""" 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 math import prod from collections import defaultdict from typing import Callable, Dict as tDict, Tuple as tTuple from sympy.core import S, Symbol, Add, Dummy from sympy.core.cache import cacheit from sympy.core.evalf import pure_complex from sympy.core.expr import Expr from sympy.core.function import Function, expand_mul from sympy.core.logic import fuzzy_not from sympy.core.mul import Mul from sympy.core.numbers import E, pi, oo, Rational, Integer from sympy.core.relational import is_le, is_gt from sympy.external.gmpy import SYMPY_INTS from sympy.functions.combinatorial.factorials import (binomial, factorial, subfactorial) from sympy.ntheory.primetest import isprime, is_square from sympy.utilities.enumerative import MultisetPartitionTraverser from sympy.utilities.exceptions import sympy_deprecation_warning from sympy.utilities.iterables import multiset, multiset_derangements, iterable from sympy.utilities.memoization import recurrence_memo from sympy.utilities.misc import as_int from mpmath import bernfrac, workprec from mpmath.libmp import ifib as _ifib def _product(a, b): return prod(range(a, b + 1)) # Dummy symbol used for computing polynomial sequences _sym = Symbol('x') #----------------------------------------------------------------------------# # # # Carmichael numbers # # # #----------------------------------------------------------------------------# def _divides(p, n): return n % p == 0 class carmichael(Function): r""" Carmichael Numbers: Certain cryptographic algorithms make use of big prime numbers. However, checking whether a big number is prime is not so easy. Randomized prime number checking tests exist that offer a high degree of confidence of accurate determination at low cost, such as the Fermat test. Let 'a' be a random number between $2$ and $n - 1$, where $n$ is the number whose primality we are testing. Then, $n$ is probably prime if it satisfies the modular arithmetic congruence relation: .. math :: a^{n-1} = 1 \pmod{n} (where mod refers to the modulo operation) If a number passes the Fermat test several times, then it is prime with a high probability. Unfortunately, certain composite numbers (non-primes) still pass the Fermat test with every number smaller than themselves. These numbers are called Carmichael numbers. A Carmichael number will pass a Fermat primality test to every base $b$ relatively prime to the number, even though it is not actually prime. This makes tests based on Fermat's Little Theorem less effective than strong probable prime tests such as the Baillie-PSW primality test and the Miller-Rabin primality test. Examples ======== >>> from sympy import carmichael >>> carmichael.find_first_n_carmichaels(5) [561, 1105, 1729, 2465, 2821] >>> carmichael.find_carmichael_numbers_in_range(0, 562) [561] >>> carmichael.find_carmichael_numbers_in_range(0,1000) [561] >>> carmichael.find_carmichael_numbers_in_range(0,2000) [561, 1105, 1729] References ========== .. [1] https://en.wikipedia.org/wiki/Carmichael_number .. [2] https://en.wikipedia.org/wiki/Fermat_primality_test .. [3] https://www.jstor.org/stable/23248683?seq=1#metadata_info_tab_contents """ @staticmethod def is_perfect_square(n): sympy_deprecation_warning( """ is_perfect_square is just a wrapper around sympy.ntheory.primetest.is_square so use that directly instead. """, deprecated_since_version="1.11", active_deprecations_target='deprecated-carmichael-static-methods', ) return is_square(n) @staticmethod def divides(p, n): sympy_deprecation_warning( """ divides can be replaced by directly testing n % p == 0. """, deprecated_since_version="1.11", active_deprecations_target='deprecated-carmichael-static-methods', ) return n % p == 0 @staticmethod def is_prime(n): sympy_deprecation_warning( """ is_prime is just a wrapper around sympy.ntheory.primetest.isprime so use that directly