[Math Game] The Four Fours

Boy this is a stretch--but the fourth Fibonacci prime is 13 (2, 3, 5, 13, 89, 233, 1597, 28657, 514229, … (sequence A005478 in OEIS).

103=!4*!4/.4+Fp(4)

Rap

104=4*4!+4+4

Rap

If you want something slightly more conventional (using only the previously defined functions):
103 = P(FF(27)) [103 is the 27th prime]
where FF(n) is the function that generates the "four fours" expression that represents n. Note that although FF(n) is a new function, it is merely a shorthand notation for previously defined solutions.

For instance, FF(27) could be "4! + 4 - 4/4" because this expression equals 27.

Then in the spirit of that idea...

... 105 = FF(7)!!

Been waiting ages to use an odd double factorial.

But writing that just rubs me the wrong way, because I'm a such a stickler for abiding by the rules that I don't even jaywalk, so I am going to add this:

105 = (4!/4 + 4/4)!!

Ah, but you broke the rules each time you posted more than one solution.
However, I did notice that mars90... jumped on the rule-breaking bandwagon and submitted a batch to get to 100.

106=44/.4-4

Many times the most elementary solutions are elegant.

Rap

107 = φ(4$) + 44/4 (with the superfactorial)



I'm thinking of making what you might call "power moves", where I post like 13 answers at once. Why? Because I've looked ahead, and this game is going to end up making it to the mid 4-digit range, at least. Power moves would speed things along pretty quickly.

A power move would work like this for the number 333:

(1) You think of some way of making 333 out of just three fours...333 = F(4,4,4) for some function F(-).
(2) You have some way to make 0, 1, 2, ..., 12 -- each out of just one four: 0 = f0(4), 1 = f1(4), 2 = f2(4), ..., 12 = f12(4), for some fi's.

(3) You add F to the f0, f1, f2, ..., f12 and get 13 numbers in a row really quickly:

333 = F(4,4,4) + f0(4)
334 = F(4,4,4) + f1(4)
335 = F(4,4,4) + f2(4)
...
345 = F(4,4,4) + f12(4)

You end up being able to write out about 13 moves in the span of a few minutes with some cutting and pasting.

(4) Every time you make another power move, you reuse the fi's. All you have to do is change the F(-,-,-).

I think that's needed, and it's related to why I stopped contributing. With all of these functions/operations we're adding, there isn't much of a challenge - at least not where we're at now.

You can add 13 to your list of numbers obtainable with a single 4 since it is the sixth prime number.

You can also subtract and do N-13 through N+13 to get 27 in a single shot.

You could also take it to the next level, and do F(4,4) + f(4,4).

You can get over 100 with just two fours, so F(4,4)+f(4,4) could get over 200 in a single shot.

Power move:

108 = (4 + 4/4)! - φ(P(Γ(4)))
109 = (4 + 4/4)! - P(P(P(√4))
110 = (4 + 4/4)! - φ(P(P(P(√4)))
111 = (4 + 4/4)! - !4
112 = (4 + 4/4)! - 4!!
113 = (4 + 4/4)! - P(4)
114 = (4 + 4/4)! - Γ(4)
115 = (4 + 4/4)! - P(P(√4))
116 = (4 + 4/4)! - 4
117 = (4 + 4/4)! - P(√4)
118 = (4 + 4/4)! - √4
119 = (4 + 4/4)! - !√4
120 = (4 + 4/4)! + !(!√4)
121 = (4 + 4/4)! + !√4
122 = (4 + 4/4)! + √4
123 = (4 + 4/4)! + P(√4)
124 = (4 + 4/4)! + 4
125 = (4 + 4/4)! + P(P(√4))
126 = (4 + 4/4)! + Γ(4)
127 = (4 + 4/4)! + P(4)
128 = (4 + 4/4)! + 4!!
129 = (4 + 4/4)! + !4
130 = (4 + 4/4)! + φ(P(P(P(√4)))
131 = (4 + 4/4)! + P(P(P(√4))
132 = (4 + 4/4)! + φ(P(Γ(4)))

(because "(4 + 4/4)!" is 120)

Give my limited basis a little guidance--you're effectively saying if a generating function is creating a unique multiple of 25 with three 4's, a second generating exists that would fill in the next digit.

One could use some permutating function of 4 into any digit ranging from -12 to +12. That is Kolyo's power move.

