"""
This module implements some special functions that commonly appear in
combinatorial contexts (e.g. in power series); in particular,
sequences of rational numbers such as Bernoulli and Fibonacci numbers.
Factorials, binomial coefficients and related functions are located in
the separate 'factorials' module.
"""
from __future__ import print_function, division
from sympy.core import S, Symbol, Rational, Integer, Add, Dummy
from sympy.core.compatibility import as_int, SYMPY_INTS, range
from sympy.core.cache import cacheit
from sympy.core.function import Function, expand_mul
from sympy.core.numbers import E, pi
from sympy.core.relational import LessThan, StrictGreaterThan
from sympy.functions.combinatorial.factorials import binomial, factorial
from sympy.functions.elementary.exponential import log
from sympy.functions.elementary.integers import floor
from sympy.functions.elementary.trigonometric import sin, cos, cot
from sympy.functions.elementary.miscellaneous import sqrt
from sympy.utilities.memoization import recurrence_memo
from mpmath import bernfrac, workprec
from mpmath.libmp import ifib as _ifib
def _product(a, b):
p = 1
for k in range(a, b + 1):
p *= k
return p
# Dummy symbol used for computing polynomial sequences
_sym = Symbol('x')
_symbols = Function('x')
#----------------------------------------------------------------------------#
# #
# Fibonacci numbers #
# #
#----------------------------------------------------------------------------#
[docs]class fibonacci(Function):
r"""
Fibonacci numbers / Fibonacci polynomials
The Fibonacci numbers are the integer sequence defined by the
initial terms F_0 = 0, F_1 = 1 and the two-term recurrence
relation F_n = F_{n-1} + F_{n-2}. This definition
extended to arbitrary real and complex arguments using
the formula
.. math :: F_z = \frac{\phi^z - \cos(\pi z) \phi^{-z}}{\sqrt 5}
The Fibonacci polynomials are defined by F_1(x) = 1,
F_2(x) = x, and F_n(x) = x*F_{n-1}(x) + F_{n-2}(x) for n > 2.
For all positive integers n, F_n(1) = F_n.
* fibonacci(n) gives the nth Fibonacci number, F_n
* fibonacci(n, x) gives the nth Fibonacci polynomial in x, F_n(x)
Examples
========
>>> from sympy import fibonacci, Symbol
>>> [fibonacci(x) for x in range(11)]
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
>>> fibonacci(5, Symbol('t'))
t**4 + 3*t**2 + 1
References
==========
.. [1] http://en.wikipedia.org/wiki/Fibonacci_number
.. [2] http://mathworld.wolfram.com/FibonacciNumber.html
See Also
========
bell, bernoulli, catalan, euler, harmonic, lucas
"""
@staticmethod
def _fib(n):
return _ifib(n)
@staticmethod
@recurrence_memo([None, S.One, _sym])
def _fibpoly(n, prev):
return (prev[-2] + _sym*prev[-1]).expand()
@classmethod
def eval(cls, n, sym=None):
if n is S.Infinity:
return S.Infinity
if n.is_Integer:
n = int(n)
if n < 0:
return S.NegativeOne**(n + 1) * fibonacci(-n)
if sym is None:
return Integer(cls._fib(n))
else:
if n < 1:
raise ValueError("Fibonacci polynomials are defined "
"only for positive integer indices.")
return cls._fibpoly(n).subs(_sym, sym)
def _eval_rewrite_as_sqrt(self, n):
return 2**(-n)*sqrt(5)*((1 + sqrt(5))**n - (-sqrt(5) + 1)**n) / 5
[docs]class lucas(Function):
"""
Lucas numbers
Lucas numbers satisfy a recurrence relation similar to that of
the Fibonacci sequence, in which each term is the sum of the
preceding two. They are generated by choosing the initial
values L_0 = 2 and L_1 = 1.
* lucas(n) gives the nth Lucas number
Examples
========
>>> from sympy import lucas
>>> [lucas(x) for x in range(11)]
[2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123]
References
==========
.. [1] http://en.wikipedia.org/wiki/Lucas_number
.. [2] http://mathworld.wolfram.com/LucasNumber.html
See Also
========
bell, bernoulli, catalan, euler, fibonacci, harmonic
"""
@classmethod
def eval(cls, n):
if n is S.Infinity:
return S.Infinity
if n.is_Integer:
return fibonacci(n + 1) + fibonacci(n - 1)
def _eval_rewrite_as_sqrt(self, n):
return 2**(-n)*((1 + sqrt(5))**n + (-sqrt(5) + 1)**n)
#----------------------------------------------------------------------------#
# #
# Bernoulli numbers #
# #
#----------------------------------------------------------------------------#
[docs]class bernoulli(Function):
r"""
Bernoulli numbers / Bernoulli polynomials
The Bernoulli numbers are a sequence of rational numbers
defined by B_0 = 1 and the recursive relation (n > 0)::
n
___
\ / n + 1 \
0 = ) | | * B .
/___ \ k / k
k = 0
They are also commonly defined by their exponential generating
function, which is x/(exp(x) - 1). For odd indices > 1, the
Bernoulli numbers are zero.
The Bernoulli polynomials satisfy the analogous formula::
n
___
\ / n \ n-k
B (x) = ) | | * B * x .
n /___ \ k / k
k = 0
Bernoulli numbers and Bernoulli polynomials are related as
B_n(0) = B_n.
We compute Bernoulli numbers using Ramanujan's formula::
/ n + 3 \
B = (A(n) - S(n)) / | |
n \ n /
where A(n) = (n+3)/3 when n = 0 or 2 (mod 6), A(n) = -(n+3)/6
when n = 4 (mod 6), and::
[n/6]
___
\ / n + 3 \
S(n) = ) | | * B
/___ \ n - 6*k / n-6*k
k = 1
This formula is similar to the sum given in the definition, but
cuts 2/3 of the terms. For Bernoulli polynomials, we use the
formula in the definition.
