Factorial

The factorial of ${\displaystyle 0}$ is ${\displaystyle 1}$, or in symbols, ${\displaystyle 0!=1}$. There are several motivations for this definition:

• For ${\displaystyle n=0}$, the definition of ${\displaystyle n!}$ as a product involves the product of no numbers at all, and so is an example of the broader convention that the empty product, a product of no factors, is equal to the multiplicative identity.[22]
• There is exactly one permutation of zero objects: with nothing to permute, the only rearrangement is to do nothing.[21]
• This convention makes many identities in combinatorics valid for all valid choices of their parameters. For instance, the number of ways to choose all ${\displaystyle n}$ elements from a set of ${\displaystyle n}$ is ${\textstyle {\tbinom {n}{n}}={\tfrac {n!}{n!0!}}=1,}$ a binomial coefficient identity that would only be valid with ${\displaystyle 0!=1}$.[23]
• With ${\displaystyle 0!=1}$, the recurrence relation for the factorial remains valid at ${\displaystyle n=1}$. Therefore, with this convention, a recursive computation of the factorial needs to have only the value for zero as a base case, simplifying the computation and avoiding the need for additional special cases.[24]
• Setting ${\displaystyle 0!=1}$ allows for the compact expression of many formulae, such as the exponential function, as a power series: ${\textstyle e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}.}$[14]
• This choice matches the gamma function ${\displaystyle 0!=\Gamma (0+1)=1}$, and the gamma function must have this value to be a continuous function.[25]

Comparison of the factorial, Stirling's approximation, and the simpler approximation ${\displaystyle (n/e)^{n}}$, on a doubly logarithmic scale

Relative error in a truncated Stirling series vs. number of terms

As a function of ${\displaystyle n}$, the factorial has faster than exponential growth, but grows more slowly than a double exponential function.[48] Its growth rate is similar to ${\displaystyle n^{n}}$, but slower by an exponential factor. One way of approaching this result is by taking the natural logarithm of the factorial, which turns its product formula into a sum, and then estimating the sum by an integral:

$${\displaystyle \ln n!=\sum _{x=1}^{n}\ln x\approx \int _{1}^{n}\ln x\,dx=n\ln n-n+1.}$$

Exponentiating the result (and ignoring the negligible ${\displaystyle +1}$ term) approximates ${\displaystyle n!}$ as ${\displaystyle (n/e)^{n}}$.[49] More carefully bounding the sum both above and below by an integral, using the trapezoid rule, shows that this estimate needs a correction term proportional to ${\displaystyle {\sqrt {n}}}$. The constant of proportionality for this correction can be found from the Wallis product, which expresses ${\displaystyle \pi }$ as a limiting ratio of factorials and powers of two. The result of these corrections is Stirling's approximation:[50]

$${\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}\,.}$$

Here, the ${\displaystyle \sim }$ symbol means that, as ${\displaystyle n}$ goes to infinity, the ratio between the left and right sides approaches one in the limit. Stirling's formula provides the first term in an asymptotic series that becomes even more accurate when taken to greater numbers of terms:[51]

$${\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}\left(1+{\frac {1}{12n}}+{\frac {1}{288n^{2}}}-{\frac {139}{51840n^{3}}}-{\frac {571}{2488320n^{4}}}+\cdots \right).}$$

An alternative version uses only odd exponents in the correction terms:[51]

$${\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}\exp \left({\frac {1}{12n}}-{\frac {1}{360n^{3}}}+{\frac {1}{1260n^{5}}}-{\frac {1}{1680n^{7}}}+\cdots \right).}$$

Many other variations of these formulas have also been developed, by Srinivasa Ramanujan, Bill Gosper, and others.[51]

The binary logarithm of the factorial, used to analyze comparison sorting, can be very accurately estimated using Stirling's approximation. In the formula below, the ${\displaystyle O(1)}$ term invokes big O notation.[45]

$${\displaystyle \log _{2}n!=n\log _{2}n-(\log _{2}e)n+{\frac {1}{2}}\log _{2}n+O(1).}$$

The product formula for the factorial implies that ${\displaystyle n!}$ is divisible by all prime numbers that are at most ${\displaystyle n}$, and by no larger prime numbers.[52] More precise information about its divisibility is given by Legendre's formula, which gives the exponent of each prime ${\displaystyle p}$ in the prime factorization of ${\displaystyle n!}$ as[53][54]

$${\displaystyle \sum _{i=1}^{\infty }\left\lfloor {\frac {n}{p^{i}}}\right\rfloor ={\frac {n-s_{p}(n)}{p-1}}.}$$

Here ${\displaystyle s_{p}(n)}$ denotes the sum of the base-${\displaystyle p}$ digits of ${\displaystyle n}$, and the exponent given by this formula can also be interpreted in advanced mathematics as the p-adic valuation of the factorial.[54] Applying Legendre's formula to the product formula for binomial coefficients produces Kummer's theorem, a similar result on the exponent of each prime in the factorization of a binomial coefficient.[55] Grouping the prime factors of the factorial into prime powers in different ways produces the multiplicative partitions of factorials.[56]

