%I #17 Mar 25 2024 06:38:53

%S 3,1,3,6,1,10,17,1,28,46,1,75,122,1,198,321,1,520,842,1,1363,2206,1,

%T 3570,5777,1,9348,15126,1,24475,39602,1,64078,103681,1,167760,271442,

%U 1,439203,710646,1,1149850,1860497,1,3010348,4870846,1,7881195,12752042,1

%N Continued fraction expansion for exp( Sum_{n>=1} 1/(n*Lucas(n)) ), where Lucas(n) = A000032(n) = ((1+sqrt(5))/2)^n + ((1-sqrt(5))/2)^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 (0,0,4,0,0,-4,0,0,1).

%F a(3n-2) = 1, a(3n-1) = A000032(2n+1) - 1, a(3n) = A000032(2n+2) - 1, for n>=1 [conjecture].

%F a(n) = 4*a(n-3)-4*a(n-6)+a(n-9) for n>9. G.f.: (x^9 -x^7 -5*x^6 +2*x^5 +3*x^4 +6*x^3 -3*x^2 -x -3) / ((x -1)*(x^2 +x +1)*(x^6 -3*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*Lucas(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 = 1/2*(sqrt(5) - 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 = 1. See the Bala link for details.

%F The theory also provides the simple continued fraction expansion of the numbers F({1/2*(sqrt(5) - 1)}^(2*k+1)), k = 1, 2, 3, ...: if [3, 1, c(3), c(4), 1, c(5), c(6), 1, ...] denotes the present sequence then the simple continued fraction expansion of F({1/2*(sqrt(5) - 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 (End)

%e Let L = Sum_{n>=1} 1/(n*A000032(n)) or, more explicitly,

%e L = 1 + 1/(2*3) + 1/(3*4) + 1/(4*7) + 1/(5*11) + 1/(6*18) +...

%e so that L = 1.3240810281350207977825663314844927483236088628781...

%e then exp(L) = 3.7587296006215531704236522952745520722012715044952...

%e equals the continued fraction given by this sequence:

%e exp(L) = [3;1,3,6,1,10,17,1,28,46,1,75,122,1,198,321,1,...]; i.e.,

%e exp(L) = 3 + 1/(1 + 1/(3 + 1/(6 + 1/(1 + 1/(10 + 1/(17 +...)))))).

%e Compare these partial quotients to A000032(n), n=1,2,3,...:

%e [1,3,4,7,11,18,29,47,76,123,199,322,521,843,1364,2207,3571,5778,...].

%o (PARI) {a(n)=local(L=sum(m=1,2*n+1000,1./(m*round(((1+sqrt(5))/2)^m+((1-sqrt(5))/2)^m))));contfrac(exp(L))[n]}

%Y Cf. A000032 (Lucas numbers), A174500, A174504, A174506.

%K cofr,nonn,easy

%O 0,1

%A _Paul D. Hanna_, Mar 21 2010