instead. """, deprecated_since_version="1.11", active_deprecations_target='deprecated-carmichael-static-methods', ) return isprime(n) @staticmethod def is_carmichael(n): if n >= 0: if (n == 1) or isprime(n) or (n % 2 == 0): return False divisors = [1, n] # get divisors divisors.extend([i for i in range(3, n // 2 + 1, 2) if n % i == 0]) for i in divisors: if is_square(i) and i != 1: return False if isprime(i): if not _divides(i - 1, n - 1): return False return True else: raise ValueError('The provided number must be greater than or equal to 0') @staticmethod def find_carmichael_numbers_in_range(x, y): if 0 <= x <= y: if x % 2 == 0: return [i for i in range(x + 1, y, 2) if carmichael.is_carmichael(i)] else: return [i for i in range(x, y, 2) if carmichael.is_carmichael(i)] else: raise ValueError('The provided range is not valid. x and y must be non-negative integers and x <= y') @staticmethod def find_first_n_carmichaels(n): i = 1 carmichaels = list() while len(carmichaels) < n: if carmichael.is_carmichael(i): carmichaels.append(i) i += 2 return carmichaels #----------------------------------------------------------------------------# # # # Fibonacci numbers # # # #----------------------------------------------------------------------------# 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 `n^{th}` Fibonacci number, `F_n` * ``fibonacci(n, x)`` gives the `n^{th}` 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 See Also ======== bell, bernoulli, catalan, euler, harmonic, lucas, genocchi, partition, tribonacci References ========== .. [1] https://en.wikipedia.org/wiki/Fibonacci_number .. [2] http://mathworld.wolfram.com/FibonacciNumber.html """ @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: if sym is None: n = int(n) if n < 0: return S.NegativeOne**(n + 1) * fibonacci(-n) else: 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, **kwargs): from sympy.functions.elementary.miscellaneous import sqrt return 2**(-n)*sqrt(5)*((1 + sqrt(5))**n - (-sqrt(5) + 1)**n) / 5 def _eval_rewrite_as_GoldenRatio(self,n, **kwargs): return (S.GoldenRatio**n - 1/(-S.GoldenRatio)**n)/(2*S.GoldenRatio-1) #----------------------------------------------------------------------------# # # # Lucas numbers # # # #----------------------------------------------------------------------------# 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 `n^{th}` 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] See Also ======== bell, bernoulli, catalan, euler, fibonacci, harmonic, genocchi, partition, tribonacci References ========== .. [1] https://en.wikipedia.org/wiki/Lucas_number .. [2] http://mathworld.wolfram.com/LucasNumber.html """ @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, **kwargs): from sympy.functions.elementary.miscellaneous import sqrt return 2**(-n)*((1 + sqrt(5))**n + (-sqrt(5) + 1)**n) #----------------------------------------------------------------------------# # # # Tribonacci numbers # # # #----------------------------------------------------------------------------# class tribonacci(Function): r""" Tribonacci numbers / Tribonacci polynomials The Tribonacci numbers are the integer sequence defined by the initial terms `T_0 = 0`, `T_1 = 1`, `T_2 = 1` and the three-term recurrence relation `T_n = T_{n-1} + T_{n-2} + T_{n-3}`. The Tribonacci polynomials are defined by `T_0(x) = 0`, `T_1(x) = 1`, `T_2(x) = x^2`, and `T_n(x) = x^2 T_{n-1}(x) + x T_{n-2}(x) + T_{n-3}(x)` for `n > 2`. For all positive integers `n`, `T_n(1) = T_n`. * ``tribonacci(n)`` gives the `n^{th}` Tribonacci number, `T_n` * ``tribonacci(n, x)`` gives the `n^{th}` Tribonacci polynomial in `x`, `T_n(x)` Examples ======== >>> from sympy import tribonacci, Symbol >>> [tribonacci(x) for x in range(11)] [0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149] >>> tribonacci(5, Symbol('t')) t**8 + 3*t**5 + 3*t**2 See Also ======== bell, bernoulli, catalan, euler, fibonacci, harmonic, lucas, genocchi, partition References ========== .. [1] https://en.wikipedia.org/wiki/Generalizations_of_Fibonacci_numbers#Tribonacci_numbers .. [2] http://mathworld.wolfram.com/TribonacciNumber.html .. [3] https://oeis.org/A000073 """ @staticmethod @recurrence_memo([S.Zero, S.One, S.One]) def _trib(n, prev): return (prev[-3] + prev[-2] + prev[-1]) @staticmethod @recurrence_memo([S.Zero, S.One, _sym**2]) def _tribpoly(n, prev): return (prev[-3] + _sym*prev[-2] + _sym**2*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: raise ValueError("Tribonacci polynomials are defined " "only for non-negative integer indices.") if sym is None: return Integer(cls._trib(n)) else: return cls._tribpoly(n).subs(_sym, sym) def _eval_rewrite_as_sqrt(self, n, **kwargs): from sympy.functions.elementary.miscellaneous import cbrt, sqrt w = (-1 + S.ImaginaryUnit * sqrt(3)) / 2 a = (1 + cbrt(19 + 3*sqrt(33)) + cbrt(19 - 3*sqrt(33))) / 3 b = (1 + w*cbrt(19 + 3*sqrt(33)) + w**2*cbrt(19 - 3*sqrt(33))) / 3 c = (1 + w**2*cbrt(19 + 3*sqrt(33)) + w*cbrt(19 - 3*sqrt(33))) / 3 Tn = (a**(n + 1)/((a - b)*(a - c)) + b**(n + 1)/((b - a)*(b - c)) + c**(n + 1)/((c - a)*(c - b))) return Tn def _eval_rewrite_as_TribonacciConstant(self, n, **kwargs): from sympy.functions.elementary.integers import floor from sympy.functions.elementary.miscellaneous import cbrt, sqrt b = cbrt(586 + 102*sqrt(33)) Tn = 3 * b * S.TribonacciConstant**n / (b**2 - 2*b + 4) return floor(Tn + S.Half) #----------------------------------------------------------------------------# # # # Bernoulli numbers # # # #----------------------------------------------------------------------------# 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`): .. math :: 0 = \sum_{k=0}^n \binom{n+1}{k} B_k They are also commonly defined by their exponential generating function, which is `\frac{x}{e^x - 1}`. For odd indices > 1, the Bernoulli numbers are zero. The Bernoulli polynomials satisfy the analogous formula: .. math :: B_n(x) = \sum_{k=0}^n \binom{n}{k} B_k x^{n-k} Bernoulli numbers and Bernoulli polynomials are related as `B_n(0) = B_n`. We compute Bernoulli numbers using Ramanujan's formula: .. math :: B_n = \frac{A(n) - S(n)}{\binom{n+3}{n}} where: .. math :: A(n) = \begin{cases} \frac{n+3}{3} & n \equiv 0\ \text{or}\ 2 \pmod{6} \\ -\frac{n+3}{6} & n \equiv 4 \pmod{6} \end{cases} and: .. math :: S(n) = \sum_{k=1}^{[n/6]} \binom{n+3}{n-6k} B_{n-6k} 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 See Also ======== bell, catalan, euler, fibonacci, harmonic, lucas, genocchi, partition, tribonacci References ========== .. [1] https://en.wikipedia.org/wiki/Bernoulli_number .. [2] https://en.wikipedia.org/wiki/Bernoulli_polynomial .. [3] http://mathworld.wolfram.com/BernoulliNumber.html .. [4] http://mathworld.wolfram.com/BernoulliPolynomial.html """ args: tTuple[Integer] # 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_zero: return S.One elif n is S.One: if sym is None: return Rational(-1, 2) 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 = int(n) result = [binomial(n, k)*cls(k)*sym**(n - k) for k in range(n + 1)] 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 # # # #----------------------------------------------------------------------------# 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 See Also ======== bernoulli, catalan, euler, fibonacci, harmonic, lucas, genocchi, partition, tribonacci References ========== .. [1] https://en.wikipedia.org/wiki/Bell_number .. [2] http://mathworld.wolfram.com/BellNumber.html .. [3] http://mathworld.wolfram.com/BellPolynomial.html """ @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 \ge 1` `B_{0,k} = 0; for k \ge 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 S.Infinity: if k_sym is None: return S.Infinity else: raise ValueError("Bell polynomial is not defined") if n.is_negative or n.is_integer is False: raise ValueError("a non-negative integer expected") 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, **kwargs): from