OFFSET
1,3
COMMENTS
The sequence A217685 gives the sequence of values n such that 10^n gets increasingly closer to a Lucas number.
Given that for sufficiently large values of n, Fibonacci(n) ~ Lucas(n)/sqrt(5) ~ (((1+sqrt(5))/2)^n)/(sqrt(5)), the intermediate differences between the terms in this sequence also need to be a member of the sequence A217685.
PROG
(PARI) default(realprecision, 1000); a=vector(100, i, (contfracpnqn(contfrac(log((1+sqrt(5))/2)/log(10), 0, i))[2, 1]))
log_fibonacci(j)=(j*log((1+sqrt(5))/2)/log(10))-(log(sqrt(5))/log(10))
deviation(k)=abs(round(log_fibonacci(k))-log_fibonacci(k))
n=6; err=deviation(n); m=3; while(n<10^20, if(deviation(n+a[m])<err, n=n+a[m]; err=deviation(n); print(round(log_fibonacci(n))), m++))
CROSSREFS
KEYWORD
nonn,base
AUTHOR
V. Raman, Oct 11 2012
STATUS
approved