[quote=grandpascorpion;98125]130 = (√4 / .4)! + 4/.4[/quote]Or, a la Andi47, 130 = √(√(√(4^4!))) * √4 + √4

[quote]131 = < snip > gamma < snip > (Am I throwing in the towel too early?)[/quote]Yes. :)

[quote]132 = (√4 / .4)! + 4!/√4[/quote]Also = √(√(√(4^4!))) * √4 + 4

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I suppose we all have our private collections of solutions with fewer than four 4s, to use as building blocks. I'll post mine in a little while.

131 = (4! / .4) / .4~ - 4
133 = (4! / .4) / .4~ -√4
134 = (4! /.4 - .4~) / .4 = 44 / .4 + 4!
135 = ((√4 * √4)! / .4) /.4~

Edit @cheesehead: Seems you like my 64 = √(√(√(4^4!))) ? :wink:

[quote=Andi47;98147]Edit @cheesehead: Seems you like my 64 = √(√(√(4^4!))) ? :wink:[/quote]I like any solution without the decimal point better than any solution with the decimal point. :-) I'll use the decimal point, but only if I can't find a way without it.

Here are building-block tables with fewer than four 4s. I haven't crosschecked them with other lists, so feel free to add entries.

Since some readers may not want such help, I've surrounded the entries with [ spoiler ] and [ /spoiler ].

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

Solutions with one "4":

- - -

[spoiler]
Using only the operators "+", "-", "*", "/", "(", ")", "!", "^"

4 = 4
24 = 4!

- - -

Using only the above operators and square-root "sqrt"

2 = sqrt4

- - -

Fractional building blocks with one "4":

Using only the decimal point (only as ".4")

2/5 = .4

- - -

Fractional building blocks with one "4":

Using only the square-root "sqrt",
decimal point and repeating decimal (only as ".4~" = 4/9 )

4/9 = .4~
2/3 = sqrt(.4~)
[/spoiler]

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

Solutions with two "4":

[If the entire solution is in brackets, that means there is another solution with fewer "4"s, using the same operator set (but not necessarily the same particular individual operators).]

- - -

[spoiler]
Using only the operators "+", "-", "*", "/", "(", ")", "!", "^",
and concatenation (only as "44")

0 = 4 - 4
1 = 4 / 4
6 = 4! / 4
8 = 4 + 4
16 = 4 * 4
20 = 4! - 4
28 = 4! + 4
44 = 44
48 = 4! + 4!
96 = 4! * 4
256 = 4 ^ 4
720 = (4! / 4)!
40320 = (4 + 4)!
2^48 = 4 ^ 4!

- - -

Using only the above operators
and square-root "sqrt"

[2 = 4 - sqrt4]
[4 = sqrt(4 * 4)]
12 = 4! / sqrt4
22 = 4! - sqrt4
[24 = (sqrt4 + sqrt4)!]
26 = 4! + sqrt4
64 = sqrt(sqrt(sqrt(4 ^ 4!)))
4096 = sqrt(sqrt(4 ^ 4!))
2^24 = sqrt(4 ^ 4!)

- - -

Using only the above operators
and decimal point (only as ".4" or ".44")

5 = sqrt4 / .4
10 = 4 / .4
60 = 4! / .4
120 = (sqrt4 / .4)!

- - -

Using only the above operators
and repeating decimal .4~ ( = 4/9 )

3 = sqrt(4 / .4~)
9 = 4 / .4~
54 = 4! / .4~

- - -

Fractional building blocks with two "4"

Using only the decimal point (only as ".4" or ".44")

11/25 = .44

- - -

Fractional building blocks with two "4"

Using only the "+", "-", "*", "/",
decimal point and repeating decimal (only as ".4~" = 4/9 )

1/9 = .4~ / 4
4/5 = .4 + .4

- - -

Fractional building blocks with two "4"

Using only the "+", "-", "*", "/". square-root "sqrt",
decimal point and repeating decimal (only as ".4~" = 4/9 )

1/6 = sqrt(.4~) / 4
2/9 = .4~ / sqrt4
1/3 = sqrt(.4~ / 4)
[/spoiler]

- - -

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

Solutions with three "4"s:

[If the entire solution is in brackets, that means there is another solution with fewer "4"s, using the same operator set (but not necessarily the same particular individual operators).]

- - -

[spoiler]
Using only the operators "+", "-", "*", "/", "(", ")", "!", "^",
and concatenation (only as "44" or "444")

[1 = 4 ^ (4 - 4)]
[[2 = (4 + 4) / 4]]
3 = 4 - 4 / 4
[[4 = 4 + 4 - 4]]
5 = 4 + 4 / 4
11 = 44 / 4
[12 = 4 + 4 + 4]
[20 = 44 - 4!]
23 = 4! - 4 / 4
[[24 = 4! * 4 / 4]]
25 = 4! + 4 / 4
30 = 4! + 4! / 4
32 = 4! + 4 + 4
40 = 44 - 4
(48 = 44 + 4)
(64 = 4 * 4 * 4)
68 = 44 + 4!
72 = 4! + 4! + 4!
92 = 4! * 4 - 4
100 = 4! * 4 + 4
120 = 4! * 4 + 4!
444 = 444

