[seqfan] Re: 3, 21, 25
M. F. Hasler
oeis at hasler.fr
Thu Aug 20 14:28:56 CEST 2020
On Mon, Aug 17, 2020 at 1:29 PM Tomasz Ordowski <tomaszordowski at gmail.com>
wrote:
> Odd numbers n = k1 2^m1 + 1 = k2 2^m2 - 1 with k1 and k2 odd
> such that k1 + 2^m1 = |k2 - 2^m2|. This is a strong condition.
>
Yes: you can easily see that the smaller of the exponents must be 1 and
then the corresponding k' is given in terms of the other k as
k' = k*2^(m-1) +- 1
and finally you get something like k = (2^m + 1)/(2^(m-1) - 1)
(similar in the "-" case),
and the only possible integers are 9/3 and 5/1 for m = 2 or 3,
so there are no other solutions than the three given ones.
Amiram Eldar only found {3, 21, 25} and no more n < 10^8.
> Is this set complete or is it worth doing further research?
>
Some details can be filled in above but there are no other solutions.
It's the same for the other variant you give at the end of your message.
- Maximilian
3 = 1*2+1 = 1*2^2-1 and 1+2 = |1-2^2| = 3,
> 21 = 5*2^2+1 = 11*2-1 and 5+2^2 = |11-2| = 9,
> 25 = 3*2^3+1 = 13*2-1 and 3+2^3 = |13-2| = 11.
>
> Best regards,
>
> Thomas Ordowski
> ___________________________
> By the way, let's define similarly:
> Odd numbers n = k1 2^m1 + 1 = k2 2^m2 - 1 with k1 and k2 odd
> such that k1 + 2^m1 = k2 + 2^m2. A slightly different condition.
> Such numbers n < 10^8 (less interesting to me) are {5, 11},
> which was also checked by Ami. Thanks!
> 5 = 1*2^2+1 = 3*2-1 and 1+2^2 = 3+2 = 5,
> 11 = 5*2+1 = 3*2^2-1 and 5+2 = 3+2^2 = 7.
>
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