Combinatorial Functions¶
This module implements some combinatorial functions, as listed below. For a more detailed description, see the relevant docstrings.
Sequences:
- Bell numbers, - bell_number()
- Catalan numbers, - catalan_number()(not to be confused with the Catalan constant)
- Narayana numbers, - narayana_number()
- Euler numbers, - euler_number()(Maxima)
- Eulerian numbers, - eulerian_number()
- Eulerian polynomial, - eulerian_polynomial()
- Fibonacci numbers, - fibonacci()(PARI) and- fibonacci_number()(GAP) The PARI version is better.
- Lucas numbers, - lucas_number1(),- lucas_number2().
- Stirling numbers, - stirling_number1(),- stirling_number2().
- Polygonal numbers, - polygonal_number()
Set-theoretic constructions:
- Derangements of a multiset, - derangements()and- number_of_derangements().
- Tuples of a multiset, - tuples()and- number_of_tuples(). An ordered tuple of length k of set S is a ordered selection with repetitions of S and is represented by a sorted list of length k containing elements from S.
- Unordered tuples of a set, - unordered_tuples()and- number_of_unordered_tuples(). An unordered tuple of length k of set S is an unordered selection with repetitions of S and is represented by a sorted list of length k containing elements from S.
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:
- Bernoulli polynomials, - bernoulli_polynomial()
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:
- number_of_partitions()(wrapped from PARI) the number of partitions:
- sage.combinat.q_analogues.gaussian_binomial()the Gaussian binomial
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:
- Rencontres numbers (Wikipedia article Rencontres_number) 
 
REFERENCES:
- Wikipedia article Twelvefold_way (general reference) 
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. 
- David Joyner (2007-11): changed permutations, added hadamard_matrix 
- 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]¶
- Bases: - CombinatorialObject,- Element- CombinatorialElementis both a- CombinatorialObjectand an- Element. So it represents a list which is an element of some parent.- A - CombinatorialElementsubclass also automatically supports the- __classcall__mechanism.- Warning - This class is slowly being deprecated. Use - ClonableListinstead.- INPUT: - parent– the- Parentclass for this element
- lst– 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]) sage: e == loads(dumps(e)) 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)]) >>> e == loads(dumps(e)) 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. - See also - CombinatorialElementif you want a combinatorial object which is an element of a parent.- Warning - This class is slowly being deprecated. Use - ClonableListinstead.- INPUT: - l– list or any object that can be converted to a list by calling- list()
- copy– boolean (default:- True); if- False, then- lmust be a- list, which is assigned to- self._listwithout copying
 - EXAMPLES: - sage: c = CombinatorialObject([1,2,3]) sage: c == loads(dumps(c)) 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)]) >>> c == loads(dumps(c)) True >>> c._list [1, 2, 3] >>> c._hash is None True - For efficiency, you can specify - copy=Falseif 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] 
- 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– 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 - precoption:- 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– boolean (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– 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- 0and- n - 1
- algorithm–- 'recursive'(default) or- 'formula'
 - OUTPUT: integer - See also - 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– integer
- algorithm–- 'derivative'(default) or- 'coeffs'
 - OUTPUT: polynomial in one variable - t- See also - 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– string; one of- '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 = startup 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 - See also - 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 = startup 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 - See also - 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– integer
- k– integer between- 0and- n - 1
 - OUTPUT: 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',- Smust be a list of objects that have string representations that can be interpreted by the GAP interpreter. If- Sconsists 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',- Smust be a list of objects that have string representations that can be interpreted by the GAP interpreter. If- Sconsists 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: 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. - See Wikipedia article Stirling_numbers_of_the_first_kind. - INPUT: - n– nonnegative machine-size integer
- k– nonnegative machine-size integer
- algorithm:- 'gap'– default; use GAP’s- Stirling1function
- 'flint'– use flint’s- arith_stirling_number_1ufunction
 
 - 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, \(k=0,...,n\). - See Wikipedia article Stirling_numbers_of_the_second_kind. - INPUT: - n– nonnegative machine-size integer
- k– nonnegative machine-size integer
- algorithm:- None– default; use native implementation
- 'flint'– use flint’s- arith_stirling_number_2function
- 'gap'– use GAP’s- Stirling2function
- 'maxima'– use Maxima’s- stirling2function
 
 - 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 - Sin the form of any iterable (list, tuple or iterator), and returns a list of \(k\)-tuples. If- Scontains 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 - Sis 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 - kof 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_replacementinstead.- 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',- Smust be a list of objects that have string representations that can be interpreted by the GAP interpreter. If- Sconsists 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 - onecan 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)]