# Combinatorial Functions¶

This module implements some combinatorial functions, as listed below. For a more detailed description, see the relevant docstrings.

Sequences:

Set-theoretic constructions:

Warning

The following function is deprecated and will soon be removed.

• Permutations of a multiset, permutations(), permutations_iterator(), number_of_permutations(). A permutation is a list that contains exactly the same elements but possibly in different order.

Related functions:

Implemented in other modules (listed for completeness):

The sage.arith.all module contains the following combinatorial functions:

• binomial the binomial coefficient (wrapped from PARI)

• factorial (wrapped from PARI)

• partition (from the Python Cookbook) Generator of the list of all the partitions of the integer $$n$$.

• number_of_partitions() (wrapped from PARI) the number of partitions:

• falling_factorial() Definition: for integer $$a \ge 0$$ we have $$x(x-1) \cdots (x-a+1)$$. In all other cases we use the GAMMA-function: $$\frac {\Gamma(x+1)} {\Gamma(x-a+1)}$$.

• rising_factorial() Definition: for integer $$a \ge 0$$ we have $$x(x+1) \cdots (x+a-1)$$. In all other cases we use the GAMMA-function: $$\frac {\Gamma(x+a)} {\Gamma(x)}$$.

• gaussian_binomial the gaussian binomial

$\binom{n}{k}_q = \frac{(1-q^m)(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^2)\cdots (1-q^r)}.$

The sage.groups.perm_gps.permgroup_elements contains the following combinatorial functions:

• matrix method of PermutationGroupElement yielding the permutation matrix of the group element.

Todo

GUAVA commands:
• VandermondeMat

• GrayMat returns a list of all different vectors of length n over the field F, using Gray ordering.

Not in GAP:

REFERENCES:

AUTHORS:

• David Joyner (2006-07): initial implementation.

• William Stein (2006-07): editing of docs and code; many optimizations, refinements, and bug fixes in corner cases

• David Joyner (2006-09): bug fix for combinations, added permutations_iterator, combinations_iterator from Python Cookbook, edited docs.

• Florent Hivert (2009-02): combinatorial class cleanup

• Fredrik Johansson (2010-07): fast implementation of stirling_number2

• Punarbasu Purkayastha (2012-12): deprecate arrangements, combinations, combinations_iterator, and clean up very old deprecated methods.

## Functions and classes¶

class sage.combinat.combinat.CombinatorialClass(category=None)

This class is deprecated, and will disappear as soon as all derived classes in Sage’s library will have been fixed. Please derive directly from Parent and use the category EnumeratedSets, FiniteEnumeratedSets, or InfiniteEnumeratedSets, as appropriate.

For examples, see:

sage: FiniteEnumeratedSets().example()
An example of a finite enumerated set: {1,2,3}
sage: InfiniteEnumeratedSets().example()
An example of an infinite enumerated set: the non negative integers

Element

alias of CombinatorialObject

cardinality()

Default implementation of cardinality which just goes through the iterator of the combinatorial class to count the number of objects.

EXAMPLES:

sage: class C(CombinatorialClass):
....:     def __iter__(self):
....:         return iter([1,2,3])
sage: C().cardinality() #indirect doctest
3

element_class()

This function is a temporary helper so that a CombinatorialClass behaves as a parent for creating elements. This will disappear when combinatorial classes will be turned into actual parents (in the category EnumeratedSets).

filter(f, name=None)

Return the combinatorial subclass of f which consists of the elements x of self such that f(x) is True.

EXAMPLES:

sage: from sage.combinat.combinat import Permutations_CC
sage: P = Permutations_CC(3).filter(lambda x: x.avoids([1,2]))
sage: P.list()
[[3, 2, 1]]

first()

Default implementation for first which uses iterator.

EXAMPLES:

sage: C = CombinatorialClass()
sage: C.list = lambda: [1,2,3]
sage: C.first() # indirect doctest
1

is_finite()

Return whether self is finite or not.

EXAMPLES:

sage: Partitions(5).is_finite()
True
sage: Permutations().is_finite()
False

last()

Default implementation for first which uses iterator.

EXAMPLES:

sage: C = CombinatorialClass()
sage: C.list = lambda: [1,2,3]
sage: C.last() # indirect doctest
3

list()

The default implementation of list which builds the list from the iterator.

EXAMPLES:

sage: class C(CombinatorialClass):
....:     def __iter__(self):
....:         return iter([1,2,3])
sage: C().list() #indirect doctest
[1, 2, 3]

map(f, name, is_injective=None)

Return the image $$\{f(x) | x \in \text{self}\}$$ of this combinatorial class by $$f$$, as a combinatorial class.

INPUT:

• is_injective – boolean (default: True) whether to assume that f is injective.

EXAMPLES:

sage: R = Permutations(3).map(attrcall('reduced_word')); R
Image of Standard permutations of 3 by The map *.reduced_word() from Standard permutations of 3
sage: R.cardinality()
6
sage: R.list()
[[], , , [1, 2], [2, 1], [2, 1, 2]]
sage: [ r for r in R]
[[], , , [1, 2], [2, 1], [2, 1, 2]]


If the function is not injective, then there may be repeated elements:

sage: P = Partitions(4)
sage: P.list()
[, [3, 1], [2, 2], [2, 1, 1], [1, 1, 1, 1]]
sage: P.map(len).list()
[1, 2, 2, 3, 4]


Use is_injective=False to get a correct result in this case:

sage: P.map(len, is_injective=False).list()
[1, 2, 3, 4]

next(obj)

Default implementation for next which uses iterator.

EXAMPLES:

sage: C = CombinatorialClass()
sage: C.list = lambda: [1,2,3]
sage: C.next(2) # indirect doctest
3

previous(obj)

Default implementation for next which uses iterator.

