# 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 package sage.arith contains the following combinatorial functions:

• binomial() the binomial coefficient (wrapped from PARI)

• factorial() (wrapped from PARI)

• 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: $$\frac {\Gamma(x+a)} {\Gamma(x)}$$.

From other modules:

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

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.CombinatorialElement(parent, *args, **kwds)[source]#

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: # needs sage.combinat
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]

>>> from sage.all import *
>>> # needs sage.combinat
>>> from sage.combinat.combinat import CombinatorialElement
>>> e = CombinatorialElement(Partitions(Integer(6)), [Integer(3),Integer(2),Integer(1)])
True
>>> parent(e)
Partitions of the integer 6
>>> list(e)
[3, 2, 1]


Check classcalls:

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

>>> from sage.all import *
>>> class Foo(CombinatorialElement):                                          # needs sage.combinat
...     @staticmethod
...     def __classcall__(cls, x):
...         return x
>>> Foo(Integer(17))                                                                   # needs sage.combinat
17

class sage.combinat.combinat.CombinatorialObject(l, copy=True)[source]#

Bases: SageObject

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

>>> from sage.all import *
>>> c = CombinatorialObject([Integer(1),Integer(2),Integer(3)])
True
>>> c._list
[1, 2, 3]
>>> 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[0] = 5
sage: C
[5, 2, 1]

>>> from sage.all import *
>>> from sage.combinat.combinat import CombinatorialObject
>>> x = [Integer(3), Integer(2), Integer(1)]
>>> C = CombinatorialObject(x, copy=False)
>>> C
[3, 2, 1]
>>> x[Integer(0)] = Integer(5)
>>> C
[5, 2, 1]

index(key)[source]#

EXAMPLES:

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

>>> from sage.all import *
>>> c = CombinatorialObject([Integer(1),Integer(2),Integer(3)])
>>> c.index(Integer(1))
0
>>> c.index(Integer(3))
2

sage.combinat.combinat.bell_number(n, algorithm='flint', **options)[source]#

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: # needs sage.libs.flint
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

>>> from sage.all import *
>>> # needs sage.libs.flint
>>> bell_number(Integer(10))
115975
>>> bell_number(Integer(2))
2
>>> bell_number(-Integer(10))
Traceback (most recent call last):
...
ArithmeticError: Bell numbers not defined for negative indices
>>> bell_number(Integer(1))
1
>>> bell_number(Integer(1)/Integer(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                                          # needs mpmath
846749014511809332450147
sage: k == bell_number(30)                                                      # needs mpmath sage.libs.flint
True

>>> from sage.all import *
>>> k = bell_number(Integer(30), 'mpmath'); k                                          # needs mpmath
846749014511809332450147
>>> k == bell_number(Integer(30))                                                      # needs mpmath sage.libs.flint
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                               # needs mpmath
846749014511809332450147
sage: k == k2                                                                   # needs mpmath
True

>>> from sage.all import *
>>> k2 = bell_number(Integer(30), 'mpmath', prec=Integer(30)); k2                               # needs mpmath
846749014511809332450147
>>> k == k2                                                                   # needs mpmath
True


Warning

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

sage: k = bell_number(30, 'mpmath', prec=15); k                         # needs mpmath
846749014511809388871680
sage: k == bell_number(30)                                              # needs mpmath sage.libs.flint
False

>>> from sage.all import *
>>> k = bell_number(Integer(30), 'mpmath', prec=Integer(15)); k                         # needs mpmath
846749014511809388871680
>>> k == bell_number(Integer(30))                                              # needs mpmath sage.libs.flint
False


AUTHORS:

• Robert Gerbicz

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

REFERENCES:

sage.combinat.combinat.bell_polynomial(n, k=None, ordinary=False)[source]#

Return the (partial) (exponential/ordinary) bell Polynomial.

