/

# Combinatorial¶

This module implements various combinatorial functions.

## bell¶

class sympy.functions.combinatorial.numbers.bell[source]

Bell numbers / Bell polynomials

The Bell numbers satisfy $$B_0 = 1$$ and

$B_n = \sum_{k=0}^{n-1} \binom{n-1}{k} B_k.$

They are also given by:

$B_n = \frac{1}{e} \sum_{k=0}^{\infty} \frac{k^n}{k!}.$

The Bell polynomials are given by $$B_0(x) = 1$$ and

$B_n(x) = x \sum_{k=1}^{n-1} \binom{n-1}{k-1} B_{k-1}(x).$

The second kind of Bell polynomials (are sometimes called “partial” Bell polynomials or incomplete Bell polynomials) are defined as

$B_{n,k}(x_1, x_2,\dotsc x_{n-k+1}) = \sum_{j_1+j_2+j_2+\dotsb=k \atop j_1+2j_2+3j_2+\dotsb=n} \frac{n!}{j_1!j_2!\dotsb j_{n-k+1}!} \left(\frac{x_1}{1!} \right)^{j_1} \left(\frac{x_2}{2!} \right)^{j_2} \dotsb \left(\frac{x_{n-k+1}}{(n-k+1)!} \right) ^{j_{n-k+1}}.$
• bell(n) gives the $$n^{th}$$ Bell number, $$B_n$$.
• bell(n, x) gives the $$n^{th}$$ Bell polynomial, $$B_n(x)$$.
• bell(n, k, (x1, x2, ...)) gives Bell polynomials of the second kind, $$B_{n,k}(x_1, x_2, \dotsc, x_{n-k+1})$$.

bernoulli, catalan, euler, fibonacci, harmonic, lucas

Notes

Not to be confused with Bernoulli numbers and Bernoulli polynomials, which use the same notation.

References

Examples

>>> from sympy import bell, Symbol, symbols

>>> [bell(n) for n in range(11)]
[1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975]
>>> bell(30)
846749014511809332450147
>>> bell(4, Symbol('t'))
t**4 + 6*t**3 + 7*t**2 + t
>>> bell(6, 2, symbols('x:6')[1:])
6*x1*x5 + 15*x2*x4 + 10*x3**2


## bernoulli¶

class sympy.functions.combinatorial.numbers.bernoulli[source]

Bernoulli numbers / Bernoulli polynomials

The Bernoulli numbers are a sequence of rational numbers defined by B_0 = 1 and the recursive relation (n > 0):

      n
___
\      / n + 1 \
0 =  )     |       | * B .
/___   \   k   /    k
k = 0

They are also commonly defined by their exponential generating function, which is x/(exp(x) - 1). For odd indices > 1, the Bernoulli numbers are zero.

The Bernoulli polynomials satisfy the analogous formula:

          n
___
\      / n \         n-k
B (x) =  )     |   | * B  * x   .
n      /___   \ k /    k
k = 0

Bernoulli numbers and Bernoulli polynomials are related as B_n(0) = B_n.

We compute Bernoulli numbers using Ramanujan’s formula:

                         / n + 3 \
B   =  (A(n) - S(n))  /  |       |
n                       \   n   /

where A(n) = (n+3)/3 when n = 0 or 2 (mod 6), A(n) = -(n+3)/6 when n = 4 (mod 6), and:

       [n/6]
___
\      /  n + 3  \
S(n) =  )     |         | * B
/___   \ n - 6*k /    n-6*k
k = 1

This formula is similar to the sum given in the definition, but cuts 2/3 of the terms. For Bernoulli polynomials, we use the formula in the definition.

• bernoulli(n) gives the nth Bernoulli number, B_n
• bernoulli(n, x) gives the nth Bernoulli polynomial in x, B_n(x)

bell, catalan, euler, fibonacci, harmonic, lucas

References

Examples

>>> from sympy import bernoulli

>>> [bernoulli(n) for n in range(11)]
[1, -1/2, 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66]
>>> bernoulli(1000001)
0


## binomial¶

class sympy.functions.combinatorial.factorials.binomial[source]

Implementation of the binomial coefficient. It can be defined in two ways depending on its desired interpretation:

C(n,k) = n!/(k!(n-k)!) or C(n, k) = ff(n, k)/k!

First, in a strict combinatorial sense it defines the number of ways we can choose ‘k’ elements from a set of ‘n’ elements. In this case both arguments are nonnegative integers and binomial is computed using an efficient algorithm based on prime factorization.