The special case of Legendre's formula for ${\displaystyle p=5}$ gives the number of trailing zeros in the decimal representation of the factorials.[57] According to this formula, the number of zeros can be obtained by subtracting the base-5 digits of ${\displaystyle n}$ from ${\displaystyle n}$, and dividing the result by four.[58] Legendre's formula implies that the exponent of the prime ${\displaystyle p=2}$ is always larger than the exponent for ${\displaystyle p=5}$, so each factor of five can be paired with a factor of two to produce one of these trailing zeros.[57] The leading digits of the factorials are distributed according to Benford's law.[59] Every sequence of digits, in any base, is the sequence of initial digits of some factorial number in that base.[60]

Another result on divisibility of factorials, Wilson's theorem, states that ${\displaystyle (n-1)!+1}$ is divisible by ${\displaystyle n}$ if and only if ${\displaystyle n}$ is a prime number.[52] For any given integer ${\displaystyle x}$, the Kempner function of ${\displaystyle x}$ is given by the smallest ${\displaystyle n}$ for which ${\displaystyle x}$ divides ${\displaystyle n!}$.[61] For almost all numbers (all but a subset of exceptions with asymptotic density zero), it coincides with the largest prime factor of ${\displaystyle x}$.[62]

The product of two factorials, ${\displaystyle m!\cdot n!}$, always evenly divides ${\displaystyle (m+n)!}$.[63] There are infinitely many factorials that equal the product of other factorials: if ${\displaystyle n}$ is itself any product of factorials, then ${\displaystyle n!}$ equals that same product multiplied by one more factorial, ${\displaystyle (n-1)!}$. The only known examples of factorials that are products of other factorials but are not of this "trivial" form are ${\displaystyle 9!=7!\cdot 3!\cdot 3!\cdot 2!}$, ${\displaystyle 10!=7!\cdot 6!=7!\cdot 5!\cdot 3!}$, and ${\displaystyle 16!=14!\cdot 5!\cdot 2!}$.[64] It would follow from the abc conjecture that there are only finitely many nontrivial examples.[65]

The greatest common divisor of the values of a primitive polynomial of degree ${\displaystyle d}$ over the integers evenly divides ${\displaystyle d!}$.[63]

The gamma function (shifted one unit left to match the factorials) continuously interpolates the factorial to non-integer values

Absolute values of the complex gamma function, showing poles at non-positive integers

There are infinitely many ways to extend the factorials to a continuous function.[66] The most widely used of these[67] uses the gamma function, which can be defined for positive real numbers as the integral

$${\displaystyle \Gamma (z)=\int _{0}^{\infty }x^{z-1}e^{-x}\,dx.}$$

The resulting function is related to the factorial of a non-negative integer ${\displaystyle n}$ by the equation

$${\displaystyle n!=\Gamma (n+1),}$$

which can be used as a definition of the factorial for non-integer arguments. At all values ${\displaystyle x}$ for which both ${\displaystyle \Gamma (x)}$ and ${\displaystyle \Gamma (x-1)}$ are defined, the gamma function obeys the functional equation

$${\displaystyle \Gamma (n)=(n-1)\Gamma (n-1),}$$

generalizing the recurrence relation for the factorials.[66]

The same integral converges more generally for any complex number ${\displaystyle z}$ whose real part is positive. It can be extended to the non-integer points in the rest of the complex plane by solving for Euler's reflection formula

$${\displaystyle \Gamma (z)\Gamma (1-z)={\frac {\pi }{\sin \pi z}}.}$$

However, this formula cannot be used at integers because, for them, the ${\displaystyle \sin \pi z}$ term would produce a division by zero. The result of this extension process is an analytic function, the analytic continuation of the integral formula for the gamma function. It has a nonzero value at all complex numbers, except for the non-positive integers where it has simple poles. Correspondingly, this provides a definition for the factorial at all complex numbers other than the negative integers.[67] One property of the gamma function, distinguishing it from other continuous interpolations of the factorials, is given by the Bohr–Mollerup theorem, which states that the gamma function (offset by one) is the only log-convex function on the positive real numbers that interpolates the factorials and obeys the same functional equation. A related uniqueness theorem of Helmut Wielandt states that the complex gamma function and its scalar multiples are the only holomorphic functions on the positive complex half-plane that obey the functional equation and remain bounded for complex numbers with real part between 1 and 2.[68]

Other complex functions that interpolate the factorial values include Hadamard's gamma function, which is an entire function over all the complex numbers, including the non-positive integers.[69][70] In the p-adic numbers, it is not possible to continuously interpolate the factorial function directly, because the factorials of large integers (a dense subset of the p-adics) converge to zero according to Legendre's formula, forcing any continuous function that is close to their values to be zero everywhere. Instead, the p-adic gamma function provides a continuous interpolation of a modified form of the factorial, omitting the factors in the factorial that are divisible by p.[71]

The digamma function is the logarithmic derivative of the gamma function. Just as the gamma function provides a continuous interpolation of the factorials, offset by one, the digamma function provides a continuous interpolation of the harmonic numbers, offset by the Euler–Mascheroni constant.[72]