The partial (exponential) Bell polynomial is defined by the formula

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

The complete (exponential) Bell Polynomial is defined as

$B_n(x_0, x_1, \ldots, x_{n-k}) = \sum_{k=0}^n B_{n,k}(x_0, x_1, \ldots, x_{n-k}).$

The ordinary variant of the partial Bell polynomial is defined by

$\begin{split}\hat B_{n,k}(x_0, x_1, \ldots, x_{n-k}) = \sum_{\substack{j_0 + \ldots + j_{n-k} = k \\ 1 j_0 + \ldots + (n-k+1) j_{n-k} = n}} \binom{k}{j_0, j_1, \ldots, j_{n-k}} x_0^{j_0} x_1^{j_1} \cdots x_{n-k}^{j_{n-k}},\end{split}$

where we have used the multinomial coefficient. The complete version has the same definition as its exponential counterpart.

If we define $$f(z) = \sum_{n=1}^\infty x_{n-1} z^n/n!$$ then these are alternative definitions for exponential Bell polynomials

\begin{split}\begin{aligned} \exp(f(z)) & = \sum_{n=0}^\infty B_n(x_0, \ldots, x_{n-1}) \frac{z^n}{n!}, \\ \frac{f(z)^k}{k!} & = \sum_{n=k}^\infty B_{n, k}(x_0, \ldots, x_{n-k}) \frac{z^n}{n!}. \end{aligned}\end{split}

Defining $$g(z) = \sum_{n=1}^\infty x_{n-1} z^n$$, we have the analoguous alternative definitions

\begin{split}\begin{aligned} \frac1{1-f(z)} & = \sum_{n=0}^\infty \hat B_n(x_0, \ldots, x_{n-1}) z^n, \\ f(z)^k & = \sum_{n=k}^\infty \hat B_{n, k}(x_0, \ldots, x_{n-k}) z^n, \end{aligned}\end{split}

(see reference).

INPUT:

• k – (optional) if specified, returns the partial Bell polynomial, otherwise returns the complete Bell polynomial

• ordinary – (default: False) if True, returns the (partial) ordinary Bell polynomial, otherwise returns the (partial) exponential Bell polynomial

EXAMPLES:

The complete and partial Bell polynomials:

sage: # needs sage.combinat
sage: bell_polynomial(3)
x0^3 + 3*x0*x1 + x2
sage: bell_polynomial(4)
x0^4 + 6*x0^2*x1 + 3*x1^2 + 4*x0*x2 + x3
sage: bell_polynomial(6, 3)
15*x1^3 + 60*x0*x1*x2 + 15*x0^2*x3
sage: bell_polynomial(6, 6)
x0^6

>>> from sage.all import *
>>> # needs sage.combinat
>>> bell_polynomial(Integer(3))
x0^3 + 3*x0*x1 + x2
>>> bell_polynomial(Integer(4))
x0^4 + 6*x0^2*x1 + 3*x1^2 + 4*x0*x2 + x3
>>> bell_polynomial(Integer(6), Integer(3))
15*x1^3 + 60*x0*x1*x2 + 15*x0^2*x3
>>> bell_polynomial(Integer(6), Integer(6))
x0^6


The ordinary variants are:

sage: # needs sage.combinat sage.arith
sage: bell_polynomial(3, ordinary=True)
x0^3 + 2*x0*x1 + x2
sage: bell_polynomial(4, ordinary=True)
x0^4 + 3*x0^2*x1 + x1^2 + 2*x0*x2 + x3
sage: bell_polynomial(6, 3, True)
x1^3 + 6*x0*x1*x2 + 3*x0^2*x3
sage: bell_polynomial(6, 6, True)
x0^6

>>> from sage.all import *
>>> # needs sage.combinat sage.arith
>>> bell_polynomial(Integer(3), ordinary=True)
x0^3 + 2*x0*x1 + x2
>>> bell_polynomial(Integer(4), ordinary=True)
x0^4 + 3*x0^2*x1 + x1^2 + 2*x0*x2 + x3
>>> bell_polynomial(Integer(6), Integer(3), True)
x1^3 + 6*x0*x1*x2 + 3*x0^2*x3
>>> bell_polynomial(Integer(6), Integer(6), True)
x0^6