EXAMPLES:

sage: C = CombinatorialClass()
sage: C.list = lambda: [1,2,3]
sage: C.previous(2) # indirect doctest
1

random_element()

Default implementation of random which uses unrank.

EXAMPLES:

sage: C = CombinatorialClass()
sage: C.list = lambda: [1,2,3]
sage: C.random_element()       # random  # indirect doctest
1

rank(obj)

Default implementation of rank which uses iterator.

EXAMPLES:

sage: C = CombinatorialClass()
sage: C.list = lambda: [1,2,3]
sage: C.rank(3) # indirect doctest
2

union(right_cc, name=None)

Return the combinatorial class representing the union of self and right_cc.

EXAMPLES:

sage: from sage.combinat.combinat import Permutations_CC
sage: P = Permutations_CC(2).union(Permutations_CC(1))
sage: P.list()
[[1, 2], [2, 1], ]

unrank(r)

Default implementation of unrank which goes through the iterator.

EXAMPLES:

sage: C = CombinatorialClass()
sage: C.list = lambda: [1,2,3]
sage: C.unrank(1) # indirect doctest
2

class sage.combinat.combinat.CombinatorialElement(parent, *args, **kwds)

CombinatorialElement is both a CombinatorialObject and an Element. So it represents a list which is an element of some parent.

A CombinatorialElement subclass also automatically supports the __classcall__ mechanism.

Warning

This class is slowly being deprecated. Use ClonableList instead.

INPUT:

• parent – the Parent class for this element.

• lst – a list or any object that can be converted to a list by calling list().

EXAMPLES:

sage: from sage.combinat.combinat import CombinatorialElement
sage: e = CombinatorialElement(Partitions(6), [3,2,1])
True
sage: parent(e)
Partitions of the integer 6
sage: list(e)
[3, 2, 1]


Check classcalls:

sage: class Foo(CombinatorialElement):
....:     @staticmethod
....:     def __classcall__(cls, x):
....:         return x
sage: Foo(17)
17

class sage.combinat.combinat.CombinatorialObject(l, copy=True)

CombinatorialObject provides a thin wrapper around a list. The main differences are that __setitem__ is disabled so that CombinatorialObjects are shallowly immutable, and the intention is that they are semantically immutable.

Because of this, CombinatorialObjects provide a __hash__ function which computes the hash of the string representation of a list and the hash of its parent’s class. Thus, each CombinatorialObject should have a unique string representation.

CombinatorialElement if you want a combinatorial object which is an element of a parent.

Warning

This class is slowly being deprecated. Use ClonableList instead.

INPUT:

• l – a list or any object that can be converted to a list by calling list().

• copy – (boolean, default True) if False, then l must be a list, which is assigned to self._list without copying.

EXAMPLES:

sage: c = CombinatorialObject([1,2,3])
True
sage: c._list
[1, 2, 3]
sage: c._hash is None
True


For efficiency, you can specify copy=False if you know what you are doing:

sage: from sage.combinat.combinat import CombinatorialObject
sage: x = [3, 2, 1]
sage: C = CombinatorialObject(x, copy=False)
sage: C
[3, 2, 1]
sage: x = 5
sage: C
[5, 2, 1]

index(key)

EXAMPLES:

sage: c = CombinatorialObject([1,2,3])
sage: c.index(1)
0
sage: c.index(3)
2

class sage.combinat.combinat.FilteredCombinatorialClass(combinatorial_class, f, name=None)

A filtered combinatorial class F is a subset of another combinatorial class C specified by a function f that takes in an element c of C and returns True if and only if c is in F.

cardinality()

EXAMPLES:

sage: from sage.combinat.combinat import Permutations_CC
sage: P = Permutations_CC(3).filter(lambda x: x.avoids([1,2]))
sage: P.cardinality()
1

class sage.combinat.combinat.InfiniteAbstractCombinatorialClass(category=None)

This is an internal class that should not be used directly. A class which inherits from InfiniteAbstractCombinatorialClass inherits the standard methods list and count.

If self._infinite_cclass_slice exists then self.__iter__ returns an iterator for self, otherwise raise NotImplementedError. The method self._infinite_cclass_slice is supposed to accept any integer as an argument and return something which is iterable.

cardinality()

Count the elements of the combinatorial class.

EXAMPLES:

sage: R = InfiniteAbstractCombinatorialClass()
doctest:warning...
DeprecationWarning: this class is deprecated, do not use
See https://trac.sagemath.org/31545 for details.

sage: R.cardinality()
+Infinity

list()

Return an error since self is an infinite combinatorial class.

EXAMPLES:

sage: R = InfiniteAbstractCombinatorialClass()
sage: R.list()
Traceback (most recent call last):
...
NotImplementedError: infinite list

class sage.combinat.combinat.MapCombinatorialClass(cc, f, name, is_injective=None)

Bases: sage.sets.image_set.ImageSubobject, sage.combinat.combinat.CombinatorialClass

The image of a combinatorial class through a function.

INPUT:

• is_injective – boolean (default: True) whether to assume that f is injective.

