Geometric/arithmetic progression of divisor, quotient, remainder

[Max A.]

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).

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.

Max