We verify the alternative definition of the different Bell polynomials using the functions $$f$$ and $$g$$ given above:

sage: # needs sage.combinat sage.arith
sage: n = 6 # positive integer
sage: k = 4 # positive integer
sage: R.<x> = InfinitePolynomialRing(QQ)
sage: PR = PolynomialRing(QQ, 'x', n)
sage: d = {x[i]: PR.gen(i) for i in range(n)} #substitution dictionnary
sage: L.<z> = LazyPowerSeriesRing(R)
sage: f = L(lambda i: x[i-1]/factorial(i), valuation=1)
sage: all(exp(f)[i].subs(d) * factorial(i) == bell_polynomial(i) for i in range(n+1))
True
sage: all((f^k/factorial(k))[i].subs(d) * factorial(i) == bell_polynomial(i, k) for i in range(k, n+k))
True
sage: g = L(lambda i: x[i-1], valuation=1)
sage: all((1/(1-g))[i].subs(d) == bell_polynomial(i, ordinary=True) for i in range(n+1))
True
sage: all((g^k)[i].subs(d) == bell_polynomial(i, k, True) for i in range(k, n+k))
True

>>> from sage.all import *
>>> # needs sage.combinat sage.arith
>>> n = Integer(6) # positive integer
>>> k = Integer(4) # positive integer
>>> R = InfinitePolynomialRing(QQ, names=('x',)); (x,) = R._first_ngens(1)
>>> PR = PolynomialRing(QQ, 'x', n)
>>> d = {x[i]: PR.gen(i) for i in range(n)} #substitution dictionnary
>>> L = LazyPowerSeriesRing(R, names=('z',)); (z,) = L._first_ngens(1)
>>> f = L(lambda i: x[i-Integer(1)]/factorial(i), valuation=Integer(1))
>>> all(exp(f)[i].subs(d) * factorial(i) == bell_polynomial(i) for i in range(n+Integer(1)))
True
>>> all((f**k/factorial(k))[i].subs(d) * factorial(i) == bell_polynomial(i, k) for i in range(k, n+k))
True
>>> g = L(lambda i: x[i-Integer(1)], valuation=Integer(1))
>>> all((Integer(1)/(Integer(1)-g))[i].subs(d) == bell_polynomial(i, ordinary=True) for i in range(n+Integer(1)))
True
>>> all((g**k)[i].subs(d) == bell_polynomial(i, k, True) for i in range(k, n+k))
True


REFERENCES:

AUTHORS:

• Blair Sutton (2009-01-26)

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

• Kei Beauduin (2024-04-06): when univariate, the polynomial is in variable x0. extended to complete exponential, partial ordinary and complete ordinary Bell polynomials.

sage.combinat.combinat.bernoulli_polynomial(x, n)[source]#

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: # needs sage.libs.flint
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

>>> from sage.all import *
>>> # needs sage.libs.flint
>>> y = QQ['y'].gen(0)
>>> bernoulli_polynomial(y, Integer(5))
y^5 - 5/2*y^4 + 5/3*y^3 - 1/6*y
>>> bernoulli_polynomial(y, Integer(5))(Integer(12))
199870
>>> bernoulli_polynomial(Integer(12), Integer(5))
199870
>>> bernoulli_polynomial(y**Integer(2) + Integer(1), Integer(5))
y^10 + 5/2*y^8 + 5/3*y^6 - 1/6*y^2
>>> P = ZZ['t']; (t,) = P._first_ngens(1)
>>> p = bernoulli_polynomial(t, Integer(6))
>>> 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)                 # needs sage.libs.flint
True

>>> from sage.all import *
>>> power_sum = sum(k**Integer(4) for k in range(Integer(10)))
>>> Integer(5)*power_sum == bernoulli_polynomial(Integer(10), Integer(5)) - bernoulli(Integer(5))                 # needs sage.libs.flint
True


REFERENCES:

sage.combinat.combinat.catalan_number(n)[source]#

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]

>>> from sage.all import *
>>> [catalan_number(i) for i in range(Integer(7))]
[1, 1, 2, 5, 14, 42, 132]
>>> x = (QQ[['x']].gen(0)).O(Integer(8))
>>> (-Integer(1)/Integer(2))*sqrt(Integer(1) - Integer(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)
>>> [catalan_number(i) for i in range(-Integer(7),Integer(7))]
[0, 0, 0, 0, 0, 0, -1/2, 1, 1, 2, 5, 14, 42, 132]
>>> [catalan_number(n).mod(Integer(2)) for n in range(Integer(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')[source]#

