This site is supported by donations to The OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A174507 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A085447(n)) ), where A085447(n) = (3+sqrt(10))^n + (3-sqrt(10))^n. 2
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 LINKS 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

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified May 24 00:38 EDT 2019. Contains 323528 sequences. (Running on oeis4.)