- - -

Using only the above operators and square-root "sqrt"

[0 = sqrt4 + sqrt4 - 4]
10 = 4 + 4 + sqrt4
13 = (4! + sqrt4) / sqrt4
[22 = 44 / sqrt4]
42 = 44 - sqrt4
46 = 44 + sqrt4
50 = 4! * sqrt4 + sqrt4
52 = 4! * sqrt4 + 4
88 = 44 * sqrt4
94 = 4! * 4 - sqrt4
98 = 4! * 4 + sqrt4

- - -

Using the above operators plus decimal point (only as ".4" or ".44" or ".444")

125 = sqrt(sqrt(sqrt((sqrt4 / .4) ^ 4!)))
240 = 4! * 4 / .4
5^6 = sqrt(sqrt((sqrt4 / .4) ^ 4!))
5^12 = sqrt((sqrt4 / .4) ^ 4!)
5^24 = (sqrt4 / .4) ^ 4!

- - -

Using only the above operators
and repeating decimal .4~ ( = 4/9 )

99 = 44 / .4~
108 = 4! * sqrt4 / .4~
135 = (4! / .4) / .4~
216 = 4! * 4 / .4~
[/spoiler]

- - -

Nobody for 136? Hint: Start with 120...

136 = (4! /.4 + .4~) / .4 = (sqrt(4)/.4)! + 4^sqrt(4)
137 = (4! /.4 ) / .4 + sqrt(4)

@Grandpascorpion: Your "137" is actually 152, a ~ is missing:

137 = (4! /.4~) / .4 + √4

138 = (4! * 4 - 4) / √(.4~)
139 = (4! / .4~) / .4 + 4
140 = (4! / √4) ^ (√4) - 4

P.S.: Your first 136 is missing a ~ too: 136 = (4! / .4 + .4~) / .4~

Ah, I got 135 from Cheesehead's post. I should have checked.

From the "Four nines" thread" the author made additional restrictions on some of the operations. It wasn't clear to me whether (for example) it was illegal to use the decimal or repeating decimal on anything besides a single 4. Also, it wasn't clear to me if concatenation was restricted to forming 44, or if things like 444, 24, or 42 were okay (where only one of the imputs in the concatenation function was 4). Could the author clarify this for me?

[QUOTE=Zeta-Flux;98244]From the "Four nines" thread" the author made additional restrictions on some of the operations. It wasn't clear to me whether (for example) it was illegal to use the decimal or repeating decimal on anything besides a single 4. Also, it wasn't clear to me if concatenation was restricted to forming 44, or if things like 444, 24, or 42 were okay (where only one of the imputs in the concatenation function was 4). Could the author clarify this for me?[/QUOTE]

THE SHORT ANSWER....
My intention was to allow:
Concatenation of digits only, not intermediate results:
VALID: 44, 444,
NOT VALID: (4!)4 = 244

Similarly for decimal/repeating(~) decimal:
VALID: .4, 4.4, .44, .444, 4.44, 44.4, etc.
VALID: .4~, 4.4~, 44.4~, etc.
NOT VALID: (√4).4 = 2.4, (√4).4~ = 2.4~

THE REASONING/HISTORY BEHIND MY LIST:
I originally found the corresponding 4-9's puzzle in a "GAMES" Magazine in the Early/Mid 1980's. Their puzzle only required answers from 1 to 20 but unless I can find it again (of course, I didn't throw it out!!) I can't say for sure what operations it allowed...but probably less than I have. I wondered if it was possible to go farther (I chose 0 to 100) and made it a challenge among friends, without answers and not knowing if we would find all or even most of them. Again, I can't remember if I added operations to the original list. If I did my intent would have been to list all operators I could think of that seemed realistic. At the time I did not know of the Gamma function and percent just didn't occur to me. Between a handful of us (and before computers were generally available) we came up with 97 of the 101 answers. I contemplated that the other 4 may not be possible. Shortly after that, one of our crew came across the same puzzle but this time with 4-4's and the addition of the .4~ function, and answers for all 101 possibilities (0-100). I considered allowing .9~ but since it equals 1 chose NOT to allow it since it would make the puzzle too easy. It never occured to me to go beyond 100. About a year ago my son, who was a much better programmer going into University than I was after 4 years of it and 20 years in the industry, wrote a program to recursively run through all the operators with 4-4's and with 4-9's. He chose to go up to 1000. He found my 4 missing answers for the 4-9's puzzle but still could not find answers for every number from 0 to 1000 for with 4's or 9's. (i.e. 22 unsolved from 100-199 with 9's and 21 unsolved from 100-199 with 4's and each getting worse the farther he went.) Mind you he did not think of using √(√(√(4^4!))) as a way to get 64....so I can't guarantee his list is complete either.

So, in a nut shell, since my list of operations was enough to solve my puzzle, 0 to 100, I had no reason to consider other operations and hence presented this puzzle to this forum with that very same list. If the forum member's collective intent is to go much beyond 100 then more operations will be required if there is a desire to solve every one of them.

Wayne....

Nobody for 141? It is possible without gamma and %...

(I could post it, but I already had my 4 solutions this turn.)