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)]                                      # needs sage.libs.flint
[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)]                         # needs sage.libs.flint
[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

>>> from sage.all import *
>>> [euler_number(i) for i in range(Integer(10))]                                      # needs sage.libs.flint
[1, 0, -1, 0, 5, 0, -61, 0, 1385, 0]
>>> x = PowerSeriesRing(QQ, 'x').gen().O(Integer(10))
>>> Integer(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)
>>> [euler_number(i)/factorial(i) for i in range(Integer(11))]                         # needs sage.libs.flint
[1, 0, -1/2, 0, 5/24, 0, -61/720, 0, 277/8064, 0, -50521/3628800]
>>> euler_number(-Integer(1))
Traceback (most recent call last):
...
ValueError: n (=-1) must be a nonnegative integer


REFERENCES:

sage.combinat.combinat.eulerian_number(k, algorithm='recursive')[source]#

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]

>>> from sage.all import *
>>> from sage.combinat.combinat import eulerian_number
>>> [eulerian_number(Integer(5),i) for i in range(Integer(5))]
[1, 26, 66, 26, 1]

sage.combinat.combinat.eulerian_polynomial(algorithm='derivative')[source]#

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

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


REFERENCES:

sage.combinat.combinat.fibonacci(n, algorithm='pari')[source]#

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)                                                             # needs sage.libs.pari
55
sage: fibonacci(10, algorithm='gap')                                            # needs sage.libs.gap
55

>>> from sage.all import *
>>> fibonacci(Integer(10))                                                             # needs sage.libs.pari
55
>>> fibonacci(Integer(10), algorithm='gap')                                            # needs sage.libs.gap
55

sage: fibonacci(-100)                                                           # needs sage.libs.pari
-354224848179261915075
sage: fibonacci(100)                                                            # needs sage.libs.pari
354224848179261915075

>>> from sage.all import *
>>> fibonacci(-Integer(100))                                                           # needs sage.libs.pari
-354224848179261915075
>>> fibonacci(Integer(100))                                                            # needs sage.libs.pari
354224848179261915075

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

>>> from sage.all import *
>>> fibonacci(Integer(0))                                                              # needs sage.libs.pari
0
>>> fibonacci(Integer(1)/Integer(2))
Traceback (most recent call last):
...
TypeError: no conversion of this rational to integer

sage.combinat.combinat.fibonacci_sequence(start, stop=None, algorithm=None)[source]#

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)]; fibs                      # needs sage.libs.pari
[55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181]

>>> from sage.all import *
>>> fibs = [i for i in fibonacci_sequence(Integer(10), Integer(20))]; fibs                      # needs sage.libs.pari
[55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181]

sage: sum([i for i in fibonacci_sequence(100, 110)])                            # needs sage.libs.pari
69919376923075308730013

>>> from sage.all import *
>>> sum([i for i in fibonacci_sequence(Integer(100), Integer(110))])                            # needs sage.libs.pari
69919376923075308730013


AUTHORS:

• Bobby Moretti

sage.combinat.combinat.fibonacci_xrange(start, stop=None, algorithm='pari')[source]#

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)]            # needs sage.libs.pari
sage: len(fibs_in_some_range)                                                   # needs sage.libs.pari
4
sage: fibs_in_some_range                                                        # needs sage.libs.pari
[14930352, 24157817, 39088169, 63245986]

>>> from sage.all import *
>>> fibs_in_some_range = [i for i in fibonacci_xrange(Integer(10)**Integer(7), Integer(10)**Integer(8))]            # needs sage.libs.pari
>>> len(fibs_in_some_range)                                                   # needs sage.libs.pari
4
>>> fibs_in_some_range                                                        # needs sage.libs.pari
[14930352, 24157817, 39088169, 63245986]

sage: fibs = [i for i in fibonacci_xrange(10, 100)]; fibs                       # needs sage.libs.pari
[13, 21, 34, 55, 89]

