A174503 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A087799(n)) ), where A087799(n) = (5+sqrt(24))^n + (5-sqrt(24))^n.

1, 8, 1, 96, 1, 968, 1, 9600, 1, 95048, 1, 940896, 1, 9313928, 1, 92198400, 1, 912670088, 1, 9034502496, 1, 89432354888, 1, 885289046400, 1, 8763458109128, 1, 86749292044896, 1, 858729462339848, 1, 8500545331353600, 1

Offset

0,2

Links

Table of n, a(n) for n=0..32.

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

Index entries for linear recurrences with constant coefficients, signature (0,11,0,-11,0,1).

Formula

a(2n-2) = 1, a(2n-1) = A087799(n) - 2, for n>=1 [conjecture].

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

a(n) = 11*a(n-2)-11*a(n-4)+a(n-6). G.f.: -(x^4+8*x^3-10*x^2+8*x+1) / ((x-1)*(x+1)*(x^4-10*x^2+1)). [Colin Barker, Jan 20 2013]

Example

Let L = Sum_{n>=1} 1/(n*A087799(n)) or, more explicitly,
L = 1/10 + 1/(2*98) + 1/(3*970) + 1/(4*9602) + 1/(5*95050) +...
so that L = 0.1054740177896236251618898675297390156061405857647...
then exp(L) = 1.1112372317482311056432125938345153306039099019639...
equals the continued fraction given by this sequence:
exp(L) = [1;8,1,96,1,968,1,9600,1,95048,1,940896,1,...]; i.e.,
exp(L) = 1 + 1/(8 + 1/(1 + 1/(96 + 1/(1 + 1/(968 + 1/(1 +...)))))).
Compare these partial quotients to A087799(n), n=1,2,3,...:
[10,98,970,9602,95050,940898,9313930,92198402,912670090,9034502498,...].

Prog

(PARI) {a(n)=local(L=sum(m=1, 2*n+1000, 1./(m*round((5+sqrt(24))^m+(5-sqrt(24))^m)))); contfrac(exp(L))[n]}

Crossrefs

Cf. A087799, A174500, A174502, A174504, A174509.

Sequence in context: A114152 A347111 A254933 * A048786 A240955 A132056

Adjacent sequences: A174500 A174501 A174502 * A174504 A174505 A174506

Keyword

cofr,nonn,easy

Author

Paul D. Hanna, Mar 21 2010

Status

approved