Variation On Harmonic Series Approximates Log

franktaw at franktaw at
Mon Jul 14 06:30:01 CEST 2008

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>

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:


When, if ever, does the sequence become negative?

Leroy Quet

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