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<DIV>
<DIV>On 1 December, Leroy Quet wrote:</DIV>
<DIV> </DIV>
<DIV>I figured (and Martin Cohen confirmed {referencing Hardy & Wright
theorem <BR>320}) that d(m), the number of positive divisors of m, was such
that<BR><BR>limit{n-> infinity} (1/n) (sum{m=1 to n} d(m)) -
ln(n)<BR><BR>= 2*c - 1, where c is Euler's constant (.5772...).<BR></DIV>
<DIV> </DIV>
<DIV>******* </DIV>
<DIV>Does anyone know, and if so could someone please tell me, whether the
"typical" nth term of sequence A018804, the sum of gcd (k,n) for 1
<= k <= n </DIV>
<DIV>(a (1) through a (10) are 1, 3, 5, 8, 9, 15, 13, 20, 21, 27 . . ;
see <A
href="http://www.research.att.com/projects/OEIS?Anum=A018804">http://www.research.att.com/projects/OEIS?Anum=A018804</A>),
</DIV>
<DIV> </DIV>
<DIV>is about the same size as the nth term of sequence A006218, sum_(k=1 . . .
n) d (k ) (corresponding terms are 1, 3, 5, 8, 10, 14, 16, 20, 23, 27
. . ; see <A
href="http://www.research.att.com/projects/OEIS?Anum=A006218">http://www.research.att.com/projects/OEIS?Anum=A006218</A>),
</DIV>
<DIV> </DIV>
<DIV>or "about" n* (ln (n ) + 2*c - 1?</DIV>
<DIV> </DIV>
<DIV>Put more technically, does </DIV>
<DIV>limit {n -> infinity} sum_(k=1 . . . n) A018804(k )/ sum_(k=1
. . n) A006218(k )</DIV>
<DIV>=1 (or possibly slightly more or less than 1)?</DIV>
<DIV> </DIV>
<DIV>I can see why this might be true (since A006218 is also sum_(k=1 . . n)
floor (n/k) ), but proving it is beyond me. Thanks in advance
to anyone who might reply to this query. </DIV>
<DIV> </DIV>
<DIV>Regards,</DIV>
<DIV>Matthew Vandermast</DIV></DIV></BODY></HTML>