[seqfan] Re: Pairs Occurring Only Once Among # Of Divisors

[franktaw at netscape.net]

One can speed this up considerably by looking at the parity of the 
number of divisors.  If n is one less than a square, it is only 
necessary to look at numbers that are one less than a square; likewise 
if n is a square, one need only look at squares.  Probably all members 
of the sequence except 2 fall into one of these two categories; and the 
search when n is not in one of these categories will usually be fairly 
quick.

Franklin T. Adams-Watters


-----Original Message-----
From: Maximilian Hasler <maximilian.hasler at gmail.com>

For what it's worth (since you seem desperate... ;-)
Using a (very stupid) brute force approach, I think the sequence starts:
1,2,3,4,8,15,16,24,35,48,63,64,80,99,...

The higher terms might be incorrect (my script simply checked all
numbers up to LIM=999999; some negative results:
? isLQ(324,1)
324 has 15 divisors; n+1 has 6 divisors.
110224 too
? isLQ(168,1)
168 has 16 divisors; n+1 has 3 divisors.
120 too)

Maximilian

(PARI)
isLQ(n,s,LIM=999999)={my(d=numdiv(n),d1=numdiv(n+1));s&print(n" has
"d" divisors; n+1 has "d1" divisors.");for(i=1,LIM,numdiv(i)==d |
next; numdiv(i+1)==d1 | next; i==n&next; s&print(i " too");return);1}

for(n=1,299,isLQ(n)&print1(n","))
1,2,3,4,8,15,16,24,35,48,63,64,80,99,288,


On Thu, Jun 11, 2009 at 9:23 AM, Leroy Quet<q1qq2qqq3qqqq at yahoo.com> 
wrote:
>
> [Sorry if this appears twice.]
>
> I wonder if someone can run a program to determine possible values 
for my
sequence.
>
> For instance, the program would make sure a pair occurs only once 
among the
number-of-divisors of all positive integers < some big number.
>
> Then the values can be independently proved to be unique, which 
shouldn't be
hard for most of them, hopefully.
>
> Thanks,
> Leroy Quet
>
>
> I wrote:
> --------------------
>
> I just submitted this sequence:
>
> %I A161460
> %S A161460 1,2,3,4,8,15,16,24
> %N A161460 Those positive integers n such that there is no m 
different than n
where both d(n) = d(m) and d(n+1) = d(m+1), where d(n) is the number of 
positive
divisors of n.
> %e A161460 d(15) = 4, and d(15+1) = 5. Any positive integers m+1 with 
exactly
5 divisors must by of the form p^4, where p is prime. So m = p^4 -1 =
(p^2+1)*(p+1)*(p-1). Now, in order for d(m) to have exactly 4 divisors, 
m must
either be of the form q^3 or q*r, where q and r are distinct primes. 
But no p is
such that (p^2+1)*(p+1)*(p-1) = q^3. And the only p where 
(p^2+1)*(p+1)*(p-1) =
q*r is when p = 2 ( and so q=5, r =3). So, there is only one m where 
both d(m) =
4 and d(m+1) = 5, which is when m=15. Therefore, 15 is in this sequence.
> %K A161460 more,nonn
> %O A161460 1,2
>
>
>
>
>
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>


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