* bernoulli(n) gives the nth Bernoulli number, B_n
* bernoulli(n, x) gives the nth Bernoulli polynomial in x, B_n(x)
Examples
========
>>> from sympy import bernoulli
>>> [bernoulli(n) for n in range(11)]
[1, -1/2, 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66]
>>> bernoulli(1000001)
0
References
==========
.. [1] http://en.wikipedia.org/wiki/Bernoulli_number
.. [2] http://en.wikipedia.org/wiki/Bernoulli_polynomial
.. [3] http://mathworld.wolfram.com/BernoulliNumber.html
.. [4] http://mathworld.wolfram.com/BernoulliPolynomial.html
See Also
========
bell, catalan, euler, fibonacci, harmonic, lucas
"""
# Calculates B_n for positive even n
@staticmethod
def _calc_bernoulli(n):
s = 0
a = int(binomial(n + 3, n - 6))
for j in range(1, n//6 + 1):
s += a * bernoulli(n - 6*j)
# Avoid computing each binomial coefficient from scratch
a *= _product(n - 6 - 6*j + 1, n - 6*j)
a //= _product(6*j + 4, 6*j + 9)
if n % 6 == 4:
s = -Rational(n + 3, 6) - s
else:
s = Rational(n + 3, 3) - s
return s / binomial(n + 3, n)
# We implement a specialized memoization scheme to handle each
# case modulo 6 separately
_cache = {0: S.One, 2: Rational(1, 6), 4: Rational(-1, 30)}
_highest = {0: 0, 2: 2, 4: 4}
@classmethod
def eval(cls, n, sym=None):
if n.is_Number:
if n.is_Integer and n.is_nonnegative:
if n is S.Zero:
return S.One
elif n is S.One:
if sym is None:
return -S.Half
else:
return sym - S.Half
# Bernoulli numbers
elif sym is None:
if n.is_odd:
return S.Zero
n = int(n)
# Use mpmath for enormous Bernoulli numbers
if n > 500:
p, q = bernfrac(n)
return Rational(int(p), int(q))
case = n % 6
highest_cached = cls._highest[case]
if n <= highest_cached:
return cls._cache[n]
# To avoid excessive recursion when, say, bernoulli(1000) is
# requested, calculate and cache the entire sequence ... B_988,
# B_994, B_1000 in increasing order
for i in range(highest_cached + 6, n + 6, 6):
b = cls._calc_bernoulli(i)
cls._cache[i] = b
cls._highest[case] = i
return b
# Bernoulli polynomials
else:
n, result = int(n), []
for k in range(n + 1):
result.append(binomial(n, k)*cls(k)*sym**(n - k))
return Add(*result)
else:
raise ValueError("Bernoulli numbers are defined only"
" for nonnegative integer indices.")
if sym is None:
if n.is_odd and (n - 1).is_positive:
return S.Zero
#----------------------------------------------------------------------------#
# #
# Bell numbers #
# #
#----------------------------------------------------------------------------#
[docs]class bell(Function):
r"""
Bell numbers / Bell polynomials
The Bell numbers satisfy `B_0 = 1` and
.. math:: B_n = \sum_{k=0}^{n-1} \binom{n-1}{k} B_k.
They are also given by:
.. math:: B_n = \frac{1}{e} \sum_{k=0}^{\infty} \frac{k^n}{k!}.
The Bell polynomials are given by `B_0(x) = 1` and
.. math:: B_n(x) = x \sum_{k=1}^{n-1} \binom{n-1}{k-1} B_{k-1}(x).
The second kind of Bell polynomials (are sometimes called "partial" Bell
polynomials or incomplete Bell polynomials) are defined as
.. math:: B_{n,k}(x_1, x_2,\dotsc x_{n-k+1}) =
\sum_{j_1+j_2+j_2+\dotsb=k \atop j_1+2j_2+3j_2+\dotsb=n}
\frac{n!}{j_1!j_2!\dotsb j_{n-k+1}!}
\left(\frac{x_1}{1!} \right)^{j_1}
\left(\frac{x_2}{2!} \right)^{j_2} \dotsb
\left(\frac{x_{n-k+1}}{(n-k+1)!} \right) ^{j_{n-k+1}}.
* bell(n) gives the `n^{th}` Bell number, `B_n`.
* bell(n, x) gives the `n^{th}` Bell polynomial, `B_n(x)`.
* bell(n, k, (x1, x2, ...)) gives Bell polynomials of the second kind,
`B_{n,k}(x_1, x_2, \dotsc, x_{n-k+1})`.
Notes
=====
Not to be confused with Bernoulli numbers and Bernoulli polynomials,
which use the same notation.
Examples
========
>>> from sympy import bell, Symbol, symbols
>>> [bell(n) for n in range(11)]
[1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975]
>>> bell(30)
846749014511809332450147
>>> bell(4, Symbol('t'))
t**4 + 6*t**3 + 7*t**2 + t
>>> bell(6, 2, symbols('x:6')[1:])
6*x1*x5 + 15*x2*x4 + 10*x3**2
References
==========
.. [1] http://en.wikipedia.org/wiki/Bell_number
.. [2] http://mathworld.wolfram.com/BellNumber.html
.. [3] http://mathworld.wolfram.com/BellPolynomial.html
See Also
========
bernoulli, catalan, euler, fibonacci, harmonic, lucas
"""
@staticmethod
@recurrence_memo([1, 1])
def _bell(n, prev):
s = 1
a = 1
for k in range(1, n):
a = a * (n - k) // k
s += a * prev[k]
return s
@staticmethod
@recurrence_memo([S.One, _sym])
def _bell_poly(n, prev):
s = 1
a = 1
for k in range(2, n + 1):
a = a * (n - k + 1) // (k - 1)
s += a * prev[k - 1]
return expand_mul(_sym * s)
@staticmethod
def _bell_incomplete_poly(n, k, symbols):
r"""
The second kind of Bell polynomials (incomplete Bell polynomials).
