problem 5.3a

[zenk...@gmail.com]

so the problem is just blatantly wrong, right?

it should read a_n = 2^2^n^2+1.

[Kenneth A. Ribet]

> so the problem is just blatantly wrong, right?
>
> it should read a_n = 2^2^n^2+1.

Seems OK to me. The first element of the sequence is 3. The next is
3+2 = 5. The next is 3.5 + 2 = 17. These are the first few Fermat
numbers. The task is to show that you keep getting the Fermat
numbers 2^{2^n}+1, and I think that you do.

It's always striking to me that students are totally stymied when
there's a misprint in a book. This is natural, but I never get used
to it. Students think that something is "blatantly wrong" when I see
a small error to be corrected in the next edition. When I last
taught 115, the students were furious with the book because it was
full of misprints. End of that book! As I started to read Erdos, I
found a bunch of misprints in the beginning and feared that it would
be deja vu all over again. In fact, the book has been pretty solid.

The most recent misprint that I saw in our book is on page 181. The
authors write 0 < a < b toward the bottom of the page, and I couldn't
figure out why r_0 was b and r_{-1} was a. In fact, it's clear from
page 183 that they intended b to be less than a.

Ken R

[zenkalia]

well, if you take it just one step further, then you won't be getting
the next fermat number but you will get a fermat number;

a_0 = 3 = 2^2^0+1
a_1 = 5 = 2^2^1+1
a_2 = 17 = 2^2^2+1
a_3 = 257 = 2^2^4+1
a_4 = 65537 = 2^2^8+1

[zenk...@gmail.com]

oops i'm an idiot... 0 1 2 4 8 is... 2^n, ish... it breaks at zero...
i don't know what i was thinking with n^2

so it should be...

a_0 = 3
a_n = 2^2^2^(n-1)+1 for n>0

[John Brooks-Jung]

Try plugging it back into the recurrence relation, or better yet, find
a recurence relation that only uses the previous term to define the
next. Hope that helps

John

[zenk...@gmail.com]

disregard this... all of it... i don't know how to use my ti-83