Koylo's function uses three 4's for 120. Kylo's power move function generates -12,-11, -10.-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12.

Neat great move it generates individual numbers (120-12) to 132 (120+12).

So, if you generate 13 from three 4's, 1 to 25 is easy. Nest generate 38 to go from 26 to 50, then 63 for 51 to 74.

The Key then is the function using the first three 4's to generate 0,13,38,63,88,113,138...(13+25n)--to make a power move using the -12 to +12 permutation for the last digit.

Markr's Extension-- two digit power move function that goes from -144 to +144.

Do I have an inkling? Forgive me I'm old and senile

Rap

Next power move

144=4*4*!4=M

132 = M - φ(P(Γ(4)))
133 = M - P(P(P(√4))
134 = M - φ(P(P(P(√4)))
135 = M - !4
136 = M - 4!!
137 = M - P(4)
138 = M - Γ(4)
139 = M - P(P(√4))
140 = M - 4
141 = M - P(√4)
142 = M - √4
143 = M - !√4
144 = M + !(!√4)
145 = M + !√4
146 = M + √4
147 = M + P(√4)
148 = M + 4
149 = M + P(P(√4))
150 = M + Γ(4)
151 = M + P(4)
152 = M + 4!!
153 = M + !4
154 = M + φ(P(P(P(√4)))
155 = M + P(P(P(√4))
156 = M + φ(P(Γ(4)))

Oops there's overlap--my fly is open M should be 145, but after that you could build around 169 (!4+4)^(Sqrt(4))

Rap

Rap,

You've got it. As I mentioned above, Kolyo could have done -13 to +13. I'm not sure what the range is for two digits. I've been typing expressions for the last hour or so, and I'm up to 100. The prime function has been extremely helpful since the primes are fairly dense in this range.

Overlap is fine. When doing big chunks, it may be difficult or impossible to avoid some overlap.

I'm not sure if I can prove this but I'm making a conjecture that you may be able to use a permutation function function, and it doesn't necessarily need to be prime as long as it is odd.

I've got n odd digits picked at random, so I can generate a series of permutation functions that picks the 4th digit out of a permutation of n digits. then each individual permutation then has a +/- spin attached doubling those permutations.

BTW 13 if the 8th fibonacci number so adding a +/-13 to the power move is easily done with +/-13 = +/-F[4!!].

Rap

I don't understand what you're doing with the permutations. Perhaps you could provide an example...

13 is also the 6th prime.

M =(!4+4)^sqrt(4) = 169

157 = M - φ(P(Γ(4)))
158 = M - P(P(P(√4))
159 = M - φ(P(P(P(√4)))
160 = M - !4
161 = M - 4!!
162 = M - P(4)
163 = M - Γ(4)
164 = M - P(P(√4))
165 = M - 4
166 = M - P(√4)
167 = M - √4
168 = M - !√4
169 = M + !(!√4)
170 = M + !√4
171 = M + √4
172 = M + P(√4)
173 = M + 4
174 = M + P(P(√4))
175 = M + Γ(4)
176 = M + P(4)
177 = M + 4!!
178 = M + !4
179 = M + φ(P(P(P(√4)))
180 = M + P(P(P(√4))
181 = M + φ(P(Γ(4)))

Here are the numbers 0-200 expressed with two fours. Henceforth, the function Z[expression containing two fours that equals N] represents N-200 through N+200.