See CombinatorialClass.map() for examples

EXAMPLES:

sage: R = SymmetricGroup(10).map(attrcall('reduced_word'))
sage: R.an_element()
[9, 8, 7, 6, 5, 4, 3, 2]
sage: R.cardinality()
3628800
sage: i = iter(R)
sage: next(i), next(i), next(i)
([], [1, 2, 3, 4, 5, 6, 7, 8, 9], )

class sage.combinat.combinat.Permutations_CC(n)

A testing class for CombinatorialClass since Permutations no longer inherits from CombinatorialClass in trac ticket #14772.

class sage.combinat.combinat.UnionCombinatorialClass(left_cc, right_cc, name=None)

A UnionCombinatorialClass is a union of two other combinatorial classes.

cardinality()

EXAMPLES:

sage: from sage.combinat.combinat import Permutations_CC
sage: P = Permutations_CC(3).union(Permutations_CC(2))
sage: P.cardinality()
8

first()

EXAMPLES:

sage: from sage.combinat.combinat import Permutations_CC
sage: P = Permutations_CC(3).union(Permutations_CC(2))
sage: P.first()
[1, 2, 3]

last()

EXAMPLES:

sage: from sage.combinat.combinat import Permutations_CC
sage: P = Permutations_CC(3).union(Permutations_CC(2))
sage: P.last()
[2, 1]

list()

EXAMPLES:

sage: from sage.combinat.combinat import Permutations_CC
sage: P = Permutations_CC(3).union(Permutations_CC(2))
sage: P.list()
[[1, 2, 3],
[1, 3, 2],
[2, 1, 3],
[2, 3, 1],
[3, 1, 2],
[3, 2, 1],
[1, 2],
[2, 1]]

rank(x)

EXAMPLES:

sage: from sage.combinat.combinat import Permutations_CC
sage: P = Permutations_CC(3).union(Permutations_CC(2))
sage: P.rank(Permutation([2,1]))
7
sage: P.rank(Permutation([1,2,3]))
0

unrank(x)

EXAMPLES:

sage: from sage.combinat.combinat import Permutations_CC
sage: P = Permutations_CC(3).union(Permutations_CC(2))
sage: P.unrank(7)
[2, 1]
sage: P.unrank(0)
[1, 2, 3]

sage.combinat.combinat.bell_number(n, algorithm='flint', **options)

Return the $$n$$-th Bell number.

This is the number of ways to partition a set of $$n$$ elements into pairwise disjoint nonempty subsets.

INPUT:

• n – a positive integer

• algorithm – (Default: 'flint') any one of the following:

• 'dobinski' – Use Dobinski’s formula implemented in Sage

• 'flint' – Wrap FLINT’s arith_bell_number

• 'gap' – Wrap GAP’s Bell

• 'mpmath' – Wrap mpmath’s bell

Warning

When using the mpmath algorithm to compute Bell numbers and you specify prec, it can return incorrect results due to low precision. See the examples section.

Let $$B_n$$ denote the $$n$$-th Bell number. Dobinski’s formula is:

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

To show our implementation of Dobinski’s method works, suppose that $$n \geq 5$$ and let $$k_0$$ be the smallest positive integer such that $$\frac{k_0^n}{k_0!} < 1$$. Note that $$k_0 > n$$ and $$k_0 \leq 2n$$ because we can prove that $$\frac{(2n)^n}{(2n)!} < 1$$ by Stirling.

If $$k > k_0$$, then we have $$\frac{k^n}{k!} < \frac{1}{2^{k-k_0}}$$. We show this by induction: let $$c_k = \frac{k^n}{k!}$$, if $$k > n$$ then

$\frac{c_{k+1}}{c_k} = \frac{(1+k^{-1})^n}{k+1} < \frac{(1+n^{-1})^n}{n} < \frac{1}{2}.$

The last inequality can easily be checked numerically for $$n \geq 5$$.

Using this, we can see that $$\frac{c_k}{c_{k_0}} < \frac{1}{2^{k-k_0}}$$ for $$k > k_0 > n$$. So summing this it gives that $$\sum_{k=k_0+1}^{\infty} \frac{k^n}{k!} < 1$$, and hence

$B_n = e^{-1} \left( \sum_{k=0}^{k_0} \frac{k^n}{k!} + E_1 \right) = e^{-1} \sum_{k=0}^{k_0} \frac{k^n}{k!} + E_2,$

where $$0 < E_1 < 1$$ and $$0 < E_2 < e^{-1}$$. Next we have for any $$q > 0$$

$\sum_{k=0}^{k_0} \frac{k^n}{k!} = \frac{1}{q} \sum_{k=0}^{k_0} \left\lfloor \frac{q k^n}{k!} \right\rfloor + \frac{E_3}{q}$

where $$0 \leq E_3 \leq k_0 + 1 \leq 2n + 1$$. Let $$E_4 = \frac{E_3}{q}$$ and let $$q = 2n + 1$$. We find $$0 \leq E_4 \leq 1$$. These two bounds give:

\begin{split}\begin{aligned} B_n & = \frac{e^{-1}}{q} \sum_{k=0}^{k_0} \left\lfloor \frac{q k^n}{k!} \right\rfloor + e^{-1} E_4 + E_2 \\ & = \frac{e^{-1}}{q} \sum_{k=0}^{k_0} \left\lfloor \frac{q k^n}{k!} \right\rfloor + E_5 \end{aligned}\end{split}

where

$0 < E_5 = e^{-1} E_4 + E_2 \leq e^{-1} + e^{-1} < \frac{3}{4}.$

It follows that

$B_n = \left\lceil \frac{e^{-1}}{q} \sum_{k=0}^{k_0} \left\lfloor \frac{q k^n}{k!} \right\rfloor \right\rceil.$

Now define

$b = \sum_{k=0}^{k_0} \left\lfloor \frac{q k^n}{k!} \right\rfloor.$

This $$b$$ can be computed exactly using integer arithmetic. To avoid the costly integer division by $$k!$$, we collect more terms and do only one division, for example with 3 terms:

$\frac{k^n}{k!} + \frac{(k+1)^n}{(k+1)!} + \frac{(k+2)^n}{(k+2)!} = \frac{k^n (k+1)(k+2) + (k+1)^n (k+2) + (k+2)^n}{(k+2)!}$

In the implementation, we collect $$\sqrt{n}/2$$ terms.

