# [seqfan] Re: Natural numbers of the form (2^m + 3)/(2^k - 9)

Max Alekseyev maxale at gmail.com
Mon May 31 23:12:29 CEST 2010

```On Fri, May 28, 2010 at 12:29 PM, Hagen von Eitzen <math at von-eitzen.de> wrote:
> Dear seqfans,
>
> I'm contemplating natural numbers of the form (2^m + 3)/(2^k - 9) and
> would like to hear some opinions:
> Does their list start
> 1, 5, 37, 149, 293, 2341, 16981, 18725, 149797, 1198373, 9586981,
> 76695845, 156180629, ...
> or am I missing a few values?
> Most of the numbers above occur in A046636, which corresponds to k=4, m
> = 2 mod 3.
> Other subsequences come from k=6, m = 13 mod 20 and k=8, m = 22 mod 36.
> There is no solution with k=10, the solutions with k=12, m = 336 mod 660

Here is a couple of further values:
k=24 with Mod(1397430, 2794836)
k=40 with Mod(26738606, 53477172)

> but that does not rule out that some big-k
> solution might be rather small -  or does it?

Notice that (2^m + 3)/(2^k - 9) ~= 2^(m-k) and it is easy to address
cases where m-k is relatively small.
For example, if m-k<k then:

0 == 2^m + 3 == 9*2^(m-k) + 3 (mod 2^k - 9)

For some positive integer q, we have
9*2^(m-k) + 3 = q*(2^k - 9)
and thus
(q*2^(2*k-m) - 9) * 2^(m-k) = 9*q + 3.

Let us consider possible values of 2*k-m:

If 2*k-m = 1, then
(2*q - 9) * 2^(m-k) = 9*q + 3
implying that (2*q - 9) | (9*q + 3) and hence (2*q - 9) | 2*(9*q + 3)
- 9*(2*q - 9) = 87. Therefore, (2*q - 9) = 1, 3, 29, or 87, thus q =
5, 6, 19, or 48 but none of these values gives (9*q + 3) / (2*q - 9)
equal a power of 2.

If 2*k-m = 2, then
(4*q - 9) * 2^(m-k) = 9*q + 3
implying that q = 3 or 10 with no values giving (9*q + 3) / (4*q - 9)
equal a power of 2.

If 2*k-m = 3, then
(8*q - 9) * 2^(m-k) = 9*q + 3
implying that q = 3 with (9*q + 3) / (8*q - 9) = 2^1. Hence, m - k =
1, resulting in k=4 and m=5.

If 2*k-m = 4, then
(16*q - 9) * 2^(m-k) = 9*q + 3
implying (16*q - 9) <=  9*q + 3, implying that q <= 12/7, i.e., q = 1
which however does not yield integer (9*q + 3)/(16*q - 9).

Larger values of 2*k-m do not yield solution by similar reasons.

Max

```