adieu - My apologies

[Gerald McGarvey]

On Mon, 15 Jan 2007 02:59:49 -0000, Max A. <maxale at gmail.com> wrote:

> On 1/14/07, Nick Hobson <nickh at qbyte.org> wrote:
>> Hi Seqfans,
>>
>> Is this sequence worth adding?
>>
>> 6, 9, 12, 20, 28, 30, 34, 42, 56, 58, 65, 72, 75, 90, 110, 126, 132,  
>> 156,
>> 182, ... .
>>
>> The sequence comprises all positive integers n such that there exist
>> positive divisor d, quotient q, and remainder r in geometric  
>> progression.
>> The order of terms in the geometric progression is left unspecified, but
>> clearly must be one of q < r < d, r < q < d, or r < d < q.  For example,
>> 58 is in the sequence because 58 = 9*6 + 4.
>
> Without loss of generality we can assume that q<d (otherwise we can
> exchange them), implying that there is only two cases to consider: q <
> r < d and r < q < d.
>
> Let m be the multiplier of the corresponding geometric progression. We  
> have:
>
> 1) if q < r < d then r = q*m and d = q*m^2, implying that
> n = q*d + r = q*m*(q*m+1)
> is the product of two consecutive integers.
>
> 2) if r < q < d then q = r*m and d = r*m^2, implying that
> n = q*d + r = r*(r*m^3 + 1)
>
>> Clearly the sequence is infinite.  There are 14, 65, 278 terms not
>> exceeding 100, 1000, 10000, respectively.  I don't know its asymptotic
>> density.
>
> The amount of such numbers below N grows proportionally to sqrt(N).
>
>> It's not difficult to show that the sequence misses the primes.
>> Interestingly, at least initially, it also hits very few perfect powers,
>> the only hits below 10^13 being 9, 10404, 16900, 97344, 576081, 6230016,
>> 7322436, 12006225, 36869184, 37344321, 70963776, 196112016, 256160025,
>> 1361388609, 1380568336, 8534988225, 9729849600, 12551169024,  
>> 13855173264,
>> 16394241600, 123383587600, 142965659664, 547674002500, 1812792960000,
>> 1882109610000, and 3602897496900, none of which are cubes or higher  
>> prime
>> powers.
>
> The first case cannot give perfect powers at all (as the product of
> two consecutive integers).
>

Thanks Max.

Yes, I got that.


> In the second case, the number n=r*(r*m^3 + 1) gives perfect powers
> quite rarely. In particular, since gcd(r,r*m^3 + 1)=1, both r and
> r*m^3 + 1 must be a perfect power (of the same degree). This can never
> give a cube, since if r is a cube, say, r=s^3, then
> r*m^3 + 1 = (s*m)^3 + 1 cannot be a cube.
>

This is true if r*m^3 is an integer.

For the case r < q < d, I got the following:

Let the common ratio be k = q/r > 1, and let k = a/b, a fraction in its  
lowest terms.

W.l.o.g., let r = s^2*t, where t is squarefree.
Then, if d = s^2*t*a^2/b^2 is integral, b^2 must divide s^2, and so b  
divides s.
Let s = bc.  We then have:

r = b^2*u,
q = a*b*u,
d = a^2*u,

where u = c^2*t.

Hence n = qd + r = a^3*b*u^2 + b^2*u = bu(u*a^3 + b)

Clearly n is composite if bu > 1.
Otherwise, n = a^3 + 1 is composite since a = k > 1.

So u is an arbitrary integer, and thus all solutions (for this case) can  
be characterised by positive integer a, b, and u, with a > b, a and b  
coprime, and arbitrary u.

I haven't yet worked out a simple form for gcd(bu, u*a^3 + b).  I can see  
that any divisor d of bu and u*a^3 + b divides u(u*a^3 + b) - bu =  
u^2*a^3, which helps a little as gcd(a, b) = 1.

Nick


> Max
>