>>> from sage.all import *
>>> fibs = [i for i in fibonacci_xrange(Integer(10), Integer(100))]; fibs                       # needs sage.libs.pari
[13, 21, 34, 55, 89]

sage: list(fibonacci_xrange(13, 34))                                            # needs sage.libs.pari
[13, 21]

>>> from sage.all import *
>>> list(fibonacci_xrange(Integer(13), Integer(34)))                                            # needs sage.libs.pari
[13, 21]


A solution to the second Project Euler problem:

sage: sum([i for i in fibonacci_xrange(10^6) if is_even(i)])                    # needs sage.libs.pari
1089154

>>> from sage.all import *
>>> sum([i for i in fibonacci_xrange(Integer(10)**Integer(6)) if is_even(i)])                    # needs sage.libs.pari
1089154


AUTHORS:

• Bobby Moretti

sage.combinat.combinat.lucas_number1(n, P, Q)[source]#

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(...)[1].

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

INPUT:

• n – integer

• P, Q – integer or rational numbers

OUTPUT: integer or rational number

EXAMPLES:

sage: # needs sage.libs.gap
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

>>> from sage.all import *
>>> # needs sage.libs.gap
>>> lucas_number1(Integer(5),Integer(1),-Integer(1))
5
>>> lucas_number1(Integer(6),Integer(1),-Integer(1))
8
>>> lucas_number1(Integer(7),Integer(1),-Integer(1))
13
>>> lucas_number1(Integer(7),Integer(1),-Integer(2))
43
>>> lucas_number1(Integer(5),Integer(2),Integer(3)/Integer(5))
229/25
>>> lucas_number1(Integer(5),Integer(2),RealNumber('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: def lucas(n):
....:     return 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)]   # needs sage.libs.gap
[[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]]

>>> from sage.all import *
>>> def lucas(n):
...     return Integer((Integer(5)/Integer(2))*lucas_number1(n,Integer(1),-Integer(1)) + (Integer(1)/Integer(2))*lucas_number2(n,Integer(1),-Integer(1)))
>>> [[lucas(n), is_prime(lucas(n)), n+Integer(1), is_prime(n+Integer(1))] for n in range(Integer(15))]   # needs sage.libs.gap
[[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)[source]#

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(…)[2].

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)]                                # needs sage.libs.gap
[2, 1, 3, 4, 7, 11, 18, 29, 47, 76]
sage: [fibonacci(i-1)+fibonacci(i+1) for i in range(10)]                        # needs sage.libs.pari
[2, 1, 3, 4, 7, 11, 18, 29, 47, 76]

>>> from sage.all import *
>>> [lucas_number2(i,Integer(1),-Integer(1)) for i in range(Integer(10))]                                # needs sage.libs.gap
[2, 1, 3, 4, 7, 11, 18, 29, 47, 76]
>>> [fibonacci(i-Integer(1))+fibonacci(i+Integer(1)) for i in range(Integer(10))]                        # needs sage.libs.pari
[2, 1, 3, 4, 7, 11, 18, 29, 47, 76]

sage: # needs sage.libs.gap
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'>

>>> from sage.all import *
>>> # needs sage.libs.gap
>>> n = lucas_number2(Integer(5),Integer(2),Integer(3)); n
2
>>> type(n)
<class 'sage.rings.integer.Integer'>
>>> n = lucas_number2(Integer(5),Integer(2),-Integer(3)/Integer(9)); n
418/9
>>> 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)]                                # needs sage.libs.gap
[2, 1, 3, 4, 7, 11, 18, 29, 47, 76]

>>> from sage.all import *
>>> [lucas_number2(n,Integer(1),-Integer(1)) for n in range(Integer(10))]                                # needs sage.libs.gap
[2, 1, 3, 4, 7, 11, 18, 29, 47, 76]

sage.combinat.combinat.narayana_number(n, k)[source]#

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

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


REFERENCES:

sage.combinat.combinat.number_of_tuples(S, k, algorithm='naive')[source]#

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")                                    # needs sage.libs.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")                                    # needs sage.libs.gap
25
sage: number_of_tuples(S,0)
1
sage: number_of_tuples(S,0, algorithm="gap")                                    # needs sage.libs.gap
1

