Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**cmowla****Member**- Registered: 2012-06-14
- Posts: 70

Besides the gamma function,

(an integral) being a link between permutations and calculus, I have found a derivative relationship **for positive integers** (with help, as I will explain shortly). I'm not sure if it's a new formula or not (probably not), but I couldn't find it on the web anywhere, so I wanted to put it on the web!

**Inputing in Mathematica, we can see that it is equivalent to n!**

**How I Found It**

In the process of analyzing cycle types for a Rubik's cube theory project (of which I have probably mentioned it on here before), I wrote this post on speedsolving.com.

It's a long post, but just click the "spoiler" button (which we call "messages" on here) entitled "Formula Derivation".

I begin it with:

2 Pieces(k = 2)

3 Pieces(k = 3)

4 Pieces (k = 4)

5 Pieces (k = 5)

6 Pieces (k = 6)

7 Pieces (k = 7)

8 Pieces (k = 8)

9 Pieces (k = 9)

10 Pieces (k = 10)

11 Pieces (k = 11)

I had to find a way to express the general term of the sequence:

{1, 1, 3, 11, 53, 309, 2119, 16687, 148329, 1468457,...} (Which is **left** fraction's numerator, "some #".)

Just inputting this sequence into Mathematica, I found this OEIS sequence page.

On that page you will find:

c = CoefficientList[Series[Exp[ -z]/(1 - z)^2, {z, 0, 30}], z] For[n = 0, n < 31, n++; Print[c[[n]]*(n - 1)! ]]

As it says right beneath this mathematica code, it was submitted by user Zerinvary Lajos in 2009. Unfortunately, he died in 2011 (the year before I even made this post on speedsolving). May he (continue to) rest in peace.

Continuing on,

If we type the above line in Mathematica, we can type the next line to get the desired sequence above.

c*Table[n!, {n, 0, 30}]

Of course, the coefficients of a Maclaurin series are found via taking the nth derivative, and thus that's where the nth derivative comes from in my formula. For convenience, here's a link to Wolfram|Alpha obtaining the number 53 in the above sequence by inputting 6 for n. **Notice that I had to take the (k-2)th derivative to obtain the correct number in the sequence which corresponds to the above pattern for 6.**

Therefore, we can express the

as (for k not equal to zero)

Since

then

The Mathematica input below illustrates that this expression gives the correct subfactorial values.

Finally, we need to take the summation of this expression. Clearly k cannot be zero (as stated when we simplified), and when k is one, the entire sum is zero. Therefore, we start with k=2. (It also makes sense to start with k=2 because we are taking the (k-2)th derivative.)

Once we evaluate the summation for a few values of n, it should be clear that it is one less than the correct corresponding factorial value, and thus we add one to achieve the (my) formula which we wanted to derive.

Thanks for reading. Comments are welcome.

*Last edited by cmowla (2016-04-02 11:29:23)*

Offline

**cmowla****Member**- Registered: 2012-06-14
- Posts: 70

I forgot to mention this, but for those who don't have Mathematica, see !6 versus !6 (with this formula) to compute subfactorial.

You can substitute any integer in place of the 6's and get the correct subfactorial value.

Offline