A174502 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A086903(n)) ), where A086903(n) = (4+sqrt(15))^n + (4-sqrt(15))^n.

1, 6, 1, 60, 1, 486, 1, 3840, 1, 30246, 1, 238140, 1, 1874886, 1, 14760960, 1, 116212806, 1, 914941500, 1, 7203319206, 1, 56711612160, 1, 446489578086, 1, 3515205012540, 1, 27675150522246, 1, 217885999165440, 1, 1715412842801286, 1

Offset

0,2

Links

Table of n, a(n) for n=0..34.

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

Index entries for linear recurrences with constant coefficients, signature (0,9,0,-9,0,1).

Formula

a(2n-2) = 1, a(2n-1) = A086903(n) - 2, for n>=1 [conjecture].

The above conjectures are correct. See the Bala link for details. - Peter Bala, Jan 08 2013

a(n) = 9*a(n-2)-9*a(n-4)+a(n-6). G.f.: -(x^4+6*x^3-8*x^2+6*x+1) / ((x-1)*(x+1)*(x^4-8*x^2+1)). [Colin Barker, Jan 20 2013]

Example

Let L = Sum_{n>=1} 1/(n*A086903(n)) or, more explicitly,
L = 1/8 + 1/(2*62) + 1/(3*488) + 1/(4*3842) + 1/(5*30248) +...
so that L = 0.1338200441271648228100625767561479963630539052200...
then exp(L) = 1.1431870779045667085973926071888878662387686835715...
equals the continued fraction given by this sequence:
exp(L) = [1;6,1,60,1,486,1,3840,1,30246,1,238140,1,...]; i.e.,
exp(L) = 1 + 1/(6 + 1/(1 + 1/(60 + 1/(1 + 1/(486 + 1/(1 +...)))))).
Compare these partial quotients to A086903(n), n=1,2,3,...:
[8,62,488,3842,30248,238142,1874888,14760962,116212808,914941502,...].

Mathematica

LinearRecurrence[{0, 9, 0, -9, 0, 1}, {1, 6, 1, 60, 1, 486}, 50] (* Harvey P. Dale, Jun 09 2013 *)

Prog

(PARI) {a(n)=local(L=sum(m=1, 2*n+1000, 1./(m*round((4+sqrt(15))^m+(4-sqrt(15))^m)))); contfrac(exp(L))[n]}

Crossrefs

Cf. A086903, A174500, A174501, A174503, A174508.

Sequence in context: A165886 A339100 A344918 * A056218 A352014 A292219

Adjacent sequences: A174499 A174500 A174501 * A174503 A174504 A174505

Keyword

cofr,nonn,easy

Author

Paul D. Hanna, Mar 21 2010

Status

approved