[seqfan] Re: Pairs Occurring Only Once Among # Of Divisors
Jack Brennen
jfb at brennen.net
Thu Jun 11 21:27:48 CEST 2009
I ran this for all numbers out to 10^8...
Here are the numbers <= 10^6 which didn't have any match within that
range:
[1, 2, 3, 4, 8, 15, 16, 24, 35, 48, 63, 64, 80, 99, 288, 528, 575, 624,
728, 960, 1023, 1024, 1088, 1295, 2303, 2400, 4095, 4096, 5328, 6399,
6723, 9408, 9999, 14640, 15624, 28223, 36863, 38415, 46655, 50175,
50624, 57121, 59048, 59049, 65535, 65536, 71824, 82944, 83520, 117648,
130320, 146688, 250000, 262143, 262144, 263168, 279840, 331775, 421200,
529983, 531440, 531441, 589824, 640000, 641600, 651249, 746495, 746496,
777924, 860624, 861183, 923520, 937024, 1000000]
Some of these aren't in the sequence, they just didn't hit a match
before 10^8. For instance, 1000000 is matched by 94^6, which didn't
take long to find.
Note that all of these are either a square or one less than a square
except for 2 and 860624. For 860624, it has 30 divisors and its
successor has 50 divisors. For a number to have 50 divisors, it
must be of one of these forms:
p^49
p^24*q
p^9*q^4
p^4*q^4*r
I'm currently searching for the smallest match for 860624, it's
probably out there somewhere... :)
franktaw at netscape.net wrote:
> 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
>>
>>
>>
>>
>>
>>
>>
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>>
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