Modifications for A069623, added formula

[David Wilson]

Yes, the last 10 should be an 11.  I computed these values using the
formula I derived, and experienced roundoff errors.  Also, I got the
lower bound in my formula wrong, it should have been 1, not 2.  Here
are the corrected lines:

%I A069623
%S A069623
1,1,1,2,2,2,2,3,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,6,6,7,7,7,7,7,8,8,8,8,9,
%T A069623
9,9,9,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,11,
%U A069623
11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,12,12,12,12,12,12,12,12
%F A069623 a(n) = n - SUM(k = 1 to [log2(n)], mu(k) [n^(1/e)-1]), where mu =
A008683

----- Original Message -----
From: "Robert G. Wilson v" <rgwv at kspaint.com>
To: "David Wilson" <davidwwilson at attbi.com>
Sent: Friday, October 11, 2002 5:31 PM
Subject: Re: Modifications for A069623, added formula


> Dear David,
>
>         I do not quite match your terms. I get:
>
> 1,1,1,2,2,2,2,3,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,6,6,7,7,7,7,7,8,8,8,8,9,
>
> 9,9,9,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,11,
>
> 11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,12,12,12,12,12,12,12,12,
>
> [etc]