Calculated by recurrence formula:
.. math:: B_{n,k}(x_1, x_2, \dotsc, x_{n-k+1}) =
\sum_{m=1}^{n-k+1}
\x_m \binom{n-1}{m-1} B_{n-m,k-1}(x_1, x_2, \dotsc, x_{n-m-k})
where
B_{0,0} = 1;
B_{n,0} = 0; for n>=1
B_{0,k} = 0; for k>=1
"""
if (n == 0) and (k == 0):
return S.One
elif (n == 0) or (k == 0):
return S.Zero
s = S.Zero
a = S.One
for m in range(1, n - k + 2):
s += a * bell._bell_incomplete_poly(
n - m, k - 1, symbols) * symbols[m - 1]
a = a * (n - m) / m
return expand_mul(s)
@classmethod
def eval(cls, n, k_sym=None, symbols=None):
if n.is_Integer and n.is_nonnegative:
if k_sym is None:
return Integer(cls._bell(int(n)))
elif symbols is None:
return cls._bell_poly(int(n)).subs(_sym, k_sym)
else:
r = cls._bell_incomplete_poly(int(n), int(k_sym), symbols)
return r
def _eval_rewrite_as_Sum(self, n, k_sym=None, symbols=None):
from sympy import Sum
if (k_sym is not None) or (symbols is not None):
return self
# Dobinski's formula
if not n.is_nonnegative:
return self
k = Dummy('k', integer=True, nonnegative=True)
return 1 / E * Sum(k**n / factorial(k), (k, 0, S.Infinity))
#----------------------------------------------------------------------------#
# #
# Harmonic numbers #
# #
#----------------------------------------------------------------------------#
[docs]class harmonic(Function):
r"""
Harmonic numbers
The nth harmonic number is given by `\operatorname{H}_{n} =
1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n}`.
More generally:
.. math:: \operatorname{H}_{n,m} = \sum_{k=1}^{n} \frac{1}{k^m}
As `n \rightarrow \infty`, `\operatorname{H}_{n,m} \rightarrow \zeta(m)`,
the Riemann zeta function.
* ``harmonic(n)`` gives the nth harmonic number, `\operatorname{H}_n`
* ``harmonic(n, m)`` gives the nth generalized harmonic number
of order `m`, `\operatorname{H}_{n,m}`, where
``harmonic(n) == harmonic(n, 1)``
Examples
========
>>> from sympy import harmonic, oo
>>> [harmonic(n) for n in range(6)]
[0, 1, 3/2, 11/6, 25/12, 137/60]
>>> [harmonic(n, 2) for n in range(6)]
[0, 1, 5/4, 49/36, 205/144, 5269/3600]
>>> harmonic(oo, 2)
pi**2/6
>>> from sympy import Symbol, Sum
>>> n = Symbol("n")
>>> harmonic(n).rewrite(Sum)
Sum(1/_k, (_k, 1, n))
We can evaluate harmonic numbers for all integral and positive
rational arguments:
>>> from sympy import S, expand_func, simplify
>>> harmonic(8)
761/280
>>> harmonic(11)
83711/27720
>>> H = harmonic(1/S(3))
>>> H
harmonic(1/3)
>>> He = expand_func(H)
>>> He
-log(6) - sqrt(3)*pi/6 + 2*Sum(log(sin(_k*pi/3))*cos(2*_k*pi/3), (_k, 1, 1))
+ 3*Sum(1/(3*_k + 1), (_k, 0, 0))
>>> He.doit()
-log(6) - sqrt(3)*pi/6 - log(sqrt(3)/2) + 3
>>> H = harmonic(25/S(7))
>>> He = simplify(expand_func(H).doit())
>>> He
log(sin(pi/7)**(-2*cos(pi/7))*sin(2*pi/7)**(2*cos(16*pi/7))*cos(pi/14)**(-2*sin(pi/14))/14)
+ pi*tan(pi/14)/2 + 30247/9900
>>> He.n(40)
1.983697455232980674869851942390639915940
>>> harmonic(25/S(7)).n(40)
1.983697455232980674869851942390639915940
We can rewrite harmonic numbers in terms of polygamma functions:
>>> from sympy import digamma, polygamma
>>> m = Symbol("m")
>>> harmonic(n).rewrite(digamma)
polygamma(0, n + 1) + EulerGamma
>>> harmonic(n).rewrite(polygamma)
polygamma(0, n + 1) + EulerGamma
>>> harmonic(n,3).rewrite(polygamma)
polygamma(2, n + 1)/2 - polygamma(2, 1)/2
>>> harmonic(n,m).rewrite(polygamma)
(-1)**m*(polygamma(m - 1, 1) - polygamma(m - 1, n + 1))/factorial(m - 1)
Integer offsets in the argument can be pulled out:
>>> from sympy import expand_func
>>> expand_func(harmonic(n+4))
harmonic(n) + 1/(n + 4) + 1/(n + 3) + 1/(n + 2) + 1/(n + 1)
>>> expand_func(harmonic(n-4))
harmonic(n) - 1/(n - 1) - 1/(n - 2) - 1/(n - 3) - 1/n
Some limits can be computed as well:
>>> from sympy import limit, oo
>>> limit(harmonic(n), n, oo)
oo
>>> limit(harmonic(n, 2), n, oo)
pi**2/6
>>> limit(harmonic(n, 3), n, oo)
-polygamma(2, 1)/2
However we can not compute the general relation yet:
>>> limit(harmonic(n, m), n, oo)
harmonic(oo, m)
which equals ``zeta(m)`` for ``m > 1``.
