[Math Game] The Four Fours

F[n] is the nth number in the Fibonacci sequence

77=F[4!/Sqrt(2)]-4!/Sqrt(2)

Rap

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@raprap,

oopsey

77=F[4!/Sqrt(4)]-4!/Sqrt(4)

Rap

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78=F[4+4]*(4+Sqrt(4))

Rap

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@raprap,

raprap wrote:

78=F[4+4]*(4+Sqrt(4))
Rap

79 = F[4!!]*Γ(4) + 4/4

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I can't resist using the Totient function on something other than a prime:

80 = φ(4*4*4) / .4

[...because φ(64) = 32.]

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81=(4-4/4)^4

Rap

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@Kolyo,

Trying to see this one.

if 79 = F[4!!]*Γ(4) + 4/4

then 78=F[4!!]*Γ(4)

and as Γ(4)=(4-1)!=6

Then F[4!!]=13

13 is the 8th Fibonacci number so somehow 4!!=8

How?

Rap

1 Reply

@raprap,

raprap wrote:

Then F[4!!]=13

13 is the 8th Fibonacci number so somehow 4!!=8

How?

Rap

I'm using the "double factorials" that markr introduced at the beginning of the thread.

See: http://en.wikipedia.org/wiki/Double_factorial

Basically they work like this:
2!! = 2
4!! = 2*4
6!! = 2*4*6
8!! = 2*4*6*8
and so on...

They are somewhat problematic in a thread like this one,
because they look just like two factorial signs back to back,
so it looks like I'm writing something that is equal to 24!.

Perhaps I should avoid them and find some other way of making 8 out of a single 4.

1 Reply

@Kolyo,

I see--the double factoral notation is for odd and even positive integers. So

4!!=2^2*2!=8

No Problem

Rap

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82=!4*!4+4/4

Rap

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@raprap,

raprap wrote:

I see--the double factoral notation is for odd and even positive integers. So

No Problem

Still, I may avoid it. I do worry people reading this thread for the first time will see "4!! = 8" and think we're complete loons. Smile

Euler's totient function also gets you to 8 in one move: φ(4!) = φ(8*3) = φ(8)φ(3) = 4*2.

Meanwhile...
83 = P(P(!4)) + 4 - sqrt(4) - sqrt(4)
84 = P(P(!4)) + sqrt(4) - 4/4
85 = P(P(!4)) * 4/4 + sqrt(4)
86 = P(P(!4)) + sqrt(4) + 4/4
87 = P(P(!4)) * 4/4 + 4

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Leaving an almost trivial

88=44+44

and

89=F[4+4+Sqrt(4)+Sqrt(4)]

Rap

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90=!4*(4+4+Sqrt(4))

Rap

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91=4*4!-(sqrt(4)/.4)

1 Reply

@timur,

92 = Superfactorial(4)/4+4!-4

1 Reply

@nacredambition,

Had to look up superfactoral

sf(1)=1!=1
sf(2)=1!*2!=1*2=2
sf(3)=1!*2!*3!=1*2*6=12
sf(4)=1!*2!*3!*4!=1*2*6*24=288

good play

Rap

1 Reply

@raprap,

93 = φ(superfactorial(4)) - 4/4 - sqrt(4)
94 = φ(superfactorial(4)) * 4/4 - sqrt(4)

Also sf's allow us to get to 96 from a single 4, which is always a bonus.
[φ(288) = φ(32*9) = φ(32)φ(9) = 16*6 = 96.]

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95=!4*!4 + Γ(4) + 4!!
96 = 4!+4!+4!+4!
97 = 4! * 4 + !4 - 4!!
98 = 4!! * (4!! + 4) + √4
99= !4*!4 + !4 + !4
100 = 4!! * (4!! + 4) + 4

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101 = φ(superfactorial(4)) + 4 + 4/4
102 = φ(superfactorial(4)) + √4 + √4 + √4

1 Reply

@Kolyo,

fyi, the notation for superfactorial(n) seems to be n$.
There is also hyperfactorial: H(n) = 1^1⋅2^2⋅3^3⋅4^4⋅…⋅n^n.
H(1), H(2), H(3), H(4), H(5) = 1, 4, 108, 27648, 86400000

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[Math Game] The Four Fours

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