To actually compute $$B_n$$ from $$b$$, we let $$p = \lfloor \log_2(b) \rfloor + 1$$ such that $$b < 2^p$$ and we compute with $$p$$ bits of precision. This implies that $$b$$ (and $$q < b$$) can be represented exactly.

We compute $$\frac{e^{-1}}{q} b$$, rounding down, and we must have an absolute error of at most $$1/4$$ (given that $$E_5 < 3/4$$). This means that we need a relative error of at most

$\frac{e q}{4 b} > \frac{(e q)/4}{2^p} > \frac{7}{2^p}$

(assuming $$n \geq 5$$). With a precision of $$p$$ bits and rounding down, every rounding has a relative error of at most $$2^{1-p} = 2/2^p$$. Since we do 3 roundings ($$b$$ and $$q$$ do not require rounding), we get a relative error of at most $$6/2^p$$. All this implies that the precision of $$p$$ bits is sufficient.

EXAMPLES:

sage: bell_number(10)
115975
sage: bell_number(2)
2
sage: bell_number(-10)
Traceback (most recent call last):
...
ArithmeticError: Bell numbers not defined for negative indices
sage: bell_number(1)
1
sage: bell_number(1/3)
Traceback (most recent call last):
...
TypeError: no conversion of this rational to integer


When using the mpmath algorithm, we are required have mpmath’s precision set to at least $$\log_2(B_n)$$ bits. If upon computing the Bell number the first time, we deem the precision too low, we use our guess to (temporarily) raise mpmath’s precision and the Bell number is recomputed.

sage: k = bell_number(30, 'mpmath'); k
846749014511809332450147
sage: k == bell_number(30)
True


If you knows what precision is necessary before computing the Bell number, you can use the prec option:

sage: k2 = bell_number(30, 'mpmath', prec=30); k2
846749014511809332450147
sage: k == k2
True


Warning

Running mpmath with the precision set too low can result in incorrect results:

sage: k = bell_number(30, 'mpmath', prec=15); k
846749014511809388871680
sage: k == bell_number(30)
False


AUTHORS:

• Robert Gerbicz

• Jeroen Demeyer: improved implementation of Dobinski formula with more accurate error estimates (trac ticket #17157)

REFERENCES:

sage.combinat.combinat.bell_polynomial(n, k)

Return the Bell Polynomial

$B_{n,k}(x_0, x_1, \ldots, x_{n-k}) = \sum_{\sum{j_i}=k, \sum{(i+1) j_i}=n} \frac{n!}{j_0!j_1!\cdots j_{n-k}!} \left(\frac{x_0}{(0+1)!}\right)^{j_0} \left(\frac{x_1}{(1+1)!}\right)^{j_1} \cdots \left(\frac{x_{n-k}}{(n-k+1)!}\right)^{j_{n-k}}.$

INPUT:

• n – integer

• k – integer

OUTPUT:

• a polynomial in $$n-k+1$$ variables over $$\ZZ$$

EXAMPLES:

sage: bell_polynomial(6,2)
10*x2^2 + 15*x1*x3 + 6*x0*x4
sage: bell_polynomial(6,3)
15*x1^3 + 60*x0*x1*x2 + 15*x0^2*x3


REFERENCES:

AUTHORS:

• Blair Sutton (2009-01-26)

• Thierry Monteil (2015-09-29): the result must always be a polynomial.

sage.combinat.combinat.bernoulli_polynomial(x, n)

Return the n-th Bernoulli polynomial evaluated at x.

The generating function for the Bernoulli polynomials is

$\frac{t e^{xt}}{e^t-1}= \sum_{n=0}^\infty B_n(x) \frac{t^n}{n!},$

and they are given directly by

$B_n(x) = \sum_{i=0}^n \binom{n}{i}B_{n-i}x^i.$

One has $$B_n(x) = - n\zeta(1 - n,x)$$, where $$\zeta(s,x)$$ is the Hurwitz zeta function. Thus, in a certain sense, the Hurwitz zeta function generalizes the Bernoulli polynomials to non-integer values of n.

EXAMPLES:

sage: y = QQ['y'].0
sage: bernoulli_polynomial(y, 5)
y^5 - 5/2*y^4 + 5/3*y^3 - 1/6*y
sage: bernoulli_polynomial(y, 5)(12)
199870
sage: bernoulli_polynomial(12, 5)
199870
sage: bernoulli_polynomial(y^2 + 1, 5)
y^10 + 5/2*y^8 + 5/3*y^6 - 1/6*y^2
sage: P.<t> = ZZ[]
sage: p = bernoulli_polynomial(t, 6)
sage: p.parent()
Univariate Polynomial Ring in t over Rational Field


We verify an instance of the formula which is the origin of the Bernoulli polynomials (and numbers):

sage: power_sum = sum(k^4 for k in range(10))
sage: 5*power_sum == bernoulli_polynomial(10, 5) - bernoulli(5)
True


REFERENCES:

sage.combinat.combinat.catalan_number(n)

Return the $$n$$-th Catalan number.

The $$n$$-th Catalan number is given directly in terms of binomial coefficients by

$C_n = \frac{1}{n+1}\binom{2n}{n} = \frac{(2n)!}{(n+1)!\,n!} \qquad\mbox{ for }\quad n\ge 0.$

Consider the set $$S = \{ 1, ..., n \}$$. A noncrossing partition of $$S$$ is a partition in which no two blocks “cross” each other, i.e., if $$a$$ and $$b$$ belong to one block and $$x$$ and $$y$$ to another, they are not arranged in the order $$axby$$. $$C_n$$ is the number of noncrossing partitions of the set $$S$$. There are many other interpretations (see REFERENCES).