References
==========
.. [1] http://en.wikipedia.org/wiki/Harmonic_number
.. [2] http://functions.wolfram.com/GammaBetaErf/HarmonicNumber/
.. [3] http://functions.wolfram.com/GammaBetaErf/HarmonicNumber2/
See Also
========
bell, bernoulli, catalan, euler, fibonacci, lucas
"""
# Generate one memoized Harmonic number-generating function for each
# order and store it in a dictionary
_functions = {}
@classmethod
def eval(cls, n, m=None):
from sympy import zeta
if m is S.One:
return cls(n)
if m is None:
m = S.One
if m.is_zero:
return n
if n is S.Infinity and m.is_Number:
# TODO: Fix for symbolic values of m
if m.is_negative:
return S.NaN
elif LessThan(m, S.One):
return S.Infinity
elif StrictGreaterThan(m, S.One):
return zeta(m)
else:
return cls
if n.is_Integer and n.is_nonnegative and m.is_Integer:
if n == 0:
return S.Zero
if not m in cls._functions:
@recurrence_memo([0])
def f(n, prev):
return prev[-1] + S.One / n**m
cls._functions[m] = f
return cls._functions[m](int(n))
def _eval_rewrite_as_polygamma(self, n, m=1):
from sympy.functions.special.gamma_functions import polygamma
return S.NegativeOne**m/factorial(m - 1) * (polygamma(m - 1, 1) - polygamma(m - 1, n + 1))
def _eval_rewrite_as_digamma(self, n, m=1):
from sympy.functions.special.gamma_functions import polygamma
return self.rewrite(polygamma)
def _eval_rewrite_as_trigamma(self, n, m=1):
from sympy.functions.special.gamma_functions import polygamma
return self.rewrite(polygamma)
def _eval_rewrite_as_Sum(self, n, m=None):
from sympy import Sum
k = Dummy("k", integer=True)
if m is None:
m = S.One
return Sum(k**(-m), (k, 1, n))
def _eval_expand_func(self, **hints):
from sympy import Sum
n = self.args[0]
m = self.args[1] if len(self.args) == 2 else 1
if m == S.One:
if n.is_Add:
off = n.args[0]
nnew = n - off
if off.is_Integer and off.is_positive:
result = [S.One/(nnew + i) for i in range(off, 0, -1)] + [harmonic(nnew)]
return Add(*result)
elif off.is_Integer and off.is_negative:
result = [-S.One/(nnew + i) for i in range(0, off, -1)] + [harmonic(nnew)]
return Add(*result)
if n.is_Rational:
# Expansions for harmonic numbers at general rational arguments (u + p/q)
# Split n as u + p/q with p < q
p, q = n.as_numer_denom()
u = p // q
p = p - u * q
if u.is_nonnegative and p.is_positive and q.is_positive and p < q:
k = Dummy("k")
t1 = q * Sum(1 / (q * k + p), (k, 0, u))
t2 = 2 * Sum(cos((2 * pi * p * k) / S(q)) *
log(sin((pi * k) / S(q))),
(k, 1, floor((q - 1) / S(2))))
t3 = (pi / 2) * cot((pi * p) / q) + log(2 * q)
return t1 + t2 - t3
return self
def _eval_rewrite_as_tractable(self, n, m=1):
from sympy import polygamma
return self.rewrite(polygamma).rewrite("tractable", deep=True)
def _eval_evalf(self, prec):
from sympy import polygamma
if all(i.is_number for i in self.args):
return self.rewrite(polygamma)._eval_evalf(prec)
#----------------------------------------------------------------------------#
# #
# Euler numbers #
# #
#----------------------------------------------------------------------------#
[docs]class euler(Function):
r"""
Euler numbers
The euler numbers are given by::
2*n+1 k
___ ___ j 2*n+1
\ \ / k \ (-1) * (k-2*j)
E = I ) ) | | --------------------
2n /___ /___ \ j / k k
k = 1 j = 0 2 * I * k
E = 0
2n+1
* euler(n) gives the n-th Euler number, E_n
Examples
========
>>> from sympy import Symbol
>>> from sympy.functions import euler
>>> [euler(n) for n in range(10)]
[1, 0, -1, 0, 5, 0, -61, 0, 1385, 0]
>>> n = Symbol("n")
>>> euler(n+2*n)
euler(3*n)
References
==========
.. [1] http://en.wikipedia.org/wiki/Euler_numbers
.. [2] http://mathworld.wolfram.com/EulerNumber.html
.. [3] http://en.wikipedia.org/wiki/Alternating_permutation
.. [4] http://mathworld.wolfram.com/AlternatingPermutation.html
See Also
========
bell, bernoulli, catalan, fibonacci, harmonic, lucas
"""
@classmethod
def eval(cls, m):
if m.is_odd:
return S.Zero
if m.is_Integer and m.is_nonnegative:
from mpmath import mp
m = m._to_mpmath(mp.prec)
res = mp.eulernum(m, exact=True)
return Integer(res)
def _eval_rewrite_as_Sum(self, arg):
from sympy import Sum
if arg.is_even:
k = Dummy("k", integer=True)
