Variation On Harmonic Series Approximates Log

[franktaw at netscape.net]

The sequence, after then initial 0, is always 1 or 2.  log(n+1) - 
log(n) =
Integral_n^{n+1} 1/t dt; compare this to 1/n.

Franklin T. Adams-Watters

-----Original Message-----
From: Leroy Quet <q1qq2qqq3qqqq at yahoo.com>

Could someone please calculate more terms of this sequence? I wonder if 
it is
already in the EIS. (I may have made a mistake with my terms.)

Each a(n) is the (+ or -) integer, given the earlier terms of the 
sequence, such
that sum{k=1 to n} a(k)/k is the best approximation to the natural log 
of n.

In other words, a(n) = the nearest integer to
n*(ln(n) - sum{k=1 to n-1} a(k)/k).

I get that the sequence begins:

0,1,2,1,1,1,1,1,1,1,1,2,

When, if ever, does the sequence become negative?

Thanks,
Leroy Quet