When $$n=-1$$, this function returns the limit value $$-1/2$$. For other $$n<0$$ it returns $$0$$.

INPUT:

• n – integer

OUTPUT:

integer

EXAMPLES:

sage: [catalan_number(i) for i in range(7)]
[1, 1, 2, 5, 14, 42, 132]
sage: x = (QQ[['x']].0).O(8)
sage: (-1/2)*sqrt(1 - 4*x)
-1/2 + x + x^2 + 2*x^3 + 5*x^4 + 14*x^5 + 42*x^6 + 132*x^7 + O(x^8)
sage: [catalan_number(i) for i in range(-7,7)]
[0, 0, 0, 0, 0, 0, -1/2, 1, 1, 2, 5, 14, 42, 132]
sage: [catalan_number(n).mod(2) for n in range(16)]
[1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1]


REFERENCES:

sage.combinat.combinat.euler_number(n, algorithm='flint')

Return the $$n$$-th Euler number.

INPUT:

• n – a positive integer

• algorithm – (Default: 'flint') any one of the following:

• 'maxima' – Wraps Maxima’s euler.

• 'flint' – Wrap FLINT’s arith_euler_number

EXAMPLES:

sage: [euler_number(i) for i in range(10)]
[1, 0, -1, 0, 5, 0, -61, 0, 1385, 0]
sage: x = PowerSeriesRing(QQ, 'x').gen().O(10)
sage: 2/(exp(x)+exp(-x))
1 - 1/2*x^2 + 5/24*x^4 - 61/720*x^6 + 277/8064*x^8 + O(x^10)
sage: [euler_number(i)/factorial(i) for i in range(11)]
[1, 0, -1/2, 0, 5/24, 0, -61/720, 0, 277/8064, 0, -50521/3628800]
sage: euler_number(-1)
Traceback (most recent call last):
...
ValueError: n (=-1) must be a nonnegative integer


REFERENCES:

sage.combinat.combinat.eulerian_number(n, k, algorithm='recursive')

Return the Eulerian number of index (n, k).

This is the coefficient of $$t^k$$ in the Eulerian polynomial $$A_n(t)$$.

INPUT:

• n – integer

• k – integer between 0 and n - 1

• algorithm"recursive" (default) or "formula"

OUTPUT:

an integer

EXAMPLES:

sage: from sage.combinat.combinat import eulerian_number
sage: [eulerian_number(5,i) for i in range(5)]
[1, 26, 66, 26, 1]

sage.combinat.combinat.eulerian_polynomial(n, algorithm='derivative')

Return the Eulerian polynomial of index n.

This is the generating polynomial counting permutations in the symmetric group $$S_n$$ according to their number of descents.

INPUT:

• n – an integer

• algorithm"derivative" (default) or "coeffs"

OUTPUT:

polynomial in one variable t

EXAMPLES:

sage: from sage.combinat.combinat import eulerian_polynomial
sage: eulerian_polynomial(5)
t^4 + 26*t^3 + 66*t^2 + 26*t + 1


REFERENCES:

sage.combinat.combinat.fibonacci(n, algorithm='pari')

Return the $$n$$-th Fibonacci number.

The Fibonacci sequence $$F_n$$ is defined by the initial conditions $$F_1 = F_2 = 1$$ and the recurrence relation $$F_{n+2} = F_{n+1} + F_n$$. For negative $$n$$ we define $$F_n = (-1)^{n+1}F_{-n}$$, which is consistent with the recurrence relation.

INPUT:

• algorithm – a string:

• "pari" - (default) use the PARI C library’s pari:fibo function

• "gap" - use GAP’s Fibonacci function

Note

PARI is tens to hundreds of times faster than GAP here. Moreover, PARI works for every large input whereas GAP does not.

EXAMPLES:

sage: fibonacci(10)
55
sage: fibonacci(10, algorithm='gap')
55

sage: fibonacci(-100)
-354224848179261915075
sage: fibonacci(100)
354224848179261915075

sage: fibonacci(0)
0
sage: fibonacci(1/2)
Traceback (most recent call last):
...
TypeError: no conversion of this rational to integer

sage.combinat.combinat.fibonacci_sequence(start, stop=None, algorithm=None)

Return an iterator over the Fibonacci sequence, for all fibonacci numbers $$f_n$$ from n = start up to (but not including) n = stop

INPUT:

• start – starting value

• stop – stopping value

• algorithm – (default: None) passed on to fibonacci function (or not passed on if None, i.e., use the default)

EXAMPLES:

sage: fibs = [i for i in fibonacci_sequence(10, 20)]
sage: fibs
[55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181]

sage: sum([i for i in fibonacci_sequence(100, 110)])
69919376923075308730013


AUTHORS:

• Bobby Moretti

sage.combinat.combinat.fibonacci_xrange(start, stop=None, algorithm='pari')

Return an iterator over all of the Fibonacci numbers in the given range, including f_n = start up to, but not including, f_n = stop.

EXAMPLES:

sage: fibs_in_some_range =  [i for i in fibonacci_xrange(10^7, 10^8)]
sage: len(fibs_in_some_range)
4
sage: fibs_in_some_range
[14930352, 24157817, 39088169, 63245986]

sage: fibs = [i for i in fibonacci_xrange(10, 100)]
sage: fibs
[13, 21, 34, 55, 89]

sage: list(fibonacci_xrange(13, 34))
[13, 21]


A solution to the second Project Euler problem:

sage: sum([i for i in fibonacci_xrange(10^6) if is_even(i)])
1089154


AUTHORS:

• Bobby Moretti

sage.combinat.combinat.lucas_number1(n, P, Q)

Return the $$n$$-th Lucas number “of the first kind” (this is not standard terminology). The Lucas sequence $$L^{(1)}_n$$ is defined by the initial conditions $$L^{(1)}_1 = 0$$, $$L^{(1)}_2 = 1$$ and the recurrence relation $$L^{(1)}_{n+2} = P \cdot L^{(1)}_{n+1} - Q \cdot L^{(1)}_n$$.

