A174507 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A085447(n)) ), where A085447(n) = (3+sqrt(10))^n + (3-sqrt(10))^n.

1, 5, 37, 1, 233, 1441, 1, 8885, 54757, 1, 337433, 2079361, 1, 12813605, 78960997, 1, 486579593, 2998438561, 1, 18477210965, 113861704357, 1, 701647437113, 4323746327041, 1, 26644125399365, 164188498723237, 1, 1011775117738793

Offset

0,2

Links

Table of n, a(n) for n=0..28.

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 (0,0,39,0,0,-39,0,0,1).

Formula

a(3n-3) = 1, a(3n-2) = A085447(2n-1) - 1, a(3n-1) = A085447(2n) - 1, for n>=1 [conjecture].

a(n) = 39*a(n-3)-39*a(n-6)+a(n-9). G.f.: -(x^2-x+1) * (x^6 -6*x^5 -6*x^4 -2*x^3 +42*x^2 +6*x +1) / ((x-1)*(x^2+x+1)*(x^6-38*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*A085447(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(10) - 3. 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 = 6. See the Bala link for details.

The theory also provides the simple continued fraction expansion of the numbers F({sqrt(10) - 3}^(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(10) - 3}^(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)

Example

Let L = Sum_{n>=1} 1/(n*A085447(n)) or, more explicitly,
L = 1/6 + 1/(2*38) + 1/(3*234) + 1/(4*1442) + 1/(5*8886) +...
so that L = 0.1814484777922995750614847484088330644558009487798...
then exp(L) = 1.1989527624251050123398509513177598419795554140316...
equals the continued fraction given by this sequence:
exp(L) = [1;5,37,1,233,1441,1,8885,54757,1,337433,...]; i.e.,
exp(L) = 1 + 1/(5 + 1/(37 + 1/(1 + 1/(233 + 1/(1441 + 1/(1 +...)))))).
Compare these partial quotients to A085447(n), n=1,2,3,...:
[6,38,234,1442,8886,54758,337434,2079362,12813606,78960998,...].

Mathematica

LinearRecurrence[{0, 0, 39, 0, 0, -39, 0, 0, 1}, {1, 5, 37, 1, 233, 1441, 1, 8885, 54757}, 30] (* Harvey P. Dale, Aug 07 2016 *)

Prog

(PARI) {a(n)=local(L=sum(m=1, 2*n+1000, 1./(m*round((3+sqrt(10))^m+(3-sqrt(10))^m)))); contfrac(exp(L))[n]}

Crossrefs

Cf. A085447, A174501, A174506, A174508.

Sequence in context: A156355 A002666 A222592 * A119483 A157809 A079339

Adjacent sequences: A174504 A174505 A174506 * A174508 A174509 A174510

Keyword

cofr,nonn,easy

Author

Paul D. Hanna, Mar 21 2010

Status

approved