j = Dummy("j", integer=True)
n = self.args[0] / 2
Em = (S.ImaginaryUnit * Sum(Sum(binomial(k, j) * ((-1)**j * (k - 2*j)**(2*n + 1)) /
(2**k*S.ImaginaryUnit**k * k), (j, 0, k)), (k, 1, 2*n + 1)))
return Em
def _eval_evalf(self, prec):
m = self.args[0]
if m.is_Integer and m.is_nonnegative:
from mpmath import mp
from sympy import Expr
m = m._to_mpmath(prec)
with workprec(prec):
res = mp.eulernum(m)
return Expr._from_mpmath(res, prec)
#----------------------------------------------------------------------------#
# #
# Catalan numbers #
# #
#----------------------------------------------------------------------------#
[docs]class catalan(Function):
r"""
Catalan numbers
The n-th catalan number is given by::
1 / 2*n \
C = ----- | |
n n + 1 \ n /
* catalan(n) gives the n-th Catalan number, C_n
Examples
========
>>> from sympy import (Symbol, binomial, gamma, hyper, polygamma,
... catalan, diff, combsimp, Rational, I)
>>> [ catalan(i) for i in range(1,10) ]
[1, 2, 5, 14, 42, 132, 429, 1430, 4862]
>>> n = Symbol("n", integer=True)
>>> catalan(n)
catalan(n)
Catalan numbers can be transformed into several other, identical
expressions involving other mathematical functions
>>> catalan(n).rewrite(binomial)
binomial(2*n, n)/(n + 1)
>>> catalan(n).rewrite(gamma)
4**n*gamma(n + 1/2)/(sqrt(pi)*gamma(n + 2))
>>> catalan(n).rewrite(hyper)
hyper((-n + 1, -n), (2,), 1)
For some non-integer values of n we can get closed form
expressions by rewriting in terms of gamma functions:
>>> catalan(Rational(1,2)).rewrite(gamma)
8/(3*pi)
We can differentiate the Catalan numbers C(n) interpreted as a
continuous real funtion in n:
>>> diff(catalan(n), n)
(polygamma(0, n + 1/2) - polygamma(0, n + 2) + log(4))*catalan(n)
As a more advanced example consider the following ratio
between consecutive numbers:
>>> combsimp((catalan(n + 1)/catalan(n)).rewrite(binomial))
2*(2*n + 1)/(n + 2)
The Catalan numbers can be generalized to complex numbers:
>>> catalan(I).rewrite(gamma)
4**I*gamma(1/2 + I)/(sqrt(pi)*gamma(2 + I))
and evaluated with arbitrary precision:
>>> catalan(I).evalf(20)
0.39764993382373624267 - 0.020884341620842555705*I
References
==========
.. [1] http://en.wikipedia.org/wiki/Catalan_number
.. [2] http://mathworld.wolfram.com/CatalanNumber.html
.. [3] http://functions.wolfram.com/GammaBetaErf/CatalanNumber/
.. [4] http://geometer.org/mathcircles/catalan.pdf
See Also
========
bell, bernoulli, euler, fibonacci, harmonic, lucas
sympy.functions.combinatorial.factorials.binomial
"""
@classmethod
def eval(cls, n):
from sympy import gamma
if (n.is_Integer and n.is_nonnegative) or \
(n.is_noninteger and n.is_negative):
return 4**n*gamma(n + S.Half)/(gamma(S.Half)*gamma(n + 2))
if (n.is_integer and n.is_negative):
if (n + 1).is_negative:
return S.Zero
if (n + 1).is_zero:
return -S.Half
def fdiff(self, argindex=1):
from sympy import polygamma, log
n = self.args[0]
return catalan(n)*(polygamma(0, n + Rational(1, 2)) - polygamma(0, n + 2) + log(4))
def _eval_rewrite_as_binomial(self, n):
return binomial(2*n, n)/(n + 1)
def _eval_rewrite_as_factorial(self, n):
return factorial(2*n) / (factorial(n+1) * factorial(n))
def _eval_rewrite_as_gamma(self, n):
from sympy import gamma
# The gamma function allows to generalize Catalan numbers to complex n
return 4**n*gamma(n + S.Half)/(gamma(S.Half)*gamma(n + 2))
def _eval_rewrite_as_hyper(self, n):
from sympy import hyper
return hyper([1 - n, -n], [2], 1)
def _eval_rewrite_as_Product(self, n):
from sympy import Product
if not (n.is_integer and n.is_nonnegative):
return self
k = Dummy('k', integer=True, positive=True)
return Product((n + k) / k, (k, 2, n))
def _eval_evalf(self, prec):
from sympy import gamma
if self.args[0].is_number:
return self.rewrite(gamma)._eval_evalf(prec)
#----------------------------------------------------------------------------#
# #
# Genocchi numbers #
# #
#----------------------------------------------------------------------------#
class genocchi(Function):
r"""
Genocchi numbers
The Genocchi numbers are a sequence of integers G_n that satisfy the
relation::
oo
____
\ `
2*t \ n
------ = \ G_n*t
t / ------
e + 1 / n!