Wraps GAP’s Lucas(...).

$$P=1$$, $$Q=-1$$ gives the Fibonacci sequence.

INPUT:

• n – integer

• P, Q – integer or rational numbers

OUTPUT: integer or rational number

EXAMPLES:

sage: lucas_number1(5,1,-1)
5
sage: lucas_number1(6,1,-1)
8
sage: lucas_number1(7,1,-1)
13
sage: lucas_number1(7,1,-2)
43
sage: lucas_number1(5,2,3/5)
229/25
sage: lucas_number1(5,2,1.5)
1/4


There was a conjecture that the sequence $$L_n$$ defined by $$L_{n+2} = L_{n+1} + L_n$$, $$L_1=1$$, $$L_2=3$$, has the property that $$n$$ prime implies that $$L_n$$ is prime.

sage: lucas = lambda n : Integer((5/2)*lucas_number1(n,1,-1)+(1/2)*lucas_number2(n,1,-1))
sage: [[lucas(n),is_prime(lucas(n)),n+1,is_prime(n+1)] for n in range(15)]
[[1, False, 1, False],
[3, True, 2, True],
[4, False, 3, True],
[7, True, 4, False],
[11, True, 5, True],
[18, False, 6, False],
[29, True, 7, True],
[47, True, 8, False],
[76, False, 9, False],
[123, False, 10, False],
[199, True, 11, True],
[322, False, 12, False],
[521, True, 13, True],
[843, False, 14, False],
[1364, False, 15, False]]


Can you use Sage to find a counterexample to the conjecture?

sage.combinat.combinat.lucas_number2(n, P, Q)

Return the $$n$$-th Lucas number “of the second kind” (this is not standard terminology). The Lucas sequence $$L^{(2)}_n$$ is defined by the initial conditions $$L^{(2)}_1 = 2$$, $$L^{(2)}_2 = P$$ and the recurrence relation $$L^{(2)}_{n+2} = P \cdot L^{(2)}_{n+1} - Q \cdot L^{(2)}_n$$.

Wraps GAP’s Lucas(…).

INPUT:

• n - integer

• P, Q - integer or rational numbers

OUTPUT: integer or rational number

EXAMPLES:

sage: [lucas_number2(i,1,-1) for i in range(10)]
[2, 1, 3, 4, 7, 11, 18, 29, 47, 76]
sage: [fibonacci(i-1)+fibonacci(i+1) for i in range(10)]
[2, 1, 3, 4, 7, 11, 18, 29, 47, 76]

sage: n = lucas_number2(5,2,3); n
2
sage: type(n)
<class 'sage.rings.integer.Integer'>
sage: n = lucas_number2(5,2,-3/9); n
418/9
sage: type(n)
<class 'sage.rings.rational.Rational'>


The case $$P=1$$, $$Q=-1$$ is the Lucas sequence in Brualdi’s Introductory Combinatorics, 4th ed., Prentice-Hall, 2004:

sage: [lucas_number2(n,1,-1) for n in range(10)]
[2, 1, 3, 4, 7, 11, 18, 29, 47, 76]

sage.combinat.combinat.narayana_number(n, k)

Return the Narayana number of index (n, k).

For every integer $$n \geq 1$$, the sum of Narayana numbers $$\sum_k N_{n,k}$$ is the Catalan number $$C_n$$.

INPUT:

• n – an integer

• k – an integer between 0 and n - 1

OUTPUT:

an integer

EXAMPLES:

sage: from sage.combinat.combinat import narayana_number
sage: [narayana_number(3, i) for i in range(3)]
[1, 3, 1]
sage: sum(narayana_number(7,i) for i in range(7)) == catalan_number(7)
True


REFERENCES:

sage.combinat.combinat.number_of_tuples(S, k, algorithm='naive')

Return the size of tuples(S, k) when $$S$$ is a set. More generally, return the size of tuples(set(S), k). (So, unlike tuples(), this method removes redundant entries from $$S$$.)

INPUT:

• S – the base set

• k – the length of the tuples

• algorithm – can be one of the following:

• 'naive' - (default) use the naive counting $$|S|^k$$

• 'gap' - wraps GAP’s NrTuples

Warning

When using algorithm='gap', S must be a list of objects that have string representations that can be interpreted by the GAP interpreter. If S consists of at all complicated Sage objects, this function might not do what you expect.

EXAMPLES:

sage: S = [1,2,3,4,5]
sage: number_of_tuples(S,2)
25
sage: number_of_tuples(S,2, algorithm="gap")
25
sage: S = [1,1,2,3,4,5]
sage: number_of_tuples(S,2)
25
sage: number_of_tuples(S,2, algorithm="gap")
25
sage: number_of_tuples(S,0)
1
sage: number_of_tuples(S,0, algorithm="gap")
1

sage.combinat.combinat.number_of_unordered_tuples(S, k, algorithm='naive')

Return the size of unordered_tuples(S, k) when $$S$$ is a set.

INPUT:

• S – the base set

• k – the length of the tuples

• algorithm – can be one of the following:

• 'naive' - (default) use the naive counting $$\binom{|S|+k-1}{k}$$

• 'gap' - wraps GAP’s NrUnorderedTuples

Warning

When using algorithm='gap', S must be a list of objects that have string representations that can be interpreted by the GAP interpreter. If S consists of at all complicated Sage objects, this function might not do what you expect.

