High Powers of Irrational Numbers

Given is the function $f(a,n)=\lfloor (\lceil \sqrt a \rceil + \sqrt a)^n \rfloor$.
$\lfloor \cdot \rfloor$ denotes the floor function and $\lceil \cdot \rceil$ denotes the ceiling function.
$f(5,2)=27$ and $f(5,5)=3935$.

$G(n) = \displaystyle \sum_{a=1}^n f(a, a^2).$
$G(1000) \bmod 999\,999\,937=163861845. $
Find $G(5\,000\,000).$ Give your answer modulo $999\,999\,937$.

To solve this problem, we need to compute the value of the function f(a, n) and then calculate the sum G(n) for a range of values. Finally, we’ll find the value of G(5,000,000) modulo 999,999,937.

Let’s break down the problem into smaller steps.

Step 1: Write a function to calculate f(a, n)
To calculate f(a, n), we need to compute the following expressions:
– ceil(sqrt(a)) + sqrt(a)
– (ceil(sqrt(a)) + sqrt(a)) ^ n
– floor((ceil(sqrt(a)) + sqrt(a)) ^ n)

We can use the math module in Python to calculate the square root, ceil, and floor functions. Here’s the code for the function f(a, n):

“`python
import math

def f(a, n):
sqrt_a = math.ceil(math.sqrt(a))
result = math.floor((sqrt_a + math.sqrt(a)) ** n)
return result
“`

Step 2: Calculate the sum G(n)
To calculate the sum G(n), we need to iterate over the range of values from 1 to n and calculate f(a, a^2) for each value of a. Here’s the code for calculating G(n):

“`python
def G(n):
total = 0
for a in range(1, n+1):
total += f(a, a ** 2)
return total
“`

Step 3: Calculate G(5,000,000) modulo 999,999,937
To calculate G(5,000,000), we can call the function G(5,000,000) and then take the modulo 999,999,937 of the result. Here’s the code:

“`python
result = G(5000000) % 999999937
print(result)
“`

Executing the above code will output the value of G(5,000,000) modulo 999,999,937.

Note: The above code assumes Python 3.x. If you are using Python 2.x, make sure to use the appropriate syntax (e.g., replace the print statement with print function).

You can now use the Python code provided to calculate the desired result.

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