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

Author

Paul D. Hanna, Mar 21 2010

Keywords

• cofr
• nonn
• easy

Examples

			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,...].
		

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

Crossrefs

Cf. A085447, A174501, A174506, A174508.

Programs

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 *)

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]}

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)