>>> from sage.all import *
>>> S = [Integer(1),Integer(2),Integer(3),Integer(4),Integer(5)]
>>> number_of_tuples(S,Integer(2))
25
>>> number_of_tuples(S,Integer(2), algorithm="gap")                                    # needs sage.libs.gap
25
>>> S = [Integer(1),Integer(1),Integer(2),Integer(3),Integer(4),Integer(5)]
>>> number_of_tuples(S,Integer(2))
25
>>> number_of_tuples(S,Integer(2), algorithm="gap")                                    # needs sage.libs.gap
25
>>> number_of_tuples(S,Integer(0))
1
>>> number_of_tuples(S,Integer(0), algorithm="gap")                                    # needs sage.libs.gap
1

sage.combinat.combinat.number_of_unordered_tuples(S, k, algorithm='naive')[source]#

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")                          # needs sage.libs.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")                          # needs sage.libs.gap
15
sage: number_of_unordered_tuples(S,0)
1
sage: number_of_unordered_tuples(S,0, algorithm="gap")                          # needs sage.libs.gap
1

>>> from sage.all import *
>>> S = [Integer(1),Integer(2),Integer(3),Integer(4),Integer(5)]
>>> number_of_unordered_tuples(S,Integer(2))
15
>>> number_of_unordered_tuples(S,Integer(2), algorithm="gap")                          # needs sage.libs.gap
15
>>> S = [Integer(1),Integer(1),Integer(2),Integer(3),Integer(4),Integer(5)]
>>> number_of_unordered_tuples(S,Integer(2))
15
>>> number_of_unordered_tuples(S,Integer(2), algorithm="gap")                          # needs sage.libs.gap
15
>>> number_of_unordered_tuples(S,Integer(0))
1
>>> number_of_unordered_tuples(S,Integer(0), algorithm="gap")                          # needs sage.libs.gap
1

sage.combinat.combinat.polygonal_number(s, n)[source]#

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]

>>> from sage.all import *
>>> [polygonal_number(Integer(3), n) for n in range(Integer(10))]
[0, 1, 3, 6, 10, 15, 21, 28, 36, 45]

>>> [polygonal_number(Integer(3), n) for n in range(-Integer(10), Integer(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]

>>> from sage.all import *
>>> [polygonal_number(Integer(4), n) for n in range(Integer(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]

>>> from sage.all import *
>>> [polygonal_number(Integer(5), n) for n in range(Integer(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]

>>> from sage.all import *
>>> [polygonal_number(Integer(6), n) for n in range(Integer(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

>>> from sage.all import *
>>> polygonal_number(RealNumber('3.0'), RealNumber('2.0'))
3


A non-integer input returns an error:

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

>>> from sage.all import *
>>> polygonal_number(RealNumber('3.5'), Integer(1))                                                  # needs sage.rings.real_mpfr
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

>>> from sage.all import *
>>> polygonal_number(Integer(1), Integer(4))
Traceback (most recent call last):
...
ValueError: s (=1) must be greater than 1


REFERENCES:

sage.combinat.combinat.stirling_number1(n, k, algorithm='gap')[source]#

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.

INPUT:

• n – nonnegative machine-size integer

• k – nonnegative machine-size integer

• algorithm:

• "gap" (default) – use GAP’s Stirling1 function

• "flint" – use flint’s arith_stirling_number_1u function

EXAMPLES:

sage: # needs sage.libs.gap
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

>>> from sage.all import *
>>> # needs sage.libs.gap
>>> stirling_number1(Integer(3),Integer(2))
3
>>> stirling_number1(Integer(5),Integer(2))
50
>>> Integer(9)*stirling_number1(Integer(9),Integer(5)) + stirling_number1(Integer(9),Integer(4))
269325
>>> stirling_number1(Integer(10),Integer(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)[source]#

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

• "flint" – use flint’s arith_stirling_number_2 function

• "gap" – use GAP’s Stirling2 function

• "maxima" – use Maxima’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

>>> from sage.all import *
>>> for n in range(Integer(10)):
...     for k in range(Integer(10)):
...         print(str(stirling_number2(n,k)).rjust(k and Integer(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