The other definition is generalization for arbitrary ‘n’, however ‘k’ must also be nonnegative. This case is very useful when evaluating summations.

For the sake of convenience for negative ‘k’ this function will return zero no matter what valued is the other argument.

To expand the binomial when n is a symbol, use either expand_func() or expand(func=True). The former will keep the polynomial in factored form while the latter will expand the polynomial itself. See examples for details.

Examples

>>> from sympy import Symbol, Rational, binomial, expand_func
>>> n = Symbol('n', integer=True)

>>> binomial(15, 8)
6435

>>> binomial(n, -1)
0

>>> [ binomial(0, i) for i in range(1)]
[1]
>>> [ binomial(1, i) for i in range(2)]
[1, 1]
>>> [ binomial(2, i) for i in range(3)]
[1, 2, 1]
>>> [ binomial(3, i) for i in range(4)]
[1, 3, 3, 1]
>>> [ binomial(4, i) for i in range(5)]
[1, 4, 6, 4, 1]

>>> binomial(Rational(5,4), 3)
-5/128

>>> binomial(n, 3)
binomial(n, 3)

>>> binomial(n, 3).expand(func=True)
n**3/6 - n**2/2 + n/3

>>> expand_func(binomial(n, 3))
n*(n - 2)*(n - 1)/6


## catalan¶

class sympy.functions.combinatorial.numbers.catalan[source]

Catalan numbers

The n-th catalan number is given by:

       1   / 2*n \
C  = ----- |     |
n   n + 1 \  n  /
• catalan(n) gives the n-th Catalan number, C_n

bell, bernoulli, euler, fibonacci, harmonic, lucas, sympy.functions.combinatorial.factorials.binomial

References

Examples

>>> from sympy import (Symbol, binomial, gamma, hyper, polygamma,
...             catalan, diff, combsimp, Rational, I)

>>> [ catalan(i) for i in range(1,10) ]
[1, 2, 5, 14, 42, 132, 429, 1430, 4862]

>>> n = Symbol("n", integer=True)

>>> catalan(n)
catalan(n)


Catalan numbers can be transformed into several other, identical expressions involving other mathematical functions

>>> catalan(n).rewrite(binomial)
binomial(2*n, n)/(n + 1)

>>> catalan(n).rewrite(gamma)
4**n*gamma(n + 1/2)/(sqrt(pi)*gamma(n + 2))

>>> catalan(n).rewrite(hyper)
hyper((-n + 1, -n), (2,), 1)


For some non-integer values of n we can get closed form expressions by rewriting in terms of gamma functions:

>>> catalan(Rational(1,2)).rewrite(gamma)
8/(3*pi)


We can differentiate the Catalan numbers C(n) interpreted as a continuous real funtion in n:

>>> diff(catalan(n), n)
(polygamma(0, n + 1/2) - polygamma(0, n + 2) + log(4))*catalan(n)


As a more advanced example consider the following ratio between consecutive numbers:

>>> combsimp((catalan(n + 1)/catalan(n)).rewrite(binomial))
2*(2*n + 1)/(n + 2)


The Catalan numbers can be generalized to complex numbers:

>>> catalan(I).rewrite(gamma)
4**I*gamma(1/2 + I)/(sqrt(pi)*gamma(2 + I))


and evaluated with arbitrary precision:

>>> catalan(I).evalf(20)
0.39764993382373624267 - 0.020884341620842555705*I


## euler¶

class sympy.functions.combinatorial.numbers.euler[source]

Euler numbers

The euler numbers are given by:

        2*n+1   k
___   ___            j          2*n+1
\     \     / k \ (-1)  * (k-2*j)
E   = I  )     )    |   | --------------------
2n     /___  /___  \ j /      k    k
k = 1 j = 0           2  * I  * k

E     = 0
2n+1
• euler(n) gives the n-th Euler number, E_n

bell, bernoulli, catalan, fibonacci, harmonic, lucas

References

Examples

>>> from sympy import Symbol
>>> from sympy.functions import euler
>>> [euler(n) for n in range(10)]
[1, 0, -1, 0, 5, 0, -61, 0, 1385, 0]
>>> n = Symbol("n")
>>> euler(n+2*n)
euler(3*n)


## factorial¶

class sympy.functions.combinatorial.factorials.factorial[source]

Implementation of factorial function over nonnegative integers. By convention (consistent with the gamma function and the binomial coefficients), factorial of a negative integer is complex infinity.

