$5$-smooth Totients

$5$-smooth numbers are numbers whose largest prime factor doesn’t exceed $5$.
$5$-smooth numbers are also called Hamming numbers.
Let $S(L)$ be the sum of the numbers $n$ not exceeding $L$ such that Euler’s totient function $\phi(n)$ is a Hamming number.

Find $S(10^{12})$. Give your answer modulo $2^{32}$.

To solve this problem, we’ll start by implementing a few helper functions.

1. `is_prime(n)`: This function will check if a number `n` is prime.

def is_prime(n):
if n <= 1: return False if n == 2: return True if n % 2 == 0: return False i = 3 while i * i <= n: if n % i == 0: return False i += 2 return True ``` 2. `get_primes(n)`: This function will return a list of all prime numbers less than `n`. ```python def get_primes(n): primes = [] for i in range(2, n): if is_prime(i): primes.append(i) return primes ``` 3. `inversion_count(sequence)`: This function will calculate the inversion count of a given sequence. ```python def inversion_count(sequence): count = 0 for i in range(len(sequence)): for j in range(i+1, len(sequence)): if sequence[i] > sequence[j]:
count += 1
return count

4. `divided_sequences(primes)`: This function will generate all possible divided sequences from a list of prime numbers.

def divided_sequences(primes):
sequences = [list(str(primes[0]))]
for p in primes[1:]:
new_sequences = []
for seq in sequences:
for d in seq:
new_seq = seq[:]
sequences = new_sequences
return sequences

Finally, we can implement the main function `F(N)` to calculate the sum of inversion counts for all possible divided sequences from G(N).

def F(N):
primes = get_primes(N)
G = ”.join(str(p) for p in primes if p != 0)
sequences = divided_sequences(primes)
total_count = 0
for seq in sequences:
divided_seq = ”.join(str(d) for d in seq)
count = inversion_count(divided_seq)
total_count += count
return total_count % (10**9 + 7)

Now we can compute the answer for F(10^8) by calling `F(10**8)`:

answer = F(10**8)

More Answers:
Integral Median
Geoboard Shapes
Dissonant Numbers

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