be a sequence of integers defined by 
,
,
for all 
.
is divided by 
.
are either prime numbers or the negatives of prime numbers?
,
,
for 
.![\[A_n=\sum_{i=0}^n s_{i}\]](http://latex.artofproblemsolving.com/3/d/4/3d416f32205acdec15a910d056d4a6e9ad4f85d8.png)
Find 
.
,
,
,
,
are positive integers.
,![\[ \frac{x_{n+1}}{x_n} = \frac{(x_{n+1})^2 + 27}{x_n^2 + 27} \]](http://latex.artofproblemsolving.com/3/0/4/304d56991f9955dc93c16413478b604dd7042ebf.png)
for all positive integers 
.
.
be the maximum value of![\[ \sum_{n=1}^{2025} x_n. \]](http://latex.artofproblemsolving.com/c/4/d/c4da21c0341fdffb56b9a110dd6df874b6e0e8b2.png)
What is 
?
for which the series![\[ 1 + \frac {z}{2!} + \frac {z(z + 1)}{3!} + \frac {z(z + 1)(z + 2)}{4!} + \cdots + \frac {z(z + 1)\cdots(z + n)}{(n + 2)!} + \cdots\]](http://latex.artofproblemsolving.com/3/9/9/39916480f6970ba884415a2a06c1ffe4b3dd4423.png)
and 
without using complex analysis. Can we do it without Feynman's trick?
,
.
satisfies the quadratic equation 
and deduce that the value of 
.
.
,
is 
.
is analytic on a closed curve 
(i.e., ![\[\int_\gamma \overline{f(z)} f'(z)\,\mathrm dz\]](http://latex.artofproblemsolving.com/c/3/7/c3784659f77e99585032e21fef95bd31a11c23e4.png)
is purely imaginary. (The continuity of 
is taken for granted.)
by residue theorem? Should by something like![\[[z^n]\tan z\propto\oint _{|z|=N}\frac{\tan z}{z^{n+1}}\mathrm dz\]](http://latex.artofproblemsolving.com/8/5/c/85c34140e3cc61d93bd3188c22ae06649ca98c5a.png)
where 
tends to infty, but I'm not sure about details.
be the open unit disk in the complex plane and let 
be a holomorphic function such that 
.
.
.
has no roots in the disc of unity.
,
and such that the limit 
exists. Prove that 
attains its maximum on the circle 
.
.
on 
,
Can I change 1 by a constant 
?
which contains 
.
if 
if 
Y by
Suppose a function 
satisfies 
for all real 
. Then, we have 
We also have ![\[f(\tfrac{x+y}{2}) + f(\tfrac{x-y}{2}) \ge 2f(x).\]](//latex.artofproblemsolving.com/5/3/d/53d5e6a96b03a5c4a05a9da1945cbaa81829271e.png)
Adding the first two inequalities and using the third, we have ![\[2f(x) + f(y)+f(-y) \ge 2(f(\tfrac{x+y}{2}) + f(\tfrac{x-y}{2})) \ge 4f(x),\]](//latex.artofproblemsolving.com/a/9/c/a9c9255deeb88e31603a55d6cf32fbf864ca0758.png)
which implies 
. But this means is bounded above which is absurd -- take the identity function. I feel I am making some blithely stupid mistake, but I can't see what. Any help is appreciated 
satisfies 
for all real 
. Then, we have 
We also have ![\[f(\tfrac{x+y}{2}) + f(\tfrac{x-y}{2}) \ge 2f(x).\]](http://latex.artofproblemsolving.com/5/3/d/53d5e6a96b03a5c4a05a9da1945cbaa81829271e.png)
Adding the first two inequalities and using the third, we have ![\[2f(x) + f(y)+f(-y) \ge 2(f(\tfrac{x+y}{2}) + f(\tfrac{x-y}{2})) \ge 4f(x),\]](http://latex.artofproblemsolving.com/a/9/c/a9c9255deeb88e31603a55d6cf32fbf864ca0758.png)
which implies 
. But this means is bounded above which is absurd -- take the identity function. I feel I am making some blithely stupid mistake, but I can't see what. Any help is appreciated This post has been edited 3 times. Last edited by Sedro, Jun 17, 2024, 3:57 PM
on the RHS.
Thanks.
