[/quote]
,
![\[ \left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor \]](http://latex.artofproblemsolving.com/5/6/0/560a9568b2c09512cac1d317a011931a439e6a02.png)
such that for any 
,
,
,
,
,
are odd primes and![\[2^ap^b=(p+2)^c-1.\]](http://latex.artofproblemsolving.com/5/8/5/58518a8ebed14836b72af743ff100eae2efba5e2.png)
![\[
2^a + 3^b + 1 = 6^c
\]](http://latex.artofproblemsolving.com/1/9/e/19eb5efa2ccdf54cd277339f8d3b99f75168779d.png)
be a triangle with 
,
be the foot of the altitude from 
.
be a point in the interior of the segment 
.
be the point on the segment 
such that 
.
be the point on the segment 
such that 
.
be the point of intersection of 
and 
.
.
.
.
is selected in the line segment 
.
and 
intersect 
,
and 
intersect (a second time) the circumcircles to triangles 
,
passes through a fixed point 
distinct from 
.
,
,
,
for all 
,
be two positive sequences defined by 
and![\[ \begin{cases} {{x}_{n+1}}{{y}_{n+1}}-{{x}_{n}}=0 \\ x_{n+1}^{2}+{{y}_{n}}=2 \end{cases} \]](http://latex.artofproblemsolving.com/2/0/1/2012dc6ff3a41478547cea4aa9b6bccbe17a3623.png)
for all 
.
be an infinite sequence of positive integers. Prove that there exists a unique integer 
such that![\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]](http://latex.artofproblemsolving.com/9/4/f/94fc5a51b7e68588123c5b527fe75183bc4c4937.png)
with 
,
and 
for all positive integers 
where 
Y by egxa, Amir Hossein, sami1618
For an integer 
, 
denotes the sum of postitive divisors of . A sequence of positive integers 
with 
is defined as follows: For each 
, 
is the smallest integer greater than 
that satisfies
Determine the number of divisors of 
amongst the sequence.
, 
denotes the sum of postitive divisors of . A sequence of positive integers 
with 
is defined as follows: For each 
, 
is the smallest integer greater than 
that satisfies
Determine the number of divisors of 
amongst the sequence.This post has been edited 3 times. Last edited by Cookierookie, Mar 18, 2024, 8:43 PM
where 
.
numbers in the sequence 
dividing the number 
.
and 
.
where 
,
,
for 
occur among 
.
where 
are distinct primes and 
.
with 
for all 
,
and claim holds.
where 
and 
.
,
,
,
.
satisfies, so 
and claim holds.
.
,
works, contradiction. Suppose 
for all 
.
and 
,
,
on both sides, we obtain 
for some 
with 
minimal, so all 
,
also satisfies the condition, contradiction.
.
![\[p^{2^{i}}\]](http://latex.artofproblemsolving.com/f/e/4/fe4c5d1854050e910fd8d8384b055f187c298ea6.png)
for a prime 
.
is precisely the nice numbers in increasing order.
to show that the numbers 
are precisely the first 
,
,
,
for some prime 
.
with 
and 
prime with 
,
is multiplied in the expression 
.
to be considered in the product.
and 
:![\[M \overset{\text{ def }}{=} a_0a_1\dots a_{k-1} = \prod \limits_{i = 1}^{v} p_i^{2^{m_i}-1}\]](http://latex.artofproblemsolving.com/1/5/2/1521d9375cf344d000ee788f025a12409ec76370.png)
Suppose that 
.
does not fit as a valid 
in the 
for some 
(from an earlier discussion). In particular, we have ![\[\frac{\sigma(MN)}{\sigma(M)} = \frac{\sigma(p_i^{2^{u+1}-1})}{\sigma(p_i^{2^{u}-1})} = p_i^{2^u}+1\]](http://latex.artofproblemsolving.com/b/6/8/b6867036963ccf0975bd8cdff565946aab624f6b.png)
In particular, we have 
,
for any 
and 
.
.
.![\[a_k = q_1^{e_1}q_2^{e_2}\dots q_x^{e_x}\]](http://latex.artofproblemsolving.com/9/2/4/924a2064545616750cbd341b63a03776fb718093.png)
for primes 
and nonnegative integers 
.
distinct from all the 
.
.![\[M' = \prod \limits_{i = 1}^{x} q_i^{2^{m_i}-1}\]](http://latex.artofproblemsolving.com/d/d/d/ddd9411e0cae9a3891ebcec5f8f77a167c182655.png)
From a similar argument as earlier, we can see that we must have 
for all 
is the same as requiring ![\[\prod \limits_{i = 1}^{x} (q_i^{2^{m_i}}-1) \mid \prod \limits_{i = 1}^{x} (q_i^{2^{m_i}+e_i}-1) \qquad (1)\]](http://latex.artofproblemsolving.com/d/a/2/da2dc914b238af1be56aee4d2f9862c95ae7c431.png)
There are two cases to consider here:
of both sides above. For a fixed ![\[\nu_2\left(q_i^{2_{m_i}}-1\right) = \nu_2\left(q_i^2-1\right) + m_i - 1\]](http://latex.artofproblemsolving.com/9/9/7/997f5ab6cdc696cca4e3efefd65591ecf5da4ad0.png)
On the other hand, 
is equal to either ![\[\nu_2(q-1) \text{ or } \nu_2(q^2-1) + \nu_2(2^{m_i}+e_i) - 1\]](http://latex.artofproblemsolving.com/1/d/b/1dbc94ba003456382a1dbc939340a1a24cc72aac.png)
In either case, we see that one has 
.
.
has a greater 
.
.
and we get 
by size, impossible yet again.
![\[2024^{2024} = 2^{6072} \cdot 11^{2024} \cdot 23^{2024}\]](http://latex.artofproblemsolving.com/f/e/e/fee9a7d460527d6e7c5a52abbbdc6c427fda12da.png)
and hence the required count is ![\[1 + \left(\lfloor \log_2(6072) \rfloor + 1\right) + 2 \left(\lfloor \log_2(2024)\rfloor + 1\right) = 1+13+22 = 36\]](http://latex.artofproblemsolving.com/1/4/d/14d11fa68c9b46e177b720a40d6a09264573270e.png)
which is the desired answer. 
and the closest next prime
.
( let that be 
.
then 
thus 
is minimal which gives the sequence of 
thus the motivation comes to prove that all terms of the sequence are of the form 
