[seqfan] Re: A sequence

Antreas Hatzipolakis anopolis72 at gmail.com
Wed Nov 26 20:28:16 CET 2014


On Wed, Nov 26, 2014 at 8:13 PM, Charles Greathouse <
charles.greathouse at case.edu> wrote:

> I think that would be nice. Antreas, you might consider submitting the
> original as well with a pointer to the simplified/generalized sequence, in
> case someone looks for that particular problem. (Or not, as you prefer.)
>



Hmmmm.... I am not familiar with submitting !!!

On the other hand the description - definition requires very good English
.... :-)

Anyway, the sequence is this (if someone of the editors would like to
submit it):

x_1 = 11, x_2 = 32, x_3 = 54,.....

x_n = 2^(n-1)  + 20n - 10n

For n>10  ==> x_n > x_(n-1) + x_(n-2)

Reference: Euclid, Athens, November 1971, pp. 98-100

PS: Just for the history of the problem.
The only one solver was Ilias Kastanas (then a high school student who
later became a good
mathematician in the USA).

He proved by induction that for every k >= 1 ==> 2^3 (2^k  - 1) >= k
(*)

Now,  x_n > x_(n-1) + x_(n-2)  is equivalent to  2^n / 8 + 70 > 20n

We have (*) ==> 2^4(2^k -1) >= 2^k ==> 2^(k+4) > 16 + 2k  > 2^(k+4) + 15 +
2k ==>  2^(k+4) + 7 > 22 + 2k

For k = n - 11 we get:   2^(n-1) + 7 > 22 + 2(n-11) ==>

2^(n-1) + 7 > 2n  ==> 2^n / 2^7  > 2^n - 7   ==> 2^n / 2^3  > 2^4 (2n - 7)
==> 2^n / 8 > 16 (2n-7)

==>  2^n / 8 > 10(2n - 7) ==> 2^n / 8 > 20n - 70 ==> 2^n / 8 + 70 > 20n

Hope there are no typos :-)

Antreas


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