[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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