forwarded message from Farideh

Thu Dec 14 15:46:33 CET 2006

>From mymontain at yahoo.com  Thu Dec 14 08:47:53 2006
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>Date: Thu, 14 Dec 2006 05:46:48 -0800 (PST)
>From: Farideh Firoozbakht <mymontain at yahoo.com>
>Subject: Re : n - digit numbers with n factors, each with a different number of digits
>To: "N. J. A. Sloane" <njas at research.att.com>,
>        Joshua Zucker <joshua.zucker at gmail.com>,
>        Jonathan Post <jvospost3 at gmail.com>,
>        David Wilson <davidwwilson at comcast.net>, franktaw at netscape.net,
>        Joerg Arndt <arndt at jjj.de>, seqfan at ext.jussieu.fr
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>Dear Neil and seqfans,
>   
>  > There is no 29-digit number with exactly 29 divisors, so the sequence ends
>  >  here unless you want one of those evil "or 0 if no such number" sequences. 
>  
>> Comments and corrections welcomed!
>   
>  If p is a prime greater than 23 then a(p) doesn't exist.    (*)
>   
>  Proof of  (*): If p is prime then a(p) must be a p-digit number of the form q^(p-1) 
>  where q is prime. But if q <= 7 number of digits of q^(p-1) is less than p and if
>  q > 7 &  p > 23 number of digits of q^(p-1) is greater than p. Hence if p is a prime
>  greater than 23, a(p) doesn't exist.
>   
>  But we can prove that for many numbers n greater than 29, a(n) exist.
>   
>  For example  10^53999 < a(54000) <= 11^2*1009^5*(10^18+9)^2999.    
>  Proof : if n = 11^2*1009^5*(10^18+9)^2999  then n has exactly 54000 divisors d_k 
>  (k=1,2, ..., 54000) and each d_k has exactly k digits. Hence a(54000) exits and 
>  a(54000) is a 54000-digit number less than n+1.
>   
>  In fact if  4 < m <= 3000 then  a(18*m) exists and a(18m) < 11^2*1009^5*(10^18+9)^(m-1)+1.
>   
>  So I think a better definition of A125315 is:
>   
>          " Smallest n-digit number that has exactly n divisors, each with a different
>            number of digits and zero if there is no such number. "
>   
>  I guess that number of n's where a(n) > 0 is infinite.
>   
>  Best wishes,
>  Farideh.
>   
>  Ps. There is a problem, I can't send message to seqfans.
>   
>
>
> 
>---------------------------------
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><DIV>Dear Neil and seqfans,</DIV>  <DIV> </DIV>  <DIV>> There is no 29-digit number with exactly 29 divisors, so the sequence ends</DIV>  <DIV>>  here unless you want one of those evil "or 0 if no such number" sequences. </DIV>  <DIV><BR>> Comments and corrections welcomed!</DIV>  <DIV> </DIV>  <DIV>If p is a prime greater than 23 then a(p) doesn't exist.    (*)</DIV>  <DIV> </DIV>  <DIV><STRONG>Proof of  (*):</STRONG> If p is prime then a(p) must be a p-digit number of the form q^(p-1) </DIV>  <DIV>where q is prime. But if q <= 7 number of digits of q^(p-1) is less than p and if</DIV>  <DIV>q > 7 &  p > 23 number of digits of q^(p-1) is greater than p. Hence if p is a prime</DIV>  <DIV>greater than 23, a(p) doesn't exist.</DIV>  <DIV> </DIV>  <DIV>But we can prove that for many numbers n greater than 29, a(n) exist.</DIV>  <DIV> </DIV>  <DIV>For example  10^53999 <
> a(54000) <= 11^2*1009^5*(10^18+9)^2999.   <DIV> </DIV>  <DIV><STRONG>Proof :</STRONG> if n = 11^2*1009^5*(10^18+9)^2999  then n has exactly 54000 divisors d_k </DIV>  <DIV>(k=1,2, ..., 54000) and each d_k has exactly k digits. Hence a(54000) exits and </DIV>  <DIV>a(54000) is a 54000-digit number less than n+1.</DIV>  <DIV> </DIV>  <DIV>In fact if  4 < m <= 3000 then  a(18*m) exists and a(18m) < 11^2*1009^5*(10^18+9)^(m-1)+1.</DIV>  <DIV> </DIV>  <DIV>So I think a better definition of A125315 is:</DIV>  <DIV> </DIV>  <DIV>        " Smallest n-digit number that has exactly n divisors, each with a different</DIV>  <DIV>          number of digits and zero if there is no such number. "</DIV>  <DIV> </DIV>  <DIV>I guess that number of n's where a(n) > 0 is <STRONG>infinite</STRONG>.</DIV>  <DIV> </DIV>  <DIV>Best
> wishes,</DIV>  <DIV>Farideh.</DIV>  <DIV> </DIV>  <DIV>Ps. There is a problem, I can't send message to seqfans.</DIV>  <DIV> </DIV></DIV><p> 
>
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