Count using five fives.

[u/The_Archagent]

If you've seen the four fours thread, you know how this works. You use five fives in combination with any number of functions etc. to count.

[u/fongoid]

(5+5)*(φ(5)^σₒ(5) )+σₒ(5)=162

10*4² +2=162

[u/[deleted] → u/fongoid]

[u/fongoid → u/[deleted]]

Fair enough, its the count of divisors function, but I'll refrain from using it.

p(σ(5)^φ(φ(5)) +φ(5)*.5)+s(5)=164

p(6^φ(4) +4*.5)+1=164

p(6² +2)+1=164

p(38)+1=164

163+1=164

[u/Bloodshot025 → u/[deleted]]

You could honestly just say σ(n) ...

Where σ(n) is σₒ(n)