EXAMPLES:

sage: S = [1,2,3,4,5]
sage: number_of_unordered_tuples(S,2)
15
sage: number_of_unordered_tuples(S,2, algorithm="gap")
15
sage: S = [1,1,2,3,4,5]
sage: number_of_unordered_tuples(S,2)
15
sage: number_of_unordered_tuples(S,2, algorithm="gap")
15
sage: number_of_unordered_tuples(S,0)
1
sage: number_of_unordered_tuples(S,0, algorithm="gap")
1

sage.combinat.combinat.polygonal_number(s, n)

Return the $$n$$-th $$s$$-gonal number.

Polygonal sequences are represented by dots forming a regular polygon. Two famous sequences are the triangular numbers (3rd column of Pascal’s Triangle) and the square numbers. The $$n$$-th term in a polygonal sequence is defined by

$P(s, n) = \frac{n^2(s-2) - n(s-4)}{2},$

where $$s$$ is the number of sides of the polygon.

INPUT:

• s – integer greater than 1; the number of sides of the polygon

• n – integer; the index of the returned $$s$$-gonal number

OUTPUT: an integer

EXAMPLES:

The triangular numbers:

sage: [polygonal_number(3, n) for n in range(10)]
[0, 1, 3, 6, 10, 15, 21, 28, 36, 45]

sage: [polygonal_number(3, n) for n in range(-10, 0)]
[45, 36, 28, 21, 15, 10, 6, 3, 1, 0]


The square numbers:

sage: [polygonal_number(4, n) for n in range(10)]
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]


The pentagonal numbers:

sage: [polygonal_number(5, n) for n in range(10)]
[0, 1, 5, 12, 22, 35, 51, 70, 92, 117]


The hexagonal numbers:

sage: [polygonal_number(6, n) for n in range(10)]
[0, 1, 6, 15, 28, 45, 66, 91, 120, 153]


The input is converted into an integer:

sage: polygonal_number(3.0, 2.0)
3


A non-integer input returns an error:

sage: polygonal_number(3.5, 1)
Traceback (most recent call last):
...
TypeError: Attempt to coerce non-integral RealNumber to Integer


$$s$$ must be greater than 1:

sage: polygonal_number(1, 4)
Traceback (most recent call last):
...
ValueError: s (=1) must be greater than 1


REFERENCES:

sage.combinat.combinat.stirling_number1(n, k)

Return the $$n$$-th Stirling number $$S_1(n,k)$$ of the first kind.

This is the number of permutations of $$n$$ points with $$k$$ cycles.

This wraps GAP’s Stirling1.

EXAMPLES:

sage: stirling_number1(3,2)
3
sage: stirling_number1(5,2)
50
sage: 9*stirling_number1(9,5)+stirling_number1(9,4)
269325
sage: stirling_number1(10,5)
269325


Indeed, $$S_1(n,k) = S_1(n-1,k-1) + (n-1)S_1(n-1,k)$$.

sage.combinat.combinat.stirling_number2(n, k, algorithm=None)

Return the $$n$$-th Stirling number $$S_2(n,k)$$ of the second kind.

This is the number of ways to partition a set of $$n$$ elements into $$k$$ pairwise disjoint nonempty subsets. The $$n$$-th Bell number is the sum of the $$S_2(n,k)$$’s, $$k=0,...,n$$.

INPUT:

• n - nonnegative machine-size integer

• k - nonnegative machine-size integer

• algorithm:

• None (default) - use native implementation

• "maxima" - use Maxima’s stirling2 function

• "gap" - use GAP’s Stirling2 function

EXAMPLES:

Print a table of the first several Stirling numbers of the second kind:

sage: for n in range(10):
....:     for k in range(10):
....:         print(str(stirling_number2(n,k)).rjust(k and 6))
1      0      0      0      0      0      0      0      0      0
0      1      0      0      0      0      0      0      0      0
0      1      1      0      0      0      0      0      0      0
0      1      3      1      0      0      0      0      0      0
0      1      7      6      1      0      0      0      0      0
0      1     15     25     10      1      0      0      0      0
0      1     31     90     65     15      1      0      0      0
0      1     63    301    350    140     21      1      0      0
0      1    127    966   1701   1050    266     28      1      0
0      1    255   3025   7770   6951   2646    462     36      1


Stirling numbers satisfy $$S_2(n,k) = S_2(n-1,k-1) + kS_2(n-1,k)$$:

sage: 5*stirling_number2(9,5) + stirling_number2(9,4)
42525
sage: stirling_number2(10,5)
42525

sage.combinat.combinat.tuples(S, k, algorithm='itertools')

Return a list of all $$k$$-tuples of elements of a given set S.

This function accepts the set S in the form of any iterable (list, tuple or iterator), and returns a list of $$k$$-tuples. If S contains duplicate entries, then you should expect the method to return tuples multiple times!

Recall that $$k$$-tuples are ordered (in the sense that two $$k$$-tuples differing in the order of their entries count as different) and can have repeated entries (even if S is a list with no repetition).

INPUT:

• S – the base set

• k – the length of the tuples

• algorithm – can be one of the following:

• 'itertools' - (default) use python’s itertools

• 'native' - use a native Sage implementation

Note

The ordering of the list of tuples depends on the algorithm.

