Source code for sympy.functions.combinatorial.numbers

"""
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 sympy import Function, S, Symbol, Rational, oo, Integer, C, Add, expand_mul

from sympy.mpmath import bernfrac
from sympy.mpmath.libmp import ifib as _ifib

def _product(a, b):
    p = 1
    for k in range(a, b+1):
        p *= k
    return p

from sympy.utilities.memoization import recurrence_memo


# Dummy symbol used for computing polynomial sequences
_sym = Symbol('x')
_symbols = Function('x')


#----------------------------------------------------------------------------#
#                                                                            #
#                           Fibonacci numbers                                #
#                                                                            #
#----------------------------------------------------------------------------#

[docs]class fibonacci(Function): """ Fibonacci numbers / Fibonacci polynomials The Fibonacci numbers are the integer sequence defined by the initial terms F_0 = 0, F_1 = 1 and the two-term recurrence relation F_n = F_{n-1} + F_{n-2}. The Fibonacci polynomials are defined by F_1(x) = 1, F_2(x) = x, and F_n(x) = x*F_{n-1}(x) + F_{n-2}(x) for n > 2. For all positive integers n, F_n(1) = F_n. * fibonacci(n) gives the nth Fibonacci number, F_n * fibonacci(n, x) gives the nth Fibonacci polynomial in x, F_n(x) 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 ========== * http://en.wikipedia.org/wiki/Fibonacci_number * http://mathworld.wolfram.com/FibonacciNumber.html See Also ======== 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_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)
[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 ========== * http://en.wikipedia.org/wiki/Lucas_number See Also ======== fibonacci """ @classmethod def eval(cls, n): if n.is_Integer: return fibonacci(n+1) + fibonacci(n-1) #----------------------------------------------------------------------------# # # # 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 ========== * http://en.wikipedia.org/wiki/Bernoulli_number * http://en.wikipedia.org/wiki/Bernoulli_polynomial See Also ======== euler, bell """ # Calculates B_n for positive even n @staticmethod def _calc_bernoulli(n): s = 0 a = int(C.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 / C.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(C.binomial(n, k)*cls(k)*sym**(n-k)) return Add(*result) else: raise ValueError("Bernoulli numbers are defined only" " for nonnegative integer indices.") #----------------------------------------------------------------------------# # # # 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 ========== * http://en.wikipedia.org/wiki/Bell_number * http://mathworld.wolfram.com/BellNumber.html * http://mathworld.wolfram.com/BellPolynomial.html See Also ======== euler, bernoulli """ @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 #----------------------------------------------------------------------------# # # # Harmonic numbers # # # #----------------------------------------------------------------------------#
[docs]class harmonic(Function): r""" Harmonic numbers The nth harmonic number is given by 1 + 1/2 + 1/3 + ... + 1/n. More generally:: n ___ \ -m H = ) k . n,m /___ k = 1 As n -> oo, H_{n,m} -> zeta(m) (the Riemann zeta function) * harmonic(n) gives the nth harmonic number, H_n * harmonic(n, m) gives the nth generalized harmonic number of order m, 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 References ========== * http://en.wikipedia.org/wiki/Harmonic_number """ # Generate one memoized Harmonic number-generating function for each # order and store it in a dictionary _functions = {} nargs = (1, 2) @classmethod def eval(cls, n, m=None): if m is None: m = S.One if n == oo: return C.zeta(m) 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)) #----------------------------------------------------------------------------# # # # 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, 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 ========== * http://en.wikipedia.org/wiki/Euler_numbers * http://mathworld.wolfram.com/EulerNumber.html * http://en.wikipedia.org/wiki/Alternating_permutation * http://mathworld.wolfram.com/AlternatingPermutation.html See Also ======== bernoulli, bell """ nargs = 1 @classmethod def eval(cls, m, evaluate=True): if not evaluate: return if m.is_odd: return S.Zero if m.is_Integer and m.is_nonnegative: from sympy.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): if arg.is_even: k = C.Dummy("k", integer=True) j = C.Dummy("j", integer=True) n = self.args[0] / 2 Em = (S.ImaginaryUnit * C.Sum( C.Sum( C.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 sympy.mpmath import mp from sympy import Expr m = m._to_mpmath(prec) oprec = mp.prec mp.prec = prec res = mp.eulernum(m) mp.prec = oprec 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) + 2*log(2))*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 ========== * http://en.wikipedia.org/wiki/Catalan_number * http://mathworld.wolfram.com/CatalanNumber.html * http://geometer.org/mathcircles/catalan.pdf See Also ======== sympy.functions.combinatorial.factorials.binomial """ @classmethod def eval(cls, n, evaluate=True): if n.is_Integer and n.is_nonnegative: return 4**n*C.gamma(n + S.Half)/(C.gamma(S.Half)*C.gamma(n+2)) def fdiff(self, argindex=1): n = self.args[0] return catalan(n)*(C.polygamma(0,n+Rational(1,2))-C.polygamma(0,n+2)+C.log(4)) def _eval_rewrite_as_binomial(self,n): return C.binomial(2*n,n)/(n + 1) def _eval_rewrite_as_gamma(self,n): # The gamma function allows to generalize Catalan numbers to complex n return 4**n*C.gamma(n + S.Half)/(C.gamma(S.Half)*C.gamma(n+2)) def _eval_rewrite_as_hyper(self,n): return C.hyper([1-n,-n],[2],1) def _eval_evalf(self, prec): return self.rewrite(C.gamma).evalf(prec)