/___,
n = 1
Examples
========
>>> from sympy import Symbol
>>> from sympy.functions import genocchi
>>> [genocchi(n) for n in range(1, 9)]
[1, -1, 0, 1, 0, -3, 0, 17]
>>> n = Symbol('n', integer=True, positive=True)
>>> genocchi(2 * n + 1)
0
References
==========
.. [1] https://en.wikipedia.org/wiki/Genocchi_number
.. [2] http://mathworld.wolfram.com/GenocchiNumber.html
See Also
========
bell, bernoulli, catalan, euler, fibonacci, harmonic, lucas
"""
@classmethod
def eval(cls, n):
if n.is_Number:
if (not n.is_Integer) or n.is_nonpositive:
raise ValueError("Genocchi numbers are defined only for " +
"positive integers")
return 2 * (1 - S(2) ** n) * bernoulli(n)
if n.is_odd and (n - 1).is_positive:
return S.Zero
if (n - 1).is_zero:
return S.One
def _eval_rewrite_as_bernoulli(self, n):
if n.is_integer and n.is_nonnegative:
return (1 - S(2) ** n) * bernoulli(n) * 2
def _eval_is_integer(self):
if self.args[0].is_integer and self.args[0].is_positive:
return True
def _eval_is_negative(self):
n = self.args[0]
if n.is_integer and n.is_positive:
if n.is_odd:
return False
return (n / 2).is_odd
def _eval_is_positive(self):
n = self.args[0]
if n.is_integer and n.is_positive:
if n.is_odd:
return fuzzy_not((n - 1).is_positive)
return (n / 2).is_even
def _eval_is_even(self):
n = self.args[0]
if n.is_integer and n.is_positive:
if n.is_even:
return False
return (n - 1).is_positive
def _eval_is_odd(self):
n = self.args[0]
if n.is_integer and n.is_positive:
if n.is_even:
return True
return fuzzy_not((n - 1).is_positive)
def _eval_is_prime(self):
n = self.args[0]
# only G_6 = -3 and G_8 = 17 are prime,
# but SymPy does not consider negatives as prime
# so only n=8 is tested
return (n - 8).is_zero
#######################################################################
###
### Functions for enumerating partitions, permutations and combinations
###
#######################################################################
class _MultisetHistogram(tuple):
pass
_N = -1
_ITEMS = -2
_M = slice(None, _ITEMS)
def _multiset_histogram(n):
"""Return tuple used in permutation and combination counting. Input
is a dictionary giving items with counts as values or a sequence of
items (which need not be sorted).
The data is stored in a class deriving from tuple so it is easily
recognized and so it can be converted easily to a list.
"""
if type(n) is dict: # item: count
if not all(isinstance(v, int) and v >= 0 for v in n.values()):
raise ValueError
tot = sum(n.values())
items = sum(1 for k in n if n[k] > 0)
return _MultisetHistogram([n[k] for k in n if n[k] > 0] + [items, tot])
else:
n = list(n)
s = set(n)
if len(s) == len(n):
n = [1]*len(n)
n.extend([len(n), len(n)])
return _MultisetHistogram(n)
m = dict(zip(s, range(len(s))))
d = dict(zip(range(len(s)), [0]*len(s)))
for i in n:
d[m[i]] += 1
return _multiset_histogram(d)
def nP(n, k=None, replacement=False):
"""Return the number of permutations of ``n`` items taken ``k`` at a time.
Possible values for ``n``::
integer - set of length ``n``
sequence - converted to a multiset internally
multiset - {element: multiplicity}
If ``k`` is None then the total of all permutations of length 0
through the number of items represented by ``n`` will be returned.
If ``replacement`` is True then a given item can appear more than once
in the ``k`` items. (For example, for 'ab' permutations of 2 would
include 'aa', 'ab', 'ba' and 'bb'.) The multiplicity of elements in
``n`` is ignored when ``replacement`` is True but the total number
of elements is considered since no element can appear more times than
the number of elements in ``n``.
Examples
========
>>> from sympy.functions.combinatorial.numbers import nP
>>> from sympy.utilities.iterables import multiset_permutations, multiset
>>> nP(3, 2)
6
>>> nP('abc', 2) == nP(multiset('abc'), 2) == 6
True
>>> nP('aab', 2)
3
>>> nP([1, 2, 2], 2)
3
>>> [nP(3, i) for i in range(4)]
[1, 3, 6, 6]
>>> nP(3) == sum(_)
True
When ``replacement`` is True, each item can have multiplicity
equal to the length represented by ``n``:
>>> nP('aabc', replacement=True)
121
>>> [len(list(multiset_permutations('aaaabbbbcccc', i))) for i in range(5)]
[1, 3, 9, 27, 81]
>>> sum(_)
121
References
==========
.. [1] http://en.wikipedia.org/wiki/Permutation
See Also
========
sympy.utilities.iterables.multiset_permutations
"""
try:
n = as_int(n)
except ValueError:
return Integer(_nP(_multiset_histogram(n), k, replacement))
return Integer(_nP(n, k, replacement))
@cacheit
def _nP(n, k=None, replacement=False):
from sympy.functions.combinatorial.factorials import factorial
from sympy.core.mul import prod
if k == 0:
return 1
if isinstance(n, SYMPY_INTS): # n different items
# assert n >= 0
if k is None:
return sum(_nP(n, i, replacement) for i in range(n + 1))
elif replacement:
return n**k
elif k > n:
return 0
elif k == n:
return factorial(k)
elif k == 1:
return n
else:
# assert k >= 0
return _product(n - k + 1, n)
elif isinstance(n, _MultisetHistogram):
if k is None:
return sum(_nP(n, i, replacement) for i in range(n[_N] + 1))
elif replacement:
return n[_ITEMS]**k
elif k == n[_N]:
return factorial(k)/prod([factorial(i) for i in n[_M] if i > 1])
elif k > n[_N]:
return 0
elif k == 1:
return n[_ITEMS]
else:
# assert k >= 0
tot = 0
n = list(n)
for i in range(len(n[_M])):
if not n[i]:
continue
n[_N] -= 1
if n[i] == 1:
n[i] = 0
n[_ITEMS] -= 1
tot += _nP(_MultisetHistogram(n), k - 1)
n[_ITEMS] += 1
n[i] = 1
else:
n[i] -= 1
tot += _nP(_MultisetHistogram(n), k - 1)
n[i] += 1
n[_N] += 1
return tot
@cacheit
def _AOP_product(n):
"""for n = (m1, m2, .., mk) return the coefficients of the polynomial,
prod(sum(x**i for i in range(nj + 1)) for nj in n); i.e. the coefficients
of the product of AOPs (all-one polynomials) or order given in n. The
resulting coefficient corresponding to x**r is the number of r-length
combinations of sum(n) elements with multiplicities given in n.
The coefficients are given as a default dictionary (so if a query is made
for a key that is not present, 0 will be returned).