0 = 4-4
1 = 4/4
2 = 4/√4
3 = 4-!√4
4 = √4*√4
5 = 4+!√4
6 = 4+√4
7 = 4+P(√4)
8 = 4+4
9 = P(√4)*P(√4)
10 = Γ(4)+4
11 = P(4)+4
12 = 4!!+4
13 = !4+4
14 = P(4)+P(4)
15 = P(4)+4!!
16 = 4*4
17 = 4!!+!4
18 = !4+!4
20 = 4!-4
21 = P(√4)*P(4)
22 = 4!-√4
23 = 4+P(4!!)
24 = P(4)+P(P(4))
25 = Γ(4)+P(4!!)
26 = 4!+√4
27 = !4*P(√4)
28 = 4!+4
29 = Γ(4)+P(!4)
30 = 4!+Γ(4)
31 = 4!+P(4)
32 = 4*4!!
33 = 4!+!4
34 = P(P(4))*√4
35 = P(P(√4))*P(4)
36 = Γ(4)*Γ(4)
37 = 4!+P(Γ(4))
38 = P(4!!)*√4
39 = P(P(Γ(4)))-√4
40 = 4*(Γ(4)+4)
41 = 4!+P(P(4))
42 = P(4!!)+P(!4)
43 = 4!+P(4!!)
44 = 44
45 = 4+P(P(Γ(4)))
46 = P(!4)*√4
47 = 4!+P(!4)
48 = 4!+4!
49 = P(4)*P(4)
50 = P(P(Γ(4)))+!4
51 = P(P(4))*P(√4)
52 = P(Γ(4))*4
53 = P(4*4)
54 = Γ(4)*!4
55 = P(P(P(4)))-4
56 = P(4)*4!!
57 = P(P(P(4)))-√4
58 = P(P(Γ(4)))+P(P(4))
59 = P(P(P(4)))+!(!√4)
60 = P(P(P(4)))+!√4
61 = P(P(P(4)))+√4
62 = P(P(P(P(√4)))*√4
63 = P(4)*!4
64 = 4!!*4!!
65 = P(P(4!!))-√4
66 = P(P(4!!))-!√4
67 = P(P(4!!))+!(!√4)
68 = P(P(4!!))+!√4
69 = P(√4)*P(!4)
70 = P(P(!4))-P(Γ(4))
71 = P(P(4!!))+4
72 = 4!*P(√4)
73 = P(P(4!!))+Γ(4)
74 = P(P(4!!))+P(4)
75 = P(P(4!!))+4!!
76 = 4*P(4!!)
77 = P(4)*P(P(P(√4))
78 = P(P(!4))-P(P(√4))
79 = P(P(!4))-4
80 = P(P(!4))-P(√4)
81 = P(√4)^4
82 = P(P(Γ(4)))*√4
83 = P(4!)-Γ(4)
84 = P(4!)-P(P(√4))
85 = P(4!)-4
86 = P(4!)-P(√4)
87 = P(4!)-√4
88 = P(4!)-!√4
89 = P(4!)+!(!√4)
90 = P(4!)+!√4
91 = P(4!)+√4
92 = P(4!)+P(√4)
93 = P(4!)+4
94 = P(4!)+P(P(√4))
95 = P(4!)+Γ(4)
96 = 4*4!
97 = P(4!)+4!!
98 = P(4!)+!4
99 = P(4!)+φ(P(P(P(√4))))
100 = P(4!)+P(P(P(√4))
101 = P(P(φ(P(P(P(√4))))))-4!!
102 = P(4!)+P(Γ(4))
103 = P(P(φ(P(P(P(√4))))))-Γ(4)
104 = P(P(φ(P(P(P(√4))))))-P(P(√4))
105 = P(P(φ(P(P(P(√4))))))-4
106 = P(4!)+P(P(4))
107 = P(P(φ(P(P(P(√4))))))-√4
108 = P(P(φ(P(P(P(√4))))))-!√4
109 = P(P(φ(P(P(P(√4))))))+!(!√4)
110 = P(P(φ(P(P(P(√4))))))+!√4
111 = P(P(φ(P(P(P(√4))))))+√4
112 = P(4!)+P(!4)
113 = P(4!)+4!
114 = P(P(φ(P(P(P(√4))))))+P(P(√4))
115 = P(P(√4))*P(!4)
116 = 4*P(φ(P(P(P(√4)))))
117 = P(P(φ(P(P(P(√4))))))+4!!
118 = Γ(Γ(4))-√4
119 = Γ(Γ(4))-!√4
120 = Γ(Γ(4))+!(!√4)
121 = Γ(Γ(4))+!√4
122 = Γ(Γ(4))+√4
123 = Γ(Γ(4))+P(√4)
124 = Γ(Γ(4))+4
125 = Γ(Γ(4))+P(P(√4))
126 = Γ(Γ(4))+Γ(4)
127 = Γ(Γ(4))+P(4)
