Rogue: Good job! Your 38 beats what I had, which was (4!sq+4sq+4sq)/4sq. I haven't had a chance to look at the rest of the solutions posted since then.

Someone could probably modify the code I wrote to solve the "Generating 2005" puzzle (which is attached to a post in there) to solve this. I haven't looked at trying yet.

P.S. That site is actually the worst of three sites with spoilers I found with Google. That site also uses the illegal floor (integer part) function.

rogue: what do you mean by ()sqsq? If I read it as (()^2)^2, 4sqsq + 4! = 65560.
BTW: 82 still lacks a solution, as 4!/.4 + 4! - 4 = 80 /= 82

Benjamin

Rogue ,

There's a problem with many of your solutions:

4sqsq isn't 64, it's 256. But where you use that with three 4's or less, you can sub your 4sqsq with 4*4sq.

62 = (4sqsq - 4 -4)/4

89 = ((4!/4)sq-.4)/.4
90 = (4!/4)sq/.4
91 = ((4!/4)sq+.4)/.4

Oops. It must have been that beer I had this evening. I don't know what I was thinking.

One little trick:

If you can't find a solution, see if 256-x is possible

113 = 4sqsq - (4!sq-4)/4
118 = 4sqsq - (4!sq-4!)/4
135 = 4sqsq - (44/4)sq
156 = 4sqsq - (4/.4)sq

That's as far as I checked.

Just recognised that you can save a 4 in my solution for 63/65 addopting this: 63 = (4^2^2-4)/4 giving 67 = (4^2^2-4)/4+4

46 = (4!-4)/.4-4
54 = 4!/4+4!+4!
66 = (4!+4)/.4-4
130 = (4!+4!+4)/.4

smallest # w/o representation: 73

Glancing over the record, I saw that 154 has a 3 4's representation, so:
158 = 4!/(.4)^2+4+4

73 = (4sqsq+(4!/4)sq)/4
127 = (4!sq-4)/4-4sq
139 = (4!sq-4)/4-4
147 = (4!sq-4)/4+4
159 = (4!sq-4)/4+4sq

Where is "75" ?

From above:
77 = (4 - 4/4)sqsq - 4
94 = (4/.4)sq - 4!/4
105 = (44/4)^2-4^2

126 = (4!sq+4!)/4-4!
129 = (4!sq+4)/4-4sq
134 = (4!sq+4!)/4 - 4sq
137 = (4!sq-4!-4)/4
141 = (4!sq-4sq+4)/4