_, Jan 20 2013]
%e Let L = Sum_{n>=1} 1/(n*A003500(n)) or, more explicitly,
%e L = 1/4 + 1/(2*14) + 1/(3*52) + 1/(4*194) + 1/(5*724) + 1/(6*2702) +...
%e so that L = 0.2937696594138291094177057532058145970820225289928...
%e then exp(L) = 1.3414748719687236691269115428250035920032300984596...
%e equals the continued fraction expansion given by this sequence:
%e exp(L) = [1;2,1,12,1,50,1,192,1,722,1,2700,1,10082,1,...]; i.e.,
%e exp(L) = 1 + 1/(2 + 1/(1 + 1/(12 + 1/(1 + 1/(50 + 1/(1 +...))))).
%e Compare these partial quotients to A003500(n), n=1,2,3,...:
%e [4,14,52,194,724,2702,10084,37634,140452,524174,1956244,...].
%t a[n_?OddQ] = 1; a[n_?EvenQ] := a[n] = 4*a[n-2] - a[n-4] + 4; a[2] = 2; a[4] = 12; Table[a[n], {n, 1, 41}] (* _Jean-François Alcover_, May 15 2014, after the first conjecture *)
%o (PARI) {a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((2+sqrt(3))^m+(2-sqrt(3))^m))));contfrac(exp(L))[n]}
%Y Cf. A003500, A174501.
%K cofr,nonn,easy
%O 1,2
%A _Paul D. Hanna_, Mar 20 2010
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