The factorial is very important in combinatorics where it gives the number of ways in which $$n$$ objects can be permuted. It also arises in calculus, probability, number theory, etc.

There is strict relation of factorial with gamma function. In fact n! = gamma(n+1) for nonnegative integers. Rewrite of this kind is very useful in case of combinatorial simplification.

Computation of the factorial is done using two algorithms. For small arguments naive product is evaluated. However for bigger input algorithm Prime-Swing is used. It is the fastest algorithm known and computes n! via prime factorization of special class of numbers, called here the ‘Swing Numbers’.

factorial2, RisingFactorial, FallingFactorial

Examples

>>> from sympy import Symbol, factorial, S
>>> n = Symbol('n', integer=True)

>>> factorial(0)
1

>>> factorial(7)
5040

>>> factorial(-2)
zoo

>>> factorial(n)
factorial(n)

>>> factorial(2*n)
factorial(2*n)

>>> factorial(S(1)/2)
factorial(1/2)


## factorial2 / double factorial¶

class sympy.functions.combinatorial.factorials.factorial2[source]

The double factorial n!!, not to be confused with (n!)!

The double factorial is defined for integers >= -1 as:

       ,
|  n*(n - 2)*(n - 4)* ... * 1    for n odd
n!! = {  n*(n - 2)*(n - 4)* ... * 2    for n even
|  1                             for n = 0, -1


factorial, RisingFactorial, FallingFactorial

Examples

>>> from sympy import factorial2, var
>>> var('n')
n
>>> factorial2(n + 1)
factorial2(n + 1)
>>> factorial2(5)
15
>>> factorial2(-1)
1


## FallingFactorial¶

class sympy.functions.combinatorial.factorials.FallingFactorial[source]

Falling factorial (related to rising factorial) is a double valued function arising in concrete mathematics, hypergeometric functions and series expansions. It is defined by

ff(x, k) = x * (x-1) * ... * (x - k+1)

where ‘x’ can be arbitrary expression and ‘k’ is an integer. For more information check “Concrete mathematics” by Graham, pp. 66 or visit http://mathworld.wolfram.com/FallingFactorial.html page.

>>> from sympy import ff
>>> from sympy.abc import x

>>> ff(x, 0)
1

>>> ff(5, 5)
120

>>> ff(x, 5) == x*(x-1)*(x-2)*(x-3)*(x-4)
True


factorial, factorial2, RisingFactorial

## fibonacci¶

class sympy.functions.combinatorial.numbers.fibonacci[source]

Fibonacci numbers / Fibonacci polynomials

The Fibonacci numbers are the integer sequence defined by the initial terms F_0 = 0, F_1 = 1 and the two-term recurrence relation F_n = F_{n-1} + F_{n-2}.

The Fibonacci polynomials are defined by F_1(x) = 1, F_2(x) = x, and F_n(x) = x*F_{n-1}(x) + F_{n-2}(x) for n > 2. For all positive integers n, F_n(1) = F_n.

• fibonacci(n) gives the nth Fibonacci number, F_n
• fibonacci(n, x) gives the nth Fibonacci polynomial in x, F_n(x)

bell, bernoulli, catalan, euler, harmonic, lucas

References

Examples

>>> from sympy import fibonacci, Symbol

>>> [fibonacci(x) for x in range(11)]
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
>>> fibonacci(5, Symbol('t'))
t**4 + 3*t**2 + 1


## harmonic¶

class sympy.functions.combinatorial.numbers.harmonic[source]

Harmonic numbers

The nth harmonic number is given by $$\operatorname{H}_{n} = 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n}$$.

More generally:

$\operatorname{H}_{n,m} = \sum_{k=1}^{n} \frac{1}{k^m}$

As $$n \rightarrow \infty$$, $$\operatorname{H}_{n,m} \rightarrow \zeta(m)$$, the Riemann zeta function.

