5-smooth Totients

Problem Statement

$5$-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.
$S(100)=3728$.

Find $S(2^N)$.

Submit Answers

You need to submit in the format: "N:problem(N)", possibly with multiple values at once, separated by commas, with $N$ between $1$ and $100$.

Top Users

🥇 icy
79.97 (86)
🥈 shs10978
79.97 (80)
🥉 byhill
59.97 (60)
4 pacome
41.97 (42)
5 mmtg
17.97 (18)
6 jonnytang
14.97 (15)
7 CandynightJ
0.03 (1)

Data

Stats

Your submissions will appear here

Recent Submissions

1
shs10978
$g(80)$, $37$ digits 1 week, 4 days ago
2
shs10978
$g(79)$, $36$ digits 1 week, 4 days ago
3
shs10978
$g(78)$, $36$ digits 1 week, 4 days ago
4
shs10978
$g(77)$, $35$ digits 1 week, 4 days ago
5
shs10978
$g(76)$, $35$ digits 1 week, 4 days ago