A174505 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.
3, 1, 3, 6, 1, 10, 17, 1, 28, 46, 1, 75, 122, 1, 198, 321, 1, 520, 842, 1, 1363, 2206, 1, 3570, 5777, 1, 9348, 15126, 1, 24475, 39602, 1, 64078, 103681, 1, 167760, 271442, 1, 439203, 710646, 1, 1149850, 1860497, 1, 3010348, 4870846, 1, 7881195, 12752042, 1
Offset: 0
Examples
Let L = Sum_{n>=1} 1/(n*A000032(n)) or, more explicitly,
L = 1 + 1/(2*3) + 1/(3*4) + 1/(4*7) + 1/(5*11) + 1/(6*18) +...
so that L = 1.3240810281350207977825663314844927483236088628781...
then exp(L) = 3.7587296006215531704236522952745520722012715044952...
equals the continued fraction given by this sequence:
exp(L) = [3;1,3,6,1,10,17,1,28,46,1,75,122,1,198,321,1,...]; i.e.,
exp(L) = 3 + 1/(1 + 1/(3 + 1/(6 + 1/(1 + 1/(10 + 1/(17 +...)))))).
Compare these partial quotients to A000032(n), n=1,2,3,...:
[1,3,4,7,11,18,29,47,76,123,199,322,521,843,1364,2207,3571,5778,...].
Links
- Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1.
- Peter Bala, Some simple continued fraction expansions for an infinite product, Part 2.
- Index entries for linear recurrences with constant coefficients, signature (0,0,4,0,0,-4,0,0,1).
Programs
-
Mathematica
LinearRecurrence[{0,0,4,0,0,-4,0,0,1},{3,1,3,6,1,10,17,1,28,46},50] (* Harvey P. Dale, Feb 02 2025 *) -
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]}
Formula
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]
From Peter Bala, Jan 25 2013: (Start)
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.
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, ...].
(End)