# Representation of numbers with three 3's

Remark: To simplify expressions, let's agree on the following notations that designated quantities expressible with a single 3:

• {1} = [3] = 1
• {2} = [3!] = 2
• {50} = [((3!)!)!] = 50

For example, {1}·{2} + 3 stands for [3]·[3!] + 3 = 5.

 1 √3·√3/3 2 3 - 3/3 = (3 + 3)/3 3 3·3/3 = 3 + 3 - 3 4 3 + 3/3 5 3!/3 + 3 6 3!·3/3 = 3/.(3) - 3 = 3<<(3/3) 7 3! + 3/3 = 3/.3 - 3 8 [3·√3 + 3] 9 3 + 3 + 3 = 33/3 10 [(√3 + √3)·3] = (3·3)^3 11 33/3 12 3! + 3 + 3 = 3·3 + 3 = 3/.(3) + 3 = (3!)!/(3!)!! - 3 13 3/.3 + 3 14 [3·(√3 + 3)] 15 3·3! - 3 = (3! - 3/3)!! = [3·3·√3] = (3! - [3/√3])!! 16 [3·3! - √3] 17 3!/.3 - 3 = 3·3! - [√3] 18 (3 + 3)·3 = (3!)!/(3!)!! + 3 19 [3!√3] - 3 = (3!)!!/3 + 3 20 (3 + 3)/.3 21 3·3! + 3 22 [[3! - √3]!! - √3] 23 [3! - √3]!! - [√3] 24 [3! - 3/√3]!! 25 [3! - √3]!! + [√3] 26 [[3! - √3]!! + √3] 27 3·3·3 28 33 + [√3] 29 [(3!)!!/√3 + √3] 30 33 + 3 = (3!)!! - 3·3! = 3/.([√3]) + 3 31 3/.[√3] + [√3] 32 33 - [√3](1) 33 33 + 3!(1) 34 33 + [√3] 35 33 + [√3!](1) 36 33 + 3 37 [√(√(3!)!]!/3 - 3(1) 38 33 + [√(√(3!)!)](1) 39 33 + 3!(1) 40 (3!)!/3!/3(1) 41 [√(√(3!)!]!/3 + [√3](1) 42 [√(√(3!)!]!/3 + [√3!](1) = (3!)!! - 3 - 3 43 [√(√(3!)!]!/3 + 3(1) 44 {50} - 3 - 3(2) 45 [√(√(3!)!]!/3 + [√(√(3!)!](1) = {50} - {2} - 3(2) 46 [√(√(3!)!]!/3 + 3!(1) = {50} - {2} - {2}(2) 47 {50} - 3! - 3(2) = {50} - {1} - {2} = (3!)!! - 3/3 48 {50} - 3!/3(2) = {50} - {1} - {1} = (3!)!!/3·3 = (3!)!! + 3 - 3 49 {50} - 3/3(2) = (3!)!! + 3/3 50 {50} - 3 + 3(2) = (3!)!! + 3!/3 51 {50} + 3/3(2) = (3!)!! + Ö3·Ö3 52 {50} +3!/3(2) = {50) + 3!/3 53 {50} + 3! - 3(2) = {50) + 3! - 3 54 {50} + {2} + {2}(2) = (3!)!! + 3 + 3 55 {50} + {2} + 3(2) 56 {50} + 3 + 3(2) 57 {50} + 3! + {1}(2) = (3!)!! + 3·3 58 {50} + 3! + {2}(2) = {50} + {2}3 = {50} + (3!)!!/3! = (3!)!! + 3/.3 59 {50} + 3·3(2) = {50) + 3! + 3 60 {50} + 3/.3(2) 61 [√((((√3!)!)!)!) - 33](3) 62 {50} + 3! + 3! 63 [{50} + √(3!)!! + √(3!)!!]

(1) By Andre Gustavo dos Santos from Brasil

(2) By Ken Nisbet, Crail, Fife, Scotland

(3) By Ken Nisbet, Crail, Fife, Scotland. Ken's formula depends on the extension of the factorial n! to arguments other than natural numbers. The extension is known as the G function. For natural x, G(x+1) = x!. In general, G(x+1) = xG(x). The G function is defined by

where x could be complex with the real part greater than 0.

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