Examples
========
>>> from sympy.functions.combinatorial.numbers import _AOP_product
>>> from sympy.abc import x
>>> n = (2, 2, 3) # e.g. aabbccc
>>> prod = ((x**2 + x + 1)*(x**2 + x + 1)*(x**3 + x**2 + x + 1)).expand()
>>> c = _AOP_product(n); dict(c)
{0: 1, 1: 3, 2: 6, 3: 8, 4: 8, 5: 6, 6: 3, 7: 1}
>>> [c[i] for i in range(8)] == [prod.coeff(x, i) for i in range(8)]
True
The generating poly used here is the same as that listed in
http://tinyurl.com/cep849r, but in a refactored form.
"""
from collections import defaultdict
n = list(n)
ord = sum(n)
need = (ord + 2)//2
rv = [1]*(n.pop() + 1)
rv.extend([0]*(need - len(rv)))
rv = rv[:need]
while n:
ni = n.pop()
N = ni + 1
was = rv[:]
for i in range(1, min(N, len(rv))):
rv[i] += rv[i - 1]
for i in range(N, need):
rv[i] += rv[i - 1] - was[i - N]
rev = list(reversed(rv))
if ord % 2:
rv = rv + rev
else:
rv[-1:] = rev
d = defaultdict(int)
for i in range(len(rv)):
d[i] = rv[i]
return d
def nC(n, k=None, replacement=False):
"""Return the number of combinations of ``n`` items taken ``k`` at a time.
Possible values for ``n``::
integer - set of length ``n``
sequence - converted to a multiset internally
multiset - {element: multiplicity}
If ``k`` is None then the total of all combinations of length 0
through the number of items represented in ``n`` will be returned.
If ``replacement`` is True then a given item can appear more than once
in the ``k`` items. (For example, for 'ab' sets of 2 would include 'aa',
'ab', and 'bb'.) The multiplicity of elements in ``n`` is ignored when
``replacement`` is True but the total number of elements is considered
since no element can appear more times than the number of elements in
``n``.
Examples
========
>>> from sympy.functions.combinatorial.numbers import nC
>>> from sympy.utilities.iterables import multiset_combinations
>>> nC(3, 2)
3
>>> nC('abc', 2)
3
>>> nC('aab', 2)
2
When ``replacement`` is True, each item can have multiplicity
equal to the length represented by ``n``:
>>> nC('aabc', replacement=True)
35
>>> [len(list(multiset_combinations('aaaabbbbcccc', i))) for i in range(5)]
[1, 3, 6, 10, 15]
>>> sum(_)
35
If there are ``k`` items with multiplicities ``m_1, m_2, ..., m_k``
then the total of all combinations of length 0 hrough ``k`` is the
product, ``(m_1 + 1)*(m_2 + 1)*...*(m_k + 1)``. When the multiplicity
of each item is 1 (i.e., k unique items) then there are 2**k
combinations. For example, if there are 4 unique items, the total number
of combinations is 16:
>>> sum(nC(4, i) for i in range(5))
16
References
==========
.. [1] http://en.wikipedia.org/wiki/Combination
.. [2] http://tinyurl.com/cep849r
See Also
========
sympy.utilities.iterables.multiset_combinations
"""
from sympy.functions.combinatorial.factorials import binomial
from sympy.core.mul import prod
if isinstance(n, SYMPY_INTS):
if k is None:
if not replacement:
return 2**n
return sum(nC(n, i, replacement) for i in range(n + 1))
if k < 0:
raise ValueError("k cannot be negative")
if replacement:
return binomial(n + k - 1, k)
return binomial(n, k)
if isinstance(n, _MultisetHistogram):
N = n[_N]
if k is None:
if not replacement:
return prod(m + 1 for m in n[_M])
return sum(nC(n, i, replacement) for i in range(N + 1))
elif replacement:
return nC(n[_ITEMS], k, replacement)
# assert k >= 0
elif k in (1, N - 1):
return n[_ITEMS]
elif k in (0, N):
return 1
return _AOP_product(tuple(n[_M]))[k]
else:
return nC(_multiset_histogram(n), k, replacement)
@cacheit
def _stirling1(n, k):
if n == k == 0:
return S.One
if 0 in (n, k):
return S.Zero
n1 = n - 1
# some special values
if n == k:
return S.One
elif k == 1:
return factorial(n1)
elif k == n1:
return binomial(n, 2)
elif k == n - 2:
return (3*n - 1)*binomial(n, 3)/4
elif k == n - 3:
return binomial(n, 2)*binomial(n, 4)
# general recurrence
return n1*_stirling1(n1, k) + _stirling1(n1, k - 1)
@cacheit
def _stirling2(n, k):
if n == k == 0:
return S.One
if 0 in (n, k):
return S.Zero
n1 = n - 1
# some special values
if k == n1:
return binomial(n, 2)
elif k == 2:
return 2**n1 - 1
# general recurrence
return k*_stirling2(n1, k) + _stirling2(n1, k - 1)
[docs]def stirling(n, k, d=None, kind=2, signed=False):
"""Return Stirling number S(n, k) of the first or second (default) kind.
The sum of all Stirling numbers of the second kind for k = 1
through n is bell(n). The recurrence relationship for these numbers
is::
{0} {n} {0} {n + 1} {n} { n }
{ } = 1; { } = { } = 0; { } = j*{ } + { }
{0} {0} {k} { k } {k} {k - 1}
where ``j`` is::
``n`` for Stirling numbers of the first kind
``-n`` for signed Stirling numbers of the first kind
``k`` for Stirling numbers of the second kind
The first kind of Stirling number counts the number of permutations of
``n`` distinct items that have ``k`` cycles; the second kind counts the
ways in which ``n`` distinct items can be partitioned into ``k`` parts.
If ``d`` is given, the "reduced Stirling number of the second kind" is
returned: ``S^{d}(n, k) = S(n - d + 1, k - d + 1)`` with ``n >= k >= d``.
