n - digit numbers with n factors, each with a different number of digits

[David Wilson]

A better bound is

a(18m) <= 11 * 101^2 * 1000001^2 * (10^18+9)^(m-1)

The right hand side is an 18m-digit number for 1 <= m <= 243040916832487184.

On the other hand, under generous assumptions about the size of prime gaps, we have

a(2^m) <=   PROD     nextPrime(10^(2^k)).
                 0 <= k < m

where the right side has 2^m digits, which would provide an infinitude of numbers with precisely one divisor of every possible length.
  ----- Original Message ----- 
  From: Farideh Firoozbakht 
  To: N. J. A. Sloane ; Joshua Zucker ; Jonathan Post ; David Wilson ; franktaw at netscape.net ; Joerg Arndt ; seqfan at ext.jussieu.fr 
  Sent: Thursday, December 14, 2006 10:58 AM
  Subject: Re: Re : n - digit numbers with n factors, each with a different number of digits


  Dear Neil and seqfans,

  Instead " 4 " in the following statement, 

                    " In fact if  4 < m <= 3000 then  a(18m) exists and 
                      a(18m) < 11^2*1009^5*(10^18+9)^(m-1)+1"  

  we can write " -1 ".

  I would like to know, what is the largest number k such that  we can write instead " 3000 " 
  in the statement.

  Best wishes,
  Farideh.
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