EXAMPLES:

sage: S = [1,2]
sage: tuples(S,3)
[(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2),
(2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)]
sage: mset = ["s","t","e","i","n"]
sage: tuples(mset, 2)
[('s', 's'), ('s', 't'), ('s', 'e'), ('s', 'i'), ('s', 'n'),
('t', 's'), ('t', 't'), ('t', 'e'), ('t', 'i'), ('t', 'n'),
('e', 's'), ('e', 't'), ('e', 'e'), ('e', 'i'), ('e', 'n'),
('i', 's'), ('i', 't'), ('i', 'e'), ('i', 'i'), ('i', 'n'),
('n', 's'), ('n', 't'), ('n', 'e'), ('n', 'i'), ('n', 'n')]

sage: K.<a> = GF(4, 'a')
sage: mset = [x for x in K if x != 0]
sage: tuples(mset, 2)
[(a, a), (a, a + 1), (a, 1), (a + 1, a), (a + 1, a + 1),
(a + 1, 1), (1, a), (1, a + 1), (1, 1)]


We check that the implementations agree (up to ordering):

sage: tuples(S, 3, 'native')
[(1, 1, 1), (2, 1, 1), (1, 2, 1), (2, 2, 1),
(1, 1, 2), (2, 1, 2), (1, 2, 2), (2, 2, 2)]


Lastly we check on a multiset:

sage: S = [1,1,2]
sage: sorted(tuples(S, 3)) == sorted(tuples(S, 3, 'native'))
True


AUTHORS:

• Jon Hanke (2006-08)

sage.combinat.combinat.unordered_tuples(S, k, algorithm='itertools')

Return a list of all unordered tuples of length k of the set S.

An unordered tuple of length $$k$$ of set $$S$$ is a unordered selection with repetitions of $$S$$ and is represented by a sorted list of length $$k$$ containing elements from $$S$$.

Unlike tuples(), the result of this method does not depend on how often an element appears in $$S$$; only the set $$S$$ is being used. For example, unordered_tuples([1, 1, 1], 2) will return [(1, 1)]. If you want it to return [(1, 1), (1, 1), (1, 1)], use Python’s itertools.combinations_with_replacement instead.

INPUT:

• S – the base set

• k – the length of the tuples

• algorithm – can be one of the following:

• 'itertools' - (default) use python’s itertools

• 'gap' - wraps GAP’s UnorderedTuples

Warning

When using algorithm='gap', S must be a list of objects that have string representations that can be interpreted by the GAP interpreter. If S consists of at all complicated Sage objects, this function might not do what you expect.

EXAMPLES:

sage: S = [1,2]
sage: unordered_tuples(S, 3)
[(1, 1, 1), (1, 1, 2), (1, 2, 2), (2, 2, 2)]


We check that this agrees with GAP:

sage: unordered_tuples(S, 3, algorithm='gap')
[(1, 1, 1), (1, 1, 2), (1, 2, 2), (2, 2, 2)]


We check the result on strings:

sage: S = ["a","b","c"]
sage: unordered_tuples(S, 2)
[('a', 'a'), ('a', 'b'), ('a', 'c'), ('b', 'b'), ('b', 'c'), ('c', 'c')]
sage: unordered_tuples(S, 2, algorithm='gap')
[('a', 'a'), ('a', 'b'), ('a', 'c'), ('b', 'b'), ('b', 'c'), ('c', 'c')]


Lastly we check on a multiset:

sage: S = [1,1,2]
sage: unordered_tuples(S, 3) == unordered_tuples(S, 3, 'gap')
True
sage: unordered_tuples(S, 3)
[(1, 1, 1), (1, 1, 2), (1, 2, 2), (2, 2, 2)]

sage.combinat.combinat.unshuffle_iterator(a, one=1)

Iterate over the unshuffles of a list (or tuple) a, also yielding the signs of the respective permutations.

If $$n$$ and $$k$$ are integers satisfying $$0 \leq k \leq n$$, then a $$(k, n-k)$$-unshuffle means a permutation $$\pi \in S_n$$ such that $$\pi(1) < \pi(2) < \cdots < \pi(k)$$ and $$\pi(k+1) < \pi(k+2) < \cdots < \pi(n)$$. This method provides, for a list $$a = (a_1, a_2, \ldots, a_n)$$ of length $$n$$, an iterator yielding all pairs:

$\Bigl( \bigl( (a_{\pi(1)}, a_{\pi(2)}, \ldots, a_{\pi(k)}), (a_{\pi(k+1)}, a_{\pi(k+2)}, \ldots, a_{\pi(n)}) \bigl), (-1)^{\pi} \Bigr)$

for all $$k \in \{0, 1, \ldots, n\}$$ and all $$(k, n-k)$$-unshuffles $$\pi$$. The optional variable one can be set to a different value which results in the $$(-1)^{\pi}$$ component being multiplied by said value.

The iterator does not yield these in order of increasing $$k$$.

EXAMPLES:

sage: from sage.combinat.combinat import unshuffle_iterator
sage: list(unshuffle_iterator([1, 3, 4]))
[(((), (1, 3, 4)), 1), (((1,), (3, 4)), 1), (((3,), (1, 4)), -1),
(((1, 3), (4,)), 1), (((4,), (1, 3)), 1), (((1, 4), (3,)), -1),
(((3, 4), (1,)), 1), (((1, 3, 4), ()), 1)]
sage: list(unshuffle_iterator([3, 1]))
[(((), (3, 1)), 1), (((3,), (1,)), 1), (((1,), (3,)), -1),
(((3, 1), ()), 1)]
sage: list(unshuffle_iterator())
[(((), (8,)), 1), (((8,), ()), 1)]
sage: list(unshuffle_iterator([]))
[(((), ()), 1)]
sage: list(unshuffle_iterator([3, 1], 3/2))
[(((), (3, 1)), 3/2), (((3,), (1,)), 3/2), (((1,), (3,)), -3/2),
(((3, 1), ()), 3/2)]