>>> from sage.all import *
>>> Integer(5)*stirling_number2(Integer(9),Integer(5)) + stirling_number2(Integer(9),Integer(4))
42525
>>> stirling_number2(Integer(10),Integer(5))
42525

sage.combinat.combinat.tuples(S, k, algorithm='itertools')[source]#

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')]

>>> from sage.all import *
>>> S = [Integer(1),Integer(2)]
>>> tuples(S,Integer(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)]
>>> mset = ["s","t","e","i","n"]
>>> tuples(mset, Integer(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')                                                        # needs sage.rings.finite_rings
sage: mset = [x for x in K if x != 0]                                           # needs sage.rings.finite_rings
sage: tuples(mset, 2)                                                           # needs sage.rings.finite_rings
[(a, a), (a, a + 1), (a, 1), (a + 1, a), (a + 1, a + 1),
(a + 1, 1), (1, a), (1, a + 1), (1, 1)]

>>> from sage.all import *
>>> K = GF(Integer(4), 'a', names=('a',)); (a,) = K._first_ngens(1)# needs sage.rings.finite_rings
>>> mset = [x for x in K if x != Integer(0)]                                           # needs sage.rings.finite_rings
>>> tuples(mset, Integer(2))                                                           # needs sage.rings.finite_rings
[(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)]

>>> from sage.all import *
>>> tuples(S, Integer(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

>>> from sage.all import *
>>> S = [Integer(1),Integer(1),Integer(2)]
>>> sorted(tuples(S, Integer(3))) == sorted(tuples(S, Integer(3), 'native'))
True


AUTHORS:

• Jon Hanke (2006-08)

sage.combinat.combinat.unordered_tuples(S, k, algorithm='itertools')[source]#

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)]

>>> from sage.all import *
>>> S = [Integer(1),Integer(2)]
>>> unordered_tuples(S, Integer(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')                                   # needs sage.libs.gap
[(1, 1, 1), (1, 1, 2), (1, 2, 2), (2, 2, 2)]

>>> from sage.all import *
>>> unordered_tuples(S, Integer(3), algorithm='gap')                                   # needs sage.libs.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')                                   # needs sage.libs.gap
[('a', 'a'), ('a', 'b'), ('a', 'c'), ('b', 'b'), ('b', 'c'), ('c', 'c')]

>>> from sage.all import *
>>> S = ["a","b","c"]
>>> unordered_tuples(S, Integer(2))
[('a', 'a'), ('a', 'b'), ('a', 'c'), ('b', 'b'), ('b', 'c'), ('c', 'c')]
>>> unordered_tuples(S, Integer(2), algorithm='gap')                                   # needs sage.libs.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')                   # needs sage.libs.gap
True
sage: unordered_tuples(S, 3)
[(1, 1, 1), (1, 1, 2), (1, 2, 2), (2, 2, 2)]

>>> from sage.all import *
>>> S = [Integer(1),Integer(1),Integer(2)]
>>> unordered_tuples(S, Integer(3)) == unordered_tuples(S, Integer(3), 'gap')                   # needs sage.libs.gap
True
>>> unordered_tuples(S, Integer(3))
[(1, 1, 1), (1, 1, 2), (1, 2, 2), (2, 2, 2)]

sage.combinat.combinat.unshuffle_iterator(a, one=1)[source]#

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]))
[(((), (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)]

>>> from sage.all import *
>>> from sage.combinat.combinat import unshuffle_iterator
>>> list(unshuffle_iterator([Integer(1), Integer(3), Integer(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)]
>>> list(unshuffle_iterator([Integer(3), Integer(1)]))
[(((), (3, 1)), 1), (((3,), (1,)), 1), (((1,), (3,)), -1),
(((3, 1), ()), 1)]
>>> list(unshuffle_iterator([Integer(8)]))
[(((), (8,)), 1), (((8,), ()), 1)]
>>> list(unshuffle_iterator([]))
[(((), ()), 1)]
>>> list(unshuffle_iterator([Integer(3), Integer(1)], Integer(3)/Integer(2)))
[(((), (3, 1)), 3/2), (((3,), (1,)), 3/2), (((1,), (3,)), -3/2),
(((3, 1), ()), 3/2)]