sympy.concrete.summations 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 # # # #----------------------------------------------------------------------------# 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(2*pi/7)**(2*cos(16*pi/7))/(14*sin(pi/7)**(2*cos(pi/7))*cos(pi/14)**(2*sin(pi/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 cannot compute the general relation yet: >>> limit(harmonic(n, m), n, oo) harmonic(oo, m) which equals ``zeta(m)`` for ``m > 1``. See Also ======== bell, bernoulli, catalan, euler, fibonacci, lucas, genocchi, partition, tribonacci References ========== .. [1] https://en.wikipedia.org/wiki/Harmonic_number .. [2] http://functions.wolfram.com/GammaBetaErf/HarmonicNumber/ .. [3] http://functions.wolfram.com/GammaBetaErf/HarmonicNumber2/ """ # Generate one memoized Harmonic number-generating function for each # order and store it in a dictionary _functions = {} # type: tDict[Integer, Callable[[int], Rational]] @classmethod def eval(cls, n, m=None): from sympy.functions.special.zeta_functions 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: if m.is_negative: return S.NaN elif is_le(m, S.One): return S.Infinity elif is_gt(m, S.One): return zeta(m) else: return if n == 0: return S.Zero if n.is_Integer and n.is_nonnegative and m.is_Integer: if m not 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, **kwargs): 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, **kwargs): from sympy.functions.special.gamma_functions import polygamma return self.rewrite(polygamma) def _eval_rewrite_as_trigamma(self, n, m=1, **kwargs): from sympy.functions.special.gamma_functions import polygamma return self.rewrite(polygamma) def _eval_rewrite_as_Sum(self, n, m=None, **kwargs): from sympy.concrete.summations 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.concrete.summations 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: from sympy.functions.elementary.exponential import log from sympy.functions.elementary.integers import floor from sympy.functions.elementary.trigonometric import sin, cos, cot 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, limitvar=None, **kwargs): from sympy.functions.special.gamma_functions import polygamma return self.rewrite(polygamma).rewrite("tractable", deep=True) def _eval_evalf(self, prec): from sympy.functions.special.gamma_functions import polygamma if all(i.is_number for i in self.args): return self.rewrite(polygamma)._eval_evalf(prec) #----------------------------------------------------------------------------# # # # Euler numbers # # # #----------------------------------------------------------------------------# class euler(Function): r""" Euler numbers / Euler polynomials The Euler numbers are given by: .. math:: E_{2n} = I \sum_{k=1}^{2n+1} \sum_{j=0}^k \binom{k}{j} \frac{(-1)^j (k-2j)^{2n+1}}{2^k I^k k} .. math:: E_{2n+1} = 0 Euler numbers and Euler polynomials are related by .. math:: E_n = 2^n E_n\left(\frac{1}{2}\right). We compute symbolic Euler polynomials using [5]_ .. math:: E_n(x) = \sum_{k=0}^n \binom{n}{k} \frac{E_k}{2^k} \left(x - \frac{1}{2}\right)^{n-k}. However, numerical evaluation of the Euler polynomial is computed more efficiently (and more accurately) using the mpmath library. * ``euler(n)`` gives the `n^{th}` Euler number, `E_n`. * ``euler(n, x)`` gives the `n^{th}` Euler polynomial, `E_n(x)`. Examples ======== >>> from sympy import euler, Symbol, S >>> [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) >>> x = Symbol("x") >>> euler(n, x) euler(n, x) >>> euler(0, x) 1 >>> euler(1, x) x - 1/2 >>> euler(2, x) x**2 - x >>> euler(3, x) x**3 - 3*x**2/2 + 1/4 >>> euler(4, x) x**4 - 2*x**3 + x >>> euler(12, S.Half) 2702765/4096 >>> euler(12) 2702765 See Also ======== bell, bernoulli, catalan, fibonacci, harmonic, lucas, genocchi, partition, tribonacci References ========== .. [1] https://en.wikipedia.org/wiki/Euler_numbers .. [2] http://mathworld.wolfram.com/EulerNumber.html .. [3] https://en.wikipedia.org/wiki/Alternating_permutation .. [4] http://mathworld.wolfram.com/AlternatingPermutation.html .. [5] http://dlmf.nist.gov/24.2#ii """ @classmethod def eval(cls, m, sym=None): if m.is_Number: if m.is_Integer and m.is_nonnegative: # Euler numbers if sym is None: if m.is_odd: return S.Zero from mpmath import mp m = m._to_mpmath(mp.prec) res = mp.eulernum(m, exact=True) return Integer(res) # Euler polynomial else: reim = pure_complex(sym, or_real=True) # Evaluate polynomial numerically using mpmath if reim and all(a.is_Float or a.is_Integer for a in reim) \ and any(a.is_Float for a in reim): from mpmath import mp m = int(m) # XXX ComplexFloat (#12192) would be nice here, above prec = min([a._prec for a in reim if a.is_Float]) with workprec(prec): res = mp.eulerpoly(m, sym) return Expr._from_mpmath(res, prec) # Construct polynomial symbolically from definition m = int(m) result = [binomial(m, k)*cls(k)/(2**k)*(sym - S.Half)**(m - k) for k in range(m + 1)] return Add(*result).expand() else: raise ValueError("Euler numbers are defined only" " for nonnegative integer indices.") if sym is None: if m.is_odd and m.is_positive: return S.Zero def _eval_rewrite_as_Sum(self, n, x=None, **kwargs): from sympy.concrete.summations import Sum if x is None and n.is_even: k = Dummy("k", integer=True) j = Dummy("j", integer=True) n = n / 2 Em = (S.ImaginaryUnit * Sum(Sum(binomial(k, j) * (S.NegativeOne**j * (k - 2*j)**(2*n + 1)) / (2**k*S.ImaginaryUnit**k * k), (j, 0, k)), (k, 1, 2*n + 1))) return Em if x: k = Dummy("k", integer=True) return Sum(binomial(n, k)*euler(k)/2**k*(x - S.Half)**(n - k), (k, 0, n)) def _eval_evalf(self, prec): m, x = (self.args[0], None) if len(self.args) == 1 else self.args if x is None and m.is_Integer and m.is_nonnegative: from mpmath import mp m = m._to_mpmath(prec) with workprec(prec): res = mp.eulernum(m) return Expr._from_mpmath(res, prec) if x and x.is_number and m.is_Integer and m.is_nonnegative: from mpmath import mp m = int(m) x = x._to_mpmath(prec) with workprec(prec): res = mp.eulerpoly(m, x) return Expr._from_mpmath(res, prec) #----------------------------------------------------------------------------# # # # Catalan numbers # # # #----------------------------------------------------------------------------# class catalan(Function): r""" Catalan numbers The `n^{th}` catalan number is given by: .. math :: C_n = \frac{1}{n+1} \binom{2n}{n} * ``catalan(n)`` gives the `n^{th}` Catalan number, `C_n` Examples ======== >>> from sympy import (Symbol, binomial, gamma, hyper, ... 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((1 - n, -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 function 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 See Also ======== bell, bernoulli, euler, fibonacci, harmonic, lucas, genocchi, partition, tribonacci sympy.functions.combinatorial.factorials.binomial References ========== .. [1] https://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 """ @classmethod def eval(cls, n): from sympy.functions.special.gamma_functions 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 Rational(-1, 2) def fdiff(self, argindex=1): from sympy.functions.elementary.exponential import log from sympy.functions.special.gamma_functions import polygamma n = self.args[0] return catalan(n)*(polygamma(0, n + S.Half) - polygamma(0, n + 2) + log(4)) def _eval_rewrite_as_binomial(self, n, **kwargs): return binomial(2*n, n)/(n + 1) def _eval_rewrite_as_factorial(self, n, **kwargs): return factorial(2*n) / (factorial(n+1) * factorial(n)) def _eval_rewrite_as_gamma(self, n, piecewise=True, **kwargs): from sympy.functions.special.gamma_functions 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, **kwargs): from sympy.functions.special.hyper import hyper return hyper([1 - n, -n], [2], 1) def _eval_rewrite_as_Product(self, n, **kwargs): from sympy.concrete.products 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_is_integer(self): if self.args[0].is_integer and self.args[0].is_nonnegative: return True def _eval_is_positive(self): if self.args[0].is_nonnegative: return True def _eval_is_composite(self): if self.args[0].is_integer and (self.args[0] - 3).is_positive: return True def _eval_evalf(self, prec): from sympy.functions.special.gamma_functions 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: .. math:: \frac{2t}{e^t + 1} = \sum_{n=1}^\infty \frac{G_n t^n}{n!