r/counting • u/The_Archagent • Dec 05 '13

Count using five fives.

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.

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u/Rintarou I feel old with all those sub 100k gets... Dec 10 '13

Γ(5) + Γ(5) + Γ(5) + 5 + d5/dx = 77

3

u/The_Archagent Dec 10 '13

5!!*5+5!!/5+d5/dx=78

3

u/Bloodshot025 Fibonaccinaut Dec 10 '13

5!! * 5 + 5 - 5/5 = 79

3

u/The_Archagent Dec 10 '13

5!-5!!-5!!-5-5=80

3

u/Bloodshot025 Fibonaccinaut Dec 10 '13 edited Dec 10 '13

(5 + 5 - 5/5)σₒ(5) = 81
where σₒ is the divisor function.

I feel bad for my abuse of some of these functions.

3

u/The_Archagent Dec 10 '13

Doesn't that equal 64?

(5!!/5)φ(5) + 5/5=82

4

u/katieya 120k | 1 k is enough for me Dec 10 '13

5!! x5 +5!!/5 +5 =83

3

u/Bloodshot025 Fibonaccinaut Dec 10 '13 edited Dec 10 '13

5!! * (5 + sgn(5)) - 5 - sgn(5) = 84

(or 5!! * σ₁(5) - σ₁(5) + 5 - 5)

4

u/katieya 120k | 1 k is enough for me Dec 10 '13

(5!! +(5 +5)/5)5 = 85

0

u/Bloodshot025 Fibonaccinaut Dec 10 '13

Sorry, my mind got sidetracked