Sums of Totients of Powers

Let $\varphi(n)$ be Euler’s totient function.
Let $f(n)=(\sum_{i=1}^{n}\varphi(n^i)) \bmod (n+1)$.
Let $g(n)=\sum_{i=1}^{n} f(i)$.
$g(100)=2007$.

Find $g(5 \times 10^8)$.

To solve this problem, we can use dynamic programming to calculate the number of ways to tile the loop. We will create a table to store the number of ways for each length and color combination.

First, let’s define a function `F(k, n)` that takes the number of colors `k` and the length of the loop `n` as input and returns the number of ways to tile the loop.

“`python
def F(k, n):
# Initialize the table with zeros
dp = [[0 for _ in range(k)] for _ in range(n+1)]

# Base case: there is only one way to tile a 2×1 loop
dp[1][0] = 1

# Iterate over each length from 2 to n
for length in range(2, n+1):
# Iterate over each color
for color in range(k):
# Consider placing a 1×1 tile at the beginning of the loop
if length >= 2:
dp[length][color] += dp[length-2][(color+1)%k]

# Consider placing a 1×2 tile at the beginning of the loop
if length >= 3:
dp[length][color] += dp[length-3][(color+1)%k]

# Consider placing a 1×3 tile at the beginning of the loop
if length >= 4:
dp[length][color] += dp[length-4][(color+1)%k]

# Consider placing a 1×2 tile at the end of the loop
if length >= 3:
dp[length][color] += dp[length-3][(color+1)%k]

# Consider placing a 1×3 tile at the end of the loop
if length >= 4:
dp[length][color] += dp[length-4][(color+1)%k]

# Take modulo to avoid large numbers
dp[length][color] %= 1000004321

# Return the number of ways to tile the loop of length n
return sum(dp[n]) % 1000004321
“`

To find `F_{10}(10,004,003,002,001) \mod 1,000,004,321`, we can simply call this function with `k=10` and `n=10,004,003,002,001`:

“`python
result = F(10, 10004003002001)
print(result)
“`

The resulting number will be the answer to the problem, modulo `1,000,004,321`.

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