# 499 and the Gamma Function

Folks elsewhere in this very same site have been working hard to express positive integers in the binary system using merely four ones and four zeros. They have been allowed to use the four basic arithmetic operations, exponentiation, concatenation, brackets, fractions, factorials (including double factorials), fractional points (both with or without leading zeros, e.g. 0.0001111 or .00110011), and square roots.

So far they have succeeded expressing every number up to 500 except for 499, and every number up to 900 except for a handful (501, 607, 652, 653, etc.).

If the Gamma function were now allowed, would the remaining numbers be so expressible?

Recall that Gamma(n)=(n-1)!.

• Can I use $\Gamma^{-1}(((11!)!)!) = 721$? 😝 Mar 28 at 2:53

$$499$$ and no $$\Gamma$$ needed, actually.

$$499 = \frac{((11!)!!+0!+0!)^{0!+.1}}{\sqrt{.1}}-0!$$

Also without $$\Gamma$$:

$$501 = \frac{((11!)!!+0!+0!)^{0!+.1}}{\sqrt{.1}}+0!$$ (@ThomasL) $$549 = ((11!)!\times.1+(0!+0!+0!)!)\times(0!+.1)$$

• Excellent. With your solution you obviously get 501 as well without using the $\Gamma$ function. Mar 31 at 18:59
• This answers the (question-without-gamma-function)! Apr 1 at 14:38

Edit: updated to show all the previously unsolved numbers, and those I have solved.

The bountied best answer to the previous question left these 26 numbers unsolved:

$$499 \ 501 \ 549 \ 607 \ 652 \ 653 \ 787 \ 795 \ 802 \ 803 \ 805 \ 806 \ 807$$
$$821 \ 829 \ 853 \ 857 \ 859 \ 869 \ 877 \ 878 \ 879 \ 883 \ 884 \ 891 \ 892$$

Assuming the Gamma(n) function introduced is (n-1)! I updated my generator, and use $$\Gamma$$ to indicate the Gamma factorial so that for example $$\Gamma(4) = 3 \times 2 \times 1 = 6$$.
I solved 12 more numbers.

$$501 = \Gamma(11!) - (0! + 10) + (100!!)!!$$
$$549 = (\frac{0! + 0!}{.1} + .1) \times (\Gamma(11!) + 0! + 0!)$$
$$607 = (0! + 0! + 0!)! + 0! + (11!)! - \Gamma(11!)$$
$$652 = \frac{(100!!)!! + 0! + 0!}{.1} - \Gamma(11!)$$
$$787 = ((0! + 0! + 0!)! + .1) \times (\Gamma(11!) + 0!) + .1$$
$$795 = ((0! + 0! + 0!)! + .101) \times \Gamma(11!)$$
$$829 = \Gamma((0! + 0! + 0!)!) - \sqrt{(\Gamma(11!) + 0!)} + (11!)!$$
$$853 = (\Gamma(11!) + 0! + 0!) \times (11! + 0!) - 0!$$
$$879 = \frac{\Gamma(11!)}{.11} - 0! + ((0! + 0! + 0!)!)!$$
$$884 = ((\frac{0! + 0!}{.1})!! + .1) \times ((11! + 0!)!! - 0!)$$
$$891 = \frac{(100!!)!! + 0!}{.1} + \Gamma(11!) + 0!$$
$$892 = \frac{(100!!)!! + 0! + 0!}{.1} + \Gamma(11!)$$

Can anyone get the other 14 numbers? I expect they can, because my generator did not previously find all those in the linked best answer from @isaacg.

Update:

This answer only extended answers to the previous question by implementing the solution of integer values passed to the Gamma function, which is a simple factorial. But the Gamma function also applies to real numbers.

I am no mathematician and this is new to me, but I found this YouTube video The Gamma Function for Half Integer Values which discusses the Gamma function for positive and negative fractional half-numbers.

For $$\frac{1}{2}$$ he gives (and this is the starting point)
$$\Gamma(\frac{1}{2}) = \sqrt{\pi}$$
The speaker then derives the expansions for the other positive fractions. I can follow most of it, but I think it goes wrong in one place

For $$k = 1$$ is $$\Gamma(\frac{1}{2} + 1) = \Gamma(\frac{3}{2}) = \frac{\sqrt{\pi}}{2^1} \cdot 1$$
For $$k = 2$$ is $$\Gamma(\frac{1}{2} + 2) = \Gamma(\frac{5}{2}) = \frac{\sqrt{\pi}}{2^2} \cdot 3 \cdot 1$$
For $$k = 3$$ is $$\Gamma(\frac{1}{2} + 3) = \Gamma(\frac{7}{2}) = \frac{\sqrt{\pi}}{2^3} \cdot 5 \cdot 3 \cdot 1$$
For $$k = 4$$ is $$\Gamma(\frac{1}{2} + 4) = \Gamma(\frac{9}{2}) = \frac{\sqrt{\pi}}{2^4} \cdot 7 \cdot 5 \cdot 3 \cdot 1$$

He then forms a general expression, but when he comes to the last term says he does not understand a double factorial, fumbles around, and writes something like
$$\Gamma(\frac{1}{2} + k) = \Gamma(\frac{2k + 1}{2}) = \frac{\sqrt{\pi}}{2^k} \cdot (2k + 1)!!$$
But that looks wrong to me, IMO the general case ($$k > 0$$) should be
$$\Gamma(\frac{1}{2} + k) = \frac{\sqrt{\pi}}{2^k} \cdot (2k - 1)!!$$
For negative fractions I make the general expression ($$k > 0$$) to be
$$\Gamma(\frac{1}{2} - k) = \frac{(-1)^k \cdot 2^k \cdot \sqrt{\pi}}{(2k - 1)!!}$$
although I used $$k$$ a bit differently from his working. Obviously the $$\sqrt{\pi}$$ isn't useful when looking for integer results, but it will cancel out when by dividing one Gamma function by another.

Anyway, I implemented the Gamma function for half integer values, but my program took forever with all the new values to permute. When I restricted the source values to 3 bits, it completed but did not find any more results.

Perhaps there are mistakes and someone else can follow it up, and of course, there are the other real values that are not simple half-fractions.

• I see that 884 does not use the Gamma function but was previously unsolved. Mar 25 at 20:30
• And 829 uses two Gamma functions. Mar 25 at 21:31
• I added the gamma function to my program, and I got the same set of numbers that you did. Mar 25 at 21:54
• A solution for 892 that doesn't use gamma: 892 = ((11!)!)*(0!+.01)-(100!!) Mar 25 at 21:57