• harmonic(n) gives the nth harmonic number, $$\operatorname{H}_n$$
• harmonic(n, m) gives the nth generalized harmonic number of order $$m$$, $$\operatorname{H}_{n,m}$$, where harmonic(n) == harmonic(n, 1)

bell, bernoulli, catalan, euler, fibonacci, lucas

References

Examples

>>> from sympy import harmonic, oo

>>> [harmonic(n) for n in range(6)]
[0, 1, 3/2, 11/6, 25/12, 137/60]
>>> [harmonic(n, 2) for n in range(6)]
[0, 1, 5/4, 49/36, 205/144, 5269/3600]
>>> harmonic(oo, 2)
pi**2/6

>>> from sympy import Symbol, Sum
>>> n = Symbol("n")

>>> harmonic(n).rewrite(Sum)
Sum(1/_k, (_k, 1, n))


We can evaluate harmonic numbers for all integral and positive rational arguments:

>>> from sympy import S, expand_func, simplify
>>> harmonic(8)
761/280
>>> harmonic(11)
83711/27720

>>> H = harmonic(1/S(3))
>>> H
harmonic(1/3)
>>> He = expand_func(H)
>>> He
-log(6) - sqrt(3)*pi/6 + 2*Sum(log(sin(_k*pi/3))*cos(2*_k*pi/3), (_k, 1, 1))
+ 3*Sum(1/(3*_k + 1), (_k, 0, 0))
>>> He.doit()
-log(6) - sqrt(3)*pi/6 - log(sqrt(3)/2) + 3
>>> H = harmonic(25/S(7))
>>> He = simplify(expand_func(H).doit())
>>> He
log(sin(pi/7)**(-2*cos(pi/7))*sin(2*pi/7)**(2*cos(16*pi/7))*cos(pi/14)**(-2*sin(pi/14))/14)
+ pi*tan(pi/14)/2 + 30247/9900
>>> He.n(40)
1.983697455232980674869851942390639915940
>>> harmonic(25/S(7)).n(40)
1.983697455232980674869851942390639915940


We can rewrite harmonic numbers in terms of polygamma functions:

>>> from sympy import digamma, polygamma
>>> m = Symbol("m")

>>> harmonic(n).rewrite(digamma)
polygamma(0, n + 1) + EulerGamma

>>> harmonic(n).rewrite(polygamma)
polygamma(0, n + 1) + EulerGamma

>>> harmonic(n,3).rewrite(polygamma)
polygamma(2, n + 1)/2 - polygamma(2, 1)/2

>>> harmonic(n,m).rewrite(polygamma)
(-1)**m*(polygamma(m - 1, 1) - polygamma(m - 1, n + 1))/factorial(m - 1)


Integer offsets in the argument can be pulled out:

>>> from sympy import expand_func

>>> expand_func(harmonic(n+4))
harmonic(n) + 1/(n + 4) + 1/(n + 3) + 1/(n + 2) + 1/(n + 1)

>>> expand_func(harmonic(n-4))
harmonic(n) - 1/(n - 1) - 1/(n - 2) - 1/(n - 3) - 1/n


Some limits can be computed as well:

>>> from sympy import limit, oo

>>> limit(harmonic(n), n, oo)
oo

>>> limit(harmonic(n, 2), n, oo)
pi**2/6

>>> limit(harmonic(n, 3), n, oo)
-polygamma(2, 1)/2


However we can not compute the general relation yet:

>>> limit(harmonic(n, m), n, oo)
harmonic(oo, m)


which equals zeta(m) for m > 1.

## lucas¶

class sympy.functions.combinatorial.numbers.lucas[source]

Lucas numbers

Lucas numbers satisfy a recurrence relation similar to that of the Fibonacci sequence, in which each term is the sum of the preceding two. They are generated by choosing the initial values L_0 = 2 and L_1 = 1.

• lucas(n) gives the nth Lucas number

bell, bernoulli, catalan, euler, fibonacci, harmonic

References

Examples

>>> from sympy import lucas

>>> [lucas(x) for x in range(11)]
[2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123]


## MultiFactorial¶

class sympy.functions.combinatorial.factorials.MultiFactorial[source]

## RisingFactorial¶

class sympy.functions.combinatorial.factorials.RisingFactorial[source]

Rising factorial (also called Pochhammer symbol) is a double valued function arising in concrete mathematics, hypergeometric functions and series expansions. It is defined by:

rf(x, k) = x * (x+1) * ... * (x + k-1)

where ‘x’ can be arbitrary expression and ‘k’ is an integer. For more information check “Concrete mathematics” by Graham, pp. 66 or visit http://mathworld.wolfram.com/RisingFactorial.html page.

factorial, factorial2, FallingFactorial

Examples

>>> from sympy import rf
>>> from sympy.abc import x

>>> rf(x, 0)
1

>>> rf(1, 5)
120

>>> rf(x, 5) == x*(1 + x)*(2 + x)*(3 + x)*(4 + x)
True


## stirling¶

sympy.functions.combinatorial.numbers.stirling(n, k, d=None, kind=2, signed=False)[source]

Return Stirling number S(n, k) of the first or second (default) kind.