128 = Γ(Γ(4))+4!!
129 = Γ(Γ(4))+!4
130 = Γ(Γ(4))+φ(P(P(P(√4))))
131 = P(P(P(P(P(√4))))+4
132 = Γ(Γ(4))+φ(P(Γ(4)))
133 = Γ(Γ(4))+P(Γ(4))
134 = P(P(P(P(P(√4))))+P(4)
135 = P(P(P(P(P(√4))))+4!!
136 = P(P(P(P(P(√4))))+!4
137 = Γ(Γ(4))+P(P(4))
138 = P(P(P(P(P(√4))))+P(P(P(√4))
139 = Γ(Γ(4))+P(4!!)
140 = P(P(P(P(P(√4))))+P(Γ(4))
141 = P(P(φ(P(Γ(4)))))-φ(P(P(4)))
142 = P(P(P(Γ(4))))-P(φ(P(Γ(4))))
143 = Γ(Γ(4))+P(!4)
144 = Γ(Γ(4))+4!
145 = P(P(φ(P(Γ(4)))))-φ(P(Γ(4)))
146 = P(P(φ(P(Γ(4)))))-P(P(P(√4))
147 = P(P(φ(P(Γ(4)))))-φ(P(P(P(√4))))
148 = 4*P(φ(P(Γ(4))))
149 = P(P(φ(P(Γ(4)))))-4!!
150 = P(P(φ(P(Γ(4)))))-P(4)
151 = P(P(φ(P(Γ(4)))))-Γ(4)
152 = P(P(φ(P(Γ(4)))))-P(P(√4))
153 = !4*P(P(4))
154 = P(4)*φ(P(!4))
155 = P(P(φ(P(Γ(4)))))-√4
156 = P(P(φ(P(Γ(4)))))-!√4
157 = P(P(φ(P(Γ(4)))))+!(!√4)
158 = P(P(φ(P(Γ(4)))))+!√4
159 = P(P(φ(P(Γ(4)))))+√4
160 = P(P(φ(P(Γ(4)))))+P(√4)
161 = P(4)*P(!4)
162 = !4*φ(P(4!!))
163 = P(P(φ(P(Γ(4)))))+Γ(4)
164 = 4*P(P(Γ(4)))
165 = P(P(φ(P(Γ(4)))))+4!!
166 = P(P(φ(P(Γ(4)))))+!4
167 = P(P(P(Γ(4))))-φ(P(Γ(4)))
168 = P(4)*4!
169 = P(P(φ(P(Γ(4)))))+φ(P(Γ(4)))
170 = P(P(P(Γ(4))))-!4
171 = !4*P(4!!)
172 = P(P(P(Γ(4))))-P(4)
173 = P(P(P(Γ(4))))-Γ(4)
174 = P(P(P(Γ(4))))-P(P(√4))
175 = P(P(P(Γ(4))))-4
176 = 4!!*φ(P(!4))
177 = P(P(P(Γ(4))))-√4
178 = P(P(P(Γ(4))))-!√4
179 = P(P(P(Γ(4))))+!(!√4)
180 = P(P(√4))*φ(P(φ(P(Γ(4)))))
181 = P(P(P(Γ(4))))+√4
182 = P(P(P(Γ(4))))+P(√4)
183 = P(P(P(Γ(4))))+4
184 = 4!!*P(!4)
185 = P(P(√4))*P(φ(P(Γ(4))))
186 = Γ(4)*P(P(P(P(√4)))
187 = P(P(P(√4))*P(P(4))
188 = P(P(P(Γ(4))))+!4
189 = P(P(P(Γ(4))))+φ(P(P(P(√4))))
190 = φ(P(P(P(√4))))*P(4!!)
191 = P(P(P(Γ(4))))+φ(P(Γ(4)))
192 = 4!!*4!
193 = P(Γ(4)!!)-φ(P(P(P(P(√4))))
194 = P(Γ(4)!!)-P(φ(P(P(P(√4)))))
195 = P(P(P(Γ(4))))+φ(P(P(4)))
196 = P(P(P(Γ(4))))+P(P(4))
197 = P(P(P(Γ(4))))+φ(P(4!!))
198 = !4*φ(P(!4))
199 = P(Γ(4)!!)-4!
200 = P(Γ(4)!!)-P(!4)

Z[φ(P(P(4))) * φ(P(!4))] = Z[16*22] = Z[352]
which covers 152-552.

Z[4! * P(P(P(P(√4)))] = Z[24 * 31] = Z[744]
which covers 544-944

see definition of Z[expression] on page 5 of this thread

Z[φ(P(φ(P(Γ(4))))) * P(P(P(P(√4)))] = Z[36 * 31] = Z[1116]
which covers 916-1316

see definition of Z[expression] on page 5 of this thread