} Examples ======== >>> from sympy import genocchi, Symbol >>> [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 See Also ======== bell, bernoulli, catalan, euler, fibonacci, harmonic, lucas, partition, tribonacci References ========== .. [1] https://en.wikipedia.org/wiki/Genocchi_number .. [2] http://mathworld.wolfram.com/GenocchiNumber.html """ @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, **kwargs): 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 #----------------------------------------------------------------------------# # # # Partition numbers # # # #----------------------------------------------------------------------------# _npartition = [1, 1] class partition(Function): r""" Partition numbers The Partition numbers are a sequence of integers `p_n` that represent the number of distinct ways of representing `n` as a sum of natural numbers (with order irrelevant). The generating function for `p_n` is given by: .. math:: \sum_{n=0}^\infty p_n x^n = \prod_{k=1}^\infty (1 - x^k)^{-1} Examples ======== >>> from sympy import partition, Symbol >>> [partition(n) for n in range(9)] [1, 1, 2, 3, 5, 7, 11, 15, 22] >>> n = Symbol('n', integer=True, negative=True) >>> partition(n) 0 See Also ======== bell, bernoulli, catalan, euler, fibonacci, harmonic, lucas, genocchi, tribonacci References ========== .. [1] https://en.wikipedia.org/wiki/Partition_(number_theory%29 .. [2] https://en.wikipedia.org/wiki/Pentagonal_number_theorem """ @staticmethod def _partition(n): L = len(_npartition) if n < L: return _npartition[n] # lengthen cache for _n in range(L, n + 1): v, p, i = 0, 0, 0 while 1: s = 0 p += 3*i + 1 # p = pentagonal number: 1, 5, 12, ... if _n >= p: s += _npartition[_n - p] i += 1 gp = p + i # gp = generalized pentagonal: 2, 7, 15, ... if _n >= gp: s += _npartition[_n - gp] if s == 0: break else: v += s if i%2 == 1 else -s _npartition.append(v) return v @classmethod def eval(cls, n): is_int = n.is_integer if is_int == False: raise ValueError("Partition numbers are defined only for " "integers") elif is_int: if n.is_negative: return S.Zero if n.is_zero or (n - 1).is_zero: return S.One if n.is_Integer: return Integer(cls._partition(n)) def _eval_is_integer(self): if self.args[0].is_integer: return True def _eval_is_negative(self): if self.args[0].is_integer: return False def _eval_is_positive(self): n = self.args[0] if n.is_nonnegative and n.is_integer: return True ####################################################################### ### ### 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 isinstance(n, 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) lens = len(s) lenn = len(n) if lens == lenn: n = [1]*lenn + [lenn, lenn] return _MultisetHistogram(n) m = dict(zip(s, range(lens))) d = dict(zip(range(lens), (0,)*lens)) 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 See Also ======== sympy.utilities.iterables.multiset_permutations References ========== .. [1] https://en.wikipedia.org/wiki/Permutation """ 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): 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. """ 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, r in enumerate(rv): d[i] = r 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 through ``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 See Also ======== sympy.utilities.iterables.multiset_combinations References ========== .. [1] https://en.wikipedia.org/wiki/Combination .. [2] http://tinyurl.com/cep849r """ 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) def _eval_stirling1(n, k): if n == k == 0: return S.One if 0 in (n, k): return S.Zero # some special values if n == k: return S.One elif k == n - 1: 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) return _stirling1(n, k) @cacheit def _stirling1(n, k): row = [0, 1]+[0]*(k-1) # for n = 1 for i in range(2, n+1): for j in range(min(k,i), 0, -1): row[j] = (i-1) * row[j] + row[j-1] return Integer(row[k]) def _eval_stirling2(n, k): if n == k == 0: return S.One if 0 in (n, k): return S.Zero # some special values if n == k: return S.One elif k == n - 1: return binomial(n, 2) elif k == 1: return S.One elif k == 2: return Integer(2**(n - 1) - 1) return _stirling2(n, k) @cacheit def _stirling2(n, k): row = [0, 1]+[0]*(k-1) # for n = 1 for i in range(2, n+1): for j in range(min(k,i), 0, -1): row[j] = j * row[j] + row[j-1] return Integer(row[k]) def stirling(n, k, d=None, kind=2, signed=False): r"""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: .. math :: {0 \brace 0} = 1; {n \brace 0} = {0 \brace k} = 0; .. math :: {{n+1} \brace k} = j {n \brace k} + {n \brace {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 $n \ge k \ge d$. (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 See Also ======== sympy.utilities.iterables.multiset_partitions References ========== .. [1] https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind .. [2] https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind """ # 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 _eval_stirling2(n - d + 1, k - d + 1) elif signed: # kind is ignored -- only kind=1 is supported return S.NegativeOne**(n - k)*_eval_stirling1(n, k) if kind == 1: return _eval_stirling1(n, k) elif kind == 2: return _eval_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.""" # really quick exits if k > n or k < 0: return 0 if k in (1, n): return 1 if k == 0: return 0 # exits that could be done below but this is quicker if k == 2: return n//2 d = n - k if d <= 3: return d # quick exit if 3*k >= n: # or, equivalently, 2*k >= d # all the information needed in this case # will be in the cache needed to calculate # partition(d), so... # update cache tot = partition._partition(d) # and correct for values not needed if d - k > 0: tot -= sum(_npartition[:d - k]) return tot # regular exit # nT(n, k) = Sum(nT(n - k, m), (m, 1, k)); # calculate needed nT(i, j) values p = [1]*d for i in range(2, k + 1): for m in range(i + 1, d): p[m] += p[m - i] d -= 1 # if p[0] were appended to the end of p then the last # k values of p are the nT(n, j) values for 0 < j < k in reverse # order p[-1] = nT(n, 1), p[-2] = nT(n, 2), etc.... Instead of # putting the 1 from p[0] there, however, it is simply added to # the sum below which is valid for 1 < k <= n//2 return (1 + sum(p[1 - k:])) 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 Partitions of an integer expressed as a sum of positive integers: >>> from sympy import partition >>> partition(4) 5 >>> nT(4, 1) + nT(4, 2) + nT(4, 3) + nT(4, 4) 5 >>> nT('1'*4) 5 See Also ======== sympy.utilities.iterables.partitions sympy.utilities.iterables.multiset_partitions sympy.functions.combinatorial.numbers.partition References ========== .. [1] http://undergraduate.csse.uwa.edu.au/units/CITS7209/partition.pdf """ if isinstance(n, SYMPY_INTS): # n identical items if k is None: return partition(n) if isinstance(k, SYMPY_INTS): n = as_int(n) k = as_int(k) return Integer(_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 #-----------------------------------------------------------------------------# # # # Motzkin numbers # # # #-----------------------------------------------------------------------------# class motzkin(Function): """ The nth Motzkin number is the number of ways of drawing non-intersecting chords between n points on a circle (not necessarily touching every point by a chord). The Motzkin