(This counts the ways to partition ``n`` consecutive integers into
``k`` groups with no pairwise difference less than ``d``. See example
below.)
To obtain the signed Stirling numbers of the first kind, use keyword
``signed=True``. Using this keyword automatically sets ``kind`` to 1.
Examples
========
>>> from sympy.functions.combinatorial.numbers import stirling, bell
>>> from sympy.combinatorics import Permutation
>>> from sympy.utilities.iterables import multiset_partitions, permutations
First kind (unsigned by default):
>>> [stirling(6, i, kind=1) for i in range(7)]
[0, 120, 274, 225, 85, 15, 1]
>>> perms = list(permutations(range(4)))
>>> [sum(Permutation(p).cycles == i for p in perms) for i in range(5)]
[0, 6, 11, 6, 1]
>>> [stirling(4, i, kind=1) for i in range(5)]
[0, 6, 11, 6, 1]
First kind (signed):
>>> [stirling(4, i, signed=True) for i in range(5)]
[0, -6, 11, -6, 1]
Second kind:
>>> [stirling(10, i) for i in range(12)]
[0, 1, 511, 9330, 34105, 42525, 22827, 5880, 750, 45, 1, 0]
>>> sum(_) == bell(10)
True
>>> len(list(multiset_partitions(range(4), 2))) == stirling(4, 2)
True
Reduced second kind:
>>> from sympy import subsets, oo
>>> def delta(p):
... if len(p) == 1:
... return oo
... return min(abs(i[0] - i[1]) for i in subsets(p, 2))
>>> parts = multiset_partitions(range(5), 3)
>>> d = 2
>>> sum(1 for p in parts if all(delta(i) >= d for i in p))
7
>>> stirling(5, 3, 2)
7
References
==========
.. [1] http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind
.. [2] http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind
See Also
========
sympy.utilities.iterables.multiset_partitions
"""
# TODO: make this a class like bell()
n = as_int(n)
k = as_int(k)
if n < 0:
raise ValueError('n must be nonnegative')
if k > n:
return S.Zero
if d:
# assert k >= d
# kind is ignored -- only kind=2 is supported
return _stirling2(n - d + 1, k - d + 1)
elif signed:
# kind is ignored -- only kind=1 is supported
return (-1)**(n - k)*_stirling1(n, k)
if kind == 1:
return _stirling1(n, k)
elif kind == 2:
return _stirling2(n, k)
else:
raise ValueError('kind must be 1 or 2, not %s' % k)
@cacheit
def _nT(n, k):
"""Return the partitions of ``n`` items into ``k`` parts. This
is used by ``nT`` for the case when ``n`` is an integer."""
if k == 0:
return 1 if k == n else 0
return sum(_nT(n - k, j) for j in range(min(k, n - k) + 1))
def nT(n, k=None):
"""Return the number of ``k``-sized partitions of ``n`` items.
Possible values for ``n``::
integer - ``n`` identical items
sequence - converted to a multiset internally
multiset - {element: multiplicity}
Note: the convention for ``nT`` is different than that of ``nC`` and
``nP`` in that
here an integer indicates ``n`` *identical* items instead of a set of
length ``n``; this is in keeping with the ``partitions`` function which
treats its integer-``n`` input like a list of ``n`` 1s. One can use
``range(n)`` for ``n`` to indicate ``n`` distinct items.
If ``k`` is None then the total number of ways to partition the elements
represented in ``n`` will be returned.
Examples
========
>>> from sympy.functions.combinatorial.numbers import nT
Partitions of the given multiset:
>>> [nT('aabbc', i) for i in range(1, 7)]
[1, 8, 11, 5, 1, 0]
>>> nT('aabbc') == sum(_)
True
>>> [nT("mississippi", i) for i in range(1, 12)]
[1, 74, 609, 1521, 1768, 1224, 579, 197, 50, 9, 1]
Partitions when all items are identical:
>>> [nT(5, i) for i in range(1, 6)]
[1, 2, 2, 1, 1]
>>> nT('1'*5) == sum(_)
True
When all items are different:
>>> [nT(range(5), i) for i in range(1, 6)]
[1, 15, 25, 10, 1]
>>> nT(range(5)) == sum(_)
True
References
==========
.. [1] http://undergraduate.csse.uwa.edu.au/units/CITS7209/partition.pdf
See Also
========
sympy.utilities.iterables.partitions
sympy.utilities.iterables.multiset_partitions
"""
from sympy.utilities.enumerative import MultisetPartitionTraverser
if isinstance(n, SYMPY_INTS):
# assert n >= 0
# all the same
if k is None:
return sum(_nT(n, k) for k in range(1, n + 1))
return _nT(n, k)
if not isinstance(n, _MultisetHistogram):
try:
# if n contains hashable items there is some
# quick handling that can be done
u = len(set(n))
if u == 1:
return nT(len(n), k)
elif u == len(n):
n = range(u)
raise TypeError
except TypeError:
n = _multiset_histogram(n)
N = n[_N]
if k is None and N == 1:
return 1
if k in (1, N):
return 1
if k == 2 or N == 2 and k is None:
m, r = divmod(N, 2)
rv = sum(nC(n, i) for i in range(1, m + 1))
if not r:
rv -= nC(n, m)//2
if k is None:
rv += 1 # for k == 1
return rv
if N == n[_ITEMS]:
# all distinct
if k is None:
return bell(N)
return stirling(N, k)
m = MultisetPartitionTraverser()
if k is None:
return m.count_partitions(n[_M])
# MultisetPartitionTraverser does not have a range-limited count
# method, so need to enumerate and count
tot = 0
for discard in m.enum_range(n[_M], k-1, k):
tot += 1
return tot