The sum of all Stirling numbers of the second kind for k = 1 through n is bell(n). The recurrence relationship for these numbers is:

{0}       {n}   {0}      {n + 1}     {n}   {  n  }
{ } = 1;  { } = { } = 0; {     } = j*{ } + {     }
{0}       {0}   {k}      {  k  }     {k}   {k - 1}
where j is::
n for Stirling numbers of the first kind -n for signed Stirling numbers of the first kind k for Stirling numbers of the second kind

The first kind of Stirling number counts the number of permutations of n distinct items that have k cycles; the second kind counts the ways in which n distinct items can be partitioned into k parts. If d is given, the “reduced Stirling number of the second kind” is returned: S^{d}(n, k) = S(n - d + 1, k - d + 1) with n >= k >= d. (This counts the ways to partition n consecutive integers into k groups with no pairwise difference less than d. See example below.)

To obtain the signed Stirling numbers of the first kind, use keyword signed=True. Using this keyword automatically sets kind to 1.

References

Examples

>>> from sympy.functions.combinatorial.numbers import stirling, bell
>>> from sympy.combinatorics import Permutation
>>> from sympy.utilities.iterables import multiset_partitions, permutations


First kind (unsigned by default):

>>> [stirling(6, i, kind=1) for i in range(7)]
[0, 120, 274, 225, 85, 15, 1]
>>> perms = list(permutations(range(4)))
>>> [sum(Permutation(p).cycles == i for p in perms) for i in range(5)]
[0, 6, 11, 6, 1]
>>> [stirling(4, i, kind=1) for i in range(5)]
[0, 6, 11, 6, 1]


First kind (signed):

>>> [stirling(4, i, signed=True) for i in range(5)]
[0, -6, 11, -6, 1]


Second kind:

>>> [stirling(10, i) for i in range(12)]
[0, 1, 511, 9330, 34105, 42525, 22827, 5880, 750, 45, 1, 0]
>>> sum(_) == bell(10)
True
>>> len(list(multiset_partitions(range(4), 2))) == stirling(4, 2)
True


Reduced second kind:

>>> from sympy import subsets, oo
>>> def delta(p):
...    if len(p) == 1:
...        return oo
...    return min(abs(i[0] - i[1]) for i in subsets(p, 2))
>>> parts = multiset_partitions(range(5), 3)
>>> d = 2
>>> sum(1 for p in parts if all(delta(i) >= d for i in p))
7
>>> stirling(5, 3, 2)
7


# Enumeration¶

Three functions are available. Each of them attempts to efficiently compute a given combinatorial quantity for a given set or multiset which can be entered as an integer, sequence or multiset (dictionary with elements as keys and multiplicities as values). The k parameter indicates the number of elements to pick (or the number of partitions to make). When k is None, the sum of the enumeration for all k (from 0 through the number of items represented by n) is returned. A replacement parameter is recognized for combinations and permutations; this indicates that any item may appear with multiplicity as high as the number of items in the original set.

>>> from sympy.functions.combinatorial.numbers import nC, nP, nT
>>> items = 'baby'


## nC¶

Calculate the number of combinations of length k.

>>> [nC(items, k) for k in range(len(items) + 1)], nC(items)
([1, 3, 4, 3, 1], 12)
>>> nC('aaa', 2)
1
>>> nC('abc', 2)
3
>>> nC(3, 2)
3


## nP¶

Calculate the number of permutations of length k.

>>> [nP(items, k) for k in range(len(items) + 1)], nP(items)
([1, 3, 7, 12, 12], 35)
>>> nC('aaa', 2)
1
>>> nC('abc', 2)
3
>>> nC(3, 2)
3


## nT¶

Calculate the number of partitions that have k parts.

>>> [nT(items, k) for k in range(len(items) + 1)], nT(items)
([0, 1, 5, 4, 1], 11)
>>> nT('aaa', 2)
1
>>> nT('abc', 2)
3
>>> nT(3, 2)
1
`

Note that the integer for n indicates identical items for nT but indicates n different items for nC and nP.