A174506 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A014448(n)) ), where A014448(n) = (2+sqrt(5))^n + (2-sqrt(5))^n.
1, 3, 17, 1, 75, 321, 1, 1363, 5777, 1, 24475, 103681, 1, 439203, 1860497, 1, 7881195, 33385281, 1, 141422323, 599074577, 1, 2537720635, 10749957121, 1, 45537549123, 192900153617, 1, 817138163595, 3461452808001, 1
Offset: 0
Examples
Let L = Sum_{n>=1} 1/(n*A014448(n)) or, more explicitly,
L = 1/4 + 1/(2*18) + 1/(3*76) + 1/(4*322) + 1/(5*1364) +...
so that L = 0.2831229765066671850017990708479258794794782639219...
then exp(L) = 1.3272683746094012523448609429829013914921330866098...
equals the continued fraction given by this sequence:
exp(L) = [1;3,17,1,75,321,1,1363,5777,1,24475,103681,1,...]; i.e.,
exp(L) = 1 + 1/(3 + 1/(17 + 1/(1 + 1/(75 + 1/(321 + 1/(1 +...)))))).
Compare these partial quotients to A014448(n), n=1,2,3,...:
[4,18,76,322,1364,5778,24476,103682,439204,1860498,7881196,33385282,...].
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 (0,0,19,0,0,-19,0,0,1).
Programs
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PARI
{a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((2+sqrt(5))^m+(2-sqrt(5))^m))));contfrac(exp(L))[n]}
Formula
a(n) = 19*a(n-3)-19*a(n-6)+a(n-9). G.f.: -(x^2 -x +1)*(x^6 -4*x^5 -4*x^4 -2*x^3 +20*x^2 +4*x +1) / ((x -1)*(x^2 -3*x +1)*(x^2 +x +1)*(x^4 +3*x^3 +8*x^2 +3*x +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*A014448(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(5) - 2. 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 = 4. See the Bala link for details.
The theory also provides the simple continued fraction expansion of the numbers F({sqrt(5) - 2}^(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(5) - 2}^(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)