## The factorial of -1, -2, -3, …

Well, n! is for integer n < 0 not defined -- as yet.

So the question is: How could a sensible generalization of the factorial for negative integers look like?

Clearly a good generalization should have a clear combinatorial meaning which combines well with the nonnegative case.

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## 7 Answers

It's not that it's not defined... Actually it has been defined more than it should have. There are plenty of functions that interpolate the factorials, some of them extend to the negative integers as well. Hadamard's Gamma function is entire, logarithmic single inflected factorial function is another example. But on the other hand, for some mysterious reason, the nice property that we want an extension of the factorial to enjoy is log-convexity. The Bohr-Mollerup-Artin Theorem tells us that the only function which is logarithmically convex on the positive real line and satisfies $f(z)=zf(z-1)$ there (also $f(1)=1$ and $f(z)>0$) is the Gamma function. Unfortunately the gamma function doesn't extend to negative integers, and that is why I guess people don't really care that much for defining them as they know that no "good" answer can be found.

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I would just add that the definition $n!=\infty$ when n<0 makes good sense; it extends the usual recursion of the factorial function and is consistent with the poles of the Gamma function. It also allows combinatorial formulas to work in degenerate cases. That is, when the factorial of a negative number appears in the denominator of a fraction, the usual convention is that the fraction is 0. – Jonas Meyer Dec 30 2009 at 12:29
Am I mistaken? Hadamars Gamma function is not logarithmic single inflected. The 'mysterious reason of log-convexity' /does not/ carry over to the general gamma function, i. e., it does not characterise the complex gamma function. This might be regarded as a hint, that there /is/ room for other sensible definitions and that log-convexity is not the last word in this matter, it is just a substitute for some analytical condition. – Bruce Arnold Dec 31 2009 at 0:22
Hadamards gamma function and the logarithmic single inflected factorial function are different functions. – Gjergji Zaimi Dec 31 2009 at 0:34
I am just referring to various functions that one can interpolate the factorials with, mostly based on playing around with the diGamma function. There is no citeable reference but here's a "pictoresque" account :) luschny.de/math/factorial/hadamard/… – Gjergji Zaimi Dec 31 2009 at 5:40
Don't blame log-convexity for making $(-1)!$ undefined. The functional equation $n!=(n-1)!n$ does that job all by itself. First (with $n=1$) it forces the standard convention that $0!=1$, and then (with $n=0$) is requires $(-1)!$ to be the reciprocal of 0. – Andreas Blass Jan 9 2012 at 1:15
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I think it's worth pointing out here that near z = -a, we have

$$\Gamma(z) = (-1)^a {1 \over a!} {1 \over {z-a}} + O(1)$$

and so it might be tempting to say that, in some sense,

$$\Gamma(-a) = (-1)^a {1 \over a!} \infty$$

where the symbol $\infty$ represents the rate at which $\Gamma$ blows up near the pole at $a = 0$. That is, $\Gamma(0) = \infty, \Gamma(-1) = -\infty, \Gamma(-2) = \infty/2, \Gamma(-3) = -\infty/6$, and so on.

In particular, this interpretation might work in some formula in which $\Gamma$ evaluated at nonpositive integers appears in both the numerator and the denominator, and the symbol $\infty$ can be canceled to yield a real number.

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mathworld.wolfram.com/RomanFactorial.html – Qiaochu Yuan Jul 17 2010 at 19:05

For a related paper see D. Loeb, Sets with a negative number of elements, Adv. Math. 91 (1992), 64–74.

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If you're wanting to compute factorials as an intermediate step to computing binomial coefficients, you may find a more satisfactory answer to your question. See this chart for determining how to compute binomial coefficients for general arguments.

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 See also "Getting results with negative thinking", D. E. Loeb et alia, especially the table on page 10 and the proposition on the six regions. arxiv.org/abs/math/9502214 – Bruce Arnold Dec 30 2009 at 23:25

I think, a better definition stems from the generalization of the triangle of eulerian numbers. For positive integer indexes the rows sum to factorials, and even if the rows are interpolated to fractional indexes the rowsums are fractional factorials or gamma values. Thus I assume the extension of the eulerian triangle to negative indexes gives the answer to a sensical definition if the factorials at negative parameters. For instance, we get -1! = 1 + 1/2 + 1/3 + 1/4 + ... -2! = (1) + (1+1/4) + (1+1/4+1/9) + ... -3! = (1) + (2+1/8) + (3+2/8+1/27) + ... and so on. The terms can be computed by the direct definition of Eulerian-numbers (see formula in wikipedia,for instance). I have discussed this in a hobby-treatize of the Eulerian-triangle in http://go.helms-net.de/math/binomial_new/01_12_Eulermatrix.pdf

Gottfried Helms

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My question was intended somewhat along the line: Assume the Gamma function is not yet invented and Goldbach asks you the question: "What is (-n)! ?" You know that Goldbach expects a combinatorial answer in the domain of integer or rational numbers. What would you answer? I will give my answer in this sens.

Looking at GKP's ConMath, Table 253, the combined Stirling triangles in their dual form, we see: If we sum the columns in this triangle for k < 0 we get the factorial numbers, if we sum the rows for k > 0 we get the Bell numbers.

What about saying the Bell numbers are the factorial numbers at negative integers? Is the answer encoded in one of the most important triangles in combinatorics?

See what Knuth says about the origin of this duality (table on page 11).

{120}
......{24}
.1,.......{6}
10, .1,......{2}
35, .6, .1,.....{1}
50, 11, 3, 1,
24, .6, 2, 1, 1,....{1}
.0, .0, 0, 0, 0, 1,
.0, .0, 0, 0, 0, 0, 1,...{1}
.0, .0, 0, 0, 0, 0, 1, .1,....{2}
.0, .0, 0, 0, 0, 0, 1, .3, .1,....{5}
.0, .0, 0, 0, 0, 0, 1, .7, .6, .1,....{15}
.0, .0, 0, 0, 0, 0, 1, 15, 25, 10, 1,....{52}

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Then you should have definitely mentioned this in the question :). To convince us that your correspondence is appropriate one has to give a combinatorial theory where the Stirling number duality makes sense. This is precisely what this thread mathoverflow.net/questions/9721/… is about. – Gjergji Zaimi Dec 31 2009 at 0:11
Thanks for the link. I just ask here another question to exemplify my main question. Looking at this triangle which was been considered by Kramp, Stanley, B. F. Logan, I. Gessel and D. Knuth, among others, the question is: Is there a simple and uniform mathematical explanation why summing rows and columns lines up the double infinite sequence ...203, 52, ,15, 5, 2, 1, 1, 1, 2, 6, 24, 120, 720,... and does this make the definition (-n)! = Bell(n) meaningful? Or do I have to conclude from your answer that there is no 'combinatorial theory where the Stirling number duality makes sense'? – Bruce Arnold Jan 1 2010 at 20:29

As 0! and 1! = 1 ,
2! = 2,
3! = 6 and so on

Can we not have

-1! = -1
-2! = 2 = -1 X -2
-3! = -6 = -1 X -2 X -3
-4! = 24 = -1 X -2 X -3 X -4
-5! = -120 = -1 X -2 X -3 X -4 X -5

...and so on?

It would produce some sort of series but would they be of any use anywhere?

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 This is known as the Roman Factorial - see Qiaochu's comment to Michael Lugo's answer. – S. Carnahan♦ Jan 9 2012 at 5:35