OFFSET
0,2
LINKS
Index entries for linear recurrences with constant coefficients, signature (0,0,67,0,0,-67,0,0,1).
FORMULA
a(n) = 67*a(n-3)-67*a(n-6)+a(n-9). G.f.: -(x^2-x+1)*(x^6-8*x^5-8*x^4-2*x^3+72*x^2+8*x+1) / ((x-1)*(x^2+x+1)*(x^6-66*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*A086594(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(17) - 4. 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 = 8. See the Bala link for details.
The theory also provides the simple continued fraction expansion of the numbers F({sqrt(17) - 4}^(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(17) - 4}^(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*A086594(n)) or, more explicitly,
L = 1/8 + 1/(2*66) + 1/(3*536) + 1/(4*4354) + 1/(5*35368) +...
so that L = 0.1332613701545977545822925541573311424901819508933...
then exp(L) = 1.1425485874089841897117810754210805471767735522069...
equals the continued fraction given by this sequence:
exp(L) = [1;7,65,1,535,4353,1,35367,287297,1,2333751,...]; i.e.,
exp(L) = 1 + 1/(7 + 1/(65 + 1/(1 + 1/(535 + 1/(4353 + 1/(1 +...)))))).
Compare these partial quotients to A086594(n), n=1,2,3,...:
[8,66,536,4354,35368,287298,2333752,18957314,153992264,...].
PROG
(PARI) {a(n)=local(L=sum(m=1, 2*n+1000, 1./(m*round((4+sqrt(17))^m+(4-sqrt(17))^m)))); contfrac(exp(L))[n]}
CROSSREFS
KEYWORD
cofr,nonn,easy
AUTHOR
Paul D. Hanna, Mar 21 2010
STATUS
approved