JavaScript, 52 bytes

h=x=>x%7||h(x/7)
for(n=0;;)~h(++n)%3||console.log(n)

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This works because \$(\mathbb{Z}/7\mathbb{Z})^{\times}/\{1,-1\}\cong C_3\$, with \$\{1,2,3\}\$ mapping to the three distinct elements of \$C_3\$, so that for \$n\$ not divisible by 7, \$\{n,2n,3n\}\$ also map to the three distinct elements of \$C_3\$.

JavaScript, 52 51 bytes

for(n=0;h=x=>x%7||h(x/7);)~h(++n)%3||console.log(n)

Try it online!

-1 thanks to Mukundan314.

The sequence contains the numbers that are congruent to 2 or 5 modulo 7 after removing all factors of 7.

This works because \$(\mathbb{Z}/7\mathbb{Z})^{\times}/\{1,-1\}\cong C_3\$, with \$\{1,2,3\}\$ mapping to the three distinct elements of \$C_3\$, so that for \$n\$ not divisible by 7, \$\{n,2n,3n\}\$ also map to the three distinct elements of \$C_3\$.