A174504 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A002203(n)) ), where A002203(n) = (1+sqrt(2))^n + (1-sqrt(2))^n.

1, 1, 5, 1, 13, 33, 1, 81, 197, 1, 477, 1153, 1, 2785, 6725, 1, 16237, 39201, 1, 94641, 228485, 1, 551613, 1331713, 1, 3215041, 7761797, 1, 18738637, 45239073, 1, 109216785, 263672645, 1, 636562077, 1536796801, 1, 3710155681, 8957108165, 1

Offset

0,3

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

P. Bala, Some simple continued fraction expansions for an infinite product, Part 1

P. Bala, Some simple continued fraction expansions for an infinite product, Part 2

Index entries for linear recurrences with constant coefficients, signature (0,0,7,0,0,-7,0,0,1).

Formula

a(3n-3) = 1, a(3n-2) = A002203(2n-1) - 1, a(3n-1) = A002203(2n) - 1, for n>=1 [conjecture].

From Colin Barker, Jan 20 2013: (Start)

a(n) = 7*a(n-3) - 7*a(n-6) + a(n-9).

G.f.: -(x^2-x+1)*(x^6-2*x^5-2*x^4-2*x^3+6*x^2+2*x+1) / ((x-1)*(x^2+x+1)*(x^6-6*x^3+1)). (End)

From Peter Bala, Jan 25 2013: (Start)

The above conjectures are correct. The real number exp( Sum {n>=1} 1/(n*A002203(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(2) - 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 = 2. See the Bala link for details.

The theory also provides the simple continued fraction expansion of the numbers F({sqrt(2) - 1}^(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(2) - 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)

Example

Let L = Sum_{n>=1} 1/(n*A002203(n)) or, more explicitly,
L = 1/2 + 1/(2*6) + 1/(3*14) + 1/(4*34) + 1/(5*82) + 1/(6*198) +...
so that L = 0.6182475539420223547415636201969551910173293917288...
then exp(L) = 1.8556732235071087707741415885016794127474675663938...
equals the continued fraction given by this sequence:
exp(L) = [1;1,5,1,13,33,1,81,197,1,477,1153,1,2785,6725,1,...]; i.e.,
exp(L) = 1 + 1/(1 + 1/(5 + 1/(1 + 1/(13 + 1/(33 + 1/(1 +...)))))).
Compare these partial quotients to A002203(n), n=1,2,3,...:
[2,6,14,34,82,198,478,1154,2786,6726,16238,39202,94642,228486,...].

Mathematica

LinearRecurrence[{0, 0, 7, 0, 0, -7, 0, 0, 1}, {1, 1, 5, 1, 13, 33, 1, 81, 197}, 40] (* Harvey P. Dale, Sep 15 2016 *)

Prog

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

Crossrefs

Cf. A002203 (companion Pell numbers), A174500, A174503, A174505.

Sequence in context: A146620 A300291 A366159 * A270654 A067558 A104792

Adjacent sequences: A174501 A174502 A174503 * A174505 A174506 A174507

Keyword

cofr,nonn,easy

Author

Paul D. Hanna, Mar 21 2010

Status

approved