TI SR-50A, a 1975 calculator with a factorial key (third row, center right)

The factorial function is a common feature in scientific calculators.[73] It is also included in scientific programming libraries such as the Python mathematical functions module[74] and the Boost C++ library.[75] If efficiency is not a concern, computing factorials is trivial: just successively multiply a variable initialized to ${\displaystyle 1}$ by the integers up to ${\displaystyle n}$. The simplicity of this computation makes it a common example in the use of different computer programming styles and methods.[76]

The computation of ${\displaystyle n!}$ can be expressed in pseudocode using iteration[77] as

define factorial(n):
  f := 1
  for i := 1, 2, 3, ..., n:
    f := f × i
  return f

or using recursion[78] based on its recurrence relation as

define factorial(n):
  if n = 0 return 1
  return n × factorial(n − 1)

Other methods suitable for its computation include memoization,[79] dynamic programming,[80] and functional programming.[81] The computational complexity of these algorithms may be analyzed using the unit-cost random-access machine model of computation, in which each arithmetic operation takes constant time and each number uses a constant amount of storage space. In this model, these methods can compute ${\displaystyle n!}$ in time ${\displaystyle O(n)}$, and the iterative version uses space ${\displaystyle O(1)}$. Unless optimized for tail recursion, the recursive version takes linear space to store its call stack.[82] However, this model of computation is only suitable when ${\displaystyle n}$ is small enough to allow ${\displaystyle n!}$ to fit into a machine word.[83] The values 12! and 20! are the largest factorials that can be stored in, respectively, the 32-bit[84] and 64-bit integers.[85] Floating point can represent larger factorials, but approximately rather than exactly, and will still overflow for factorials larger than ${\displaystyle 170!}$.[84]

The exact computation of larger factorials involves arbitrary-precision arithmetic, because of fast growth and integer overflow. Time of computation can be analyzed as a function of the number of digits or bits in the result.[85] By Stirling's formula, ${\displaystyle n!}$ has ${\displaystyle b=O(n\log n)}$ bits.[86] The Schönhage–Strassen algorithm can produce a ${\displaystyle b}$-bit product in time ${\displaystyle O(b\log b\log \log b)}$, and faster multiplication algorithms taking time ${\displaystyle O(b\log b)}$ are known.[87] However, computing the factorial involves repeated products, rather than a single multiplication, so these time bounds do not apply directly. In this setting, computing ${\displaystyle n!}$ by multiplying the numbers from 1 to ${\displaystyle n}$ in sequence is inefficient, because it involves ${\displaystyle n}$ multiplications, a constant fraction of which take time ${\displaystyle O(n\log ^{2}n)}$ each, giving total time ${\displaystyle O(n^{2}\log ^{2}n)}$. A better approach is to perform the multiplications as a divide-and-conquer algorithm that multiplies a sequence of ${\displaystyle i}$ numbers by splitting it into two subsequences of ${\displaystyle i/2}$ numbers, multiplies each subsequence, and combines the results with one last multiplication. This approach to the factorial takes total time ${\displaystyle O(n\log ^{3}n)}$: one logarithm comes from the number of bits in the factorial, a second comes from the multiplication algorithm, and a third comes from the divide and conquer.[88]

Even better efficiency is obtained by computing n! from its prime factorization, based on the principle that exponentiation by squaring is faster than expanding an exponent into a product.[86][89] An algorithm for this by Arnold Schönhage begins by finding the list of the primes up to ${\displaystyle n}$, for instance using the sieve of Eratosthenes, and uses Legendre's formula to compute the exponent for each prime. Then it computes the product of the prime powers with these exponents, using a recursive algorithm, as follows:

• Use divide and conquer to compute the product of the primes whose exponents are odd
• Divide all of the exponents by two (rounding down to an integer), recursively compute the product of the prime powers with these smaller exponents, and square the result
• Multiply together the results of the two previous steps

The product of all primes up to ${\displaystyle n}$ is an ${\displaystyle O(n)}$-bit number, by the prime number theorem, so the time for the first step is ${\displaystyle O(n\log ^{2}n)}$, with one logarithm coming from the divide and conquer and another coming from the multiplication algorithm. In the recursive calls to the algorithm, the prime number theorem can again be invoked to prove that the numbers of bits in the corresponding products decrease by a constant factor at each level of recursion, so the total time for these steps at all levels of recursion adds in a geometric series to ${\displaystyle O(n\log ^{2}n)}$. The time for the squaring in the second step and the multiplication in the third step are again ${\displaystyle O(n\log ^{2}n)}$, because each is a single multiplication of a number with ${\displaystyle O(n\log n)}$ bits. Again, at each level of recursion the numbers involved have a constant fraction as many bits (because otherwise repeatedly squaring them would produce too large a final result) so again the amounts of time for these steps in the recursive calls add in a geometric series to ${\displaystyle O(n\log ^{2}n)}$. Consequentially, the whole algorithm takes time ${\displaystyle O(n\log ^{2}n)}$, proportional to a single multiplication with the same number of bits in its result.[89]