numbers are named after Theodore Motzkin and have diverse applications in geometry, combinatorics and number theory. Motzkin numbers are the integer sequence defined by the initial terms `M_0 = 1`, `M_1 = 1` and the two-term recurrence relation `M_n = \frac{2*n + 1}{n + 2} * M_{n-1} + \frac{3n - 3}{n + 2} * M_{n-2}`. Examples ======== >>> from sympy import motzkin >>> motzkin.is_motzkin(5) False >>> motzkin.find_motzkin_numbers_in_range(2,300) [2, 4, 9, 21, 51, 127] >>> motzkin.find_motzkin_numbers_in_range(2,900) [2, 4, 9, 21, 51, 127, 323, 835] >>> motzkin.find_first_n_motzkins(10) [1, 1, 2, 4, 9, 21, 51, 127, 323, 835] References ========== .. [1] https://en.wikipedia.org/wiki/Motzkin_number .. [2] https://mathworld.wolfram.com/MotzkinNumber.html """ @staticmethod def is_motzkin(n): try: n = as_int(n) except ValueError: return False if n > 0: if n in (1, 2): return True tn1 = 1 tn = 2 i = 3 while tn < n: a = ((2*i + 1)*tn + (3*i - 3)*tn1)/(i + 2) i += 1 tn1 = tn tn = a if tn == n: return True else: return False else: return False @staticmethod def find_motzkin_numbers_in_range(x, y): if 0 <= x <= y: motzkins = list() if x <= 1 <= y: motzkins.append(1) tn1 = 1 tn = 2 i = 3 while tn <= y: if tn >= x: motzkins.append(tn) a = ((2*i + 1)*tn + (3*i - 3)*tn1)/(i + 2) i += 1 tn1 = tn tn = int(a) return motzkins else: raise ValueError('The provided range is not valid. This condition should satisfy x <= y') @staticmethod def find_first_n_motzkins(n): try: n = as_int(n) except ValueError: raise ValueError('The provided number must be a positive integer') if n < 0: raise ValueError('The provided number must be a positive integer') motzkins = [1] if n >= 1: motzkins.append(1) tn1 = 1 tn = 2 i = 3 while i <= n: motzkins.append(tn) a = ((2*i + 1)*tn + (3*i - 3)*tn1)/(i + 2) i += 1 tn1 = tn tn = int(a) return motzkins @staticmethod @recurrence_memo([S.One, S.One]) def _motzkin(n, prev): return ((2*n + 1)*prev[-1] + (3*n - 3)*prev[-2]) // (n + 2) @classmethod def eval(cls, n): try: n = as_int(n) except ValueError: raise ValueError('The provided number must be a positive integer') if n < 0: raise ValueError('The provided number must be a positive integer') return Integer(cls._motzkin(n - 1)) def nD(i=None, brute=None, *, n=None, m=None): """return the number of derangements for: ``n`` unique items, ``i`` items (as a sequence or multiset), or multiplicities, ``m`` given as a sequence or multiset. Examples ======== >>> from sympy.utilities.iterables import generate_derangements as enum >>> from sympy.functions.combinatorial.numbers import nD A derangement ``d`` of sequence ``s`` has all ``d[i] != s[i]``: >>> set([''.join(i) for i in enum('abc')]) {'bca', 'cab'} >>> nD('abc') 2 Input as iterable or dictionary (multiset form) is accepted: >>> assert nD([1, 2, 2, 3, 3, 3]) == nD({1: 1, 2: 2, 3: 3}) By default, a brute-force enumeration and count of multiset permutations is only done if there are fewer than 9 elements. There may be cases when there is high multiplicty with few unique elements that will benefit from a brute-force enumeration, too. For this reason, the `brute` keyword (default None) is provided. When False, the brute-force enumeration will never be used. When True, it will always be used. >>> nD('1111222233', brute=True) 44 For convenience, one may specify ``n`` distinct items using the ``n`` keyword: >>> assert nD(n=3) == nD('abc') == 2 Since the number of derangments depends on the multiplicity of the elements and not the elements themselves, it may be more convenient to give a list or multiset of multiplicities using keyword ``m``: >>> assert nD('abc') == nD(m=(1,1,1)) == nD(m={1:3}) == 2 """ from sympy.integrals.integrals import integrate from sympy.functions.special.polynomials import laguerre from sympy.abc import x def ok(x): if not isinstance(x, SYMPY_INTS): raise TypeError('expecting integer values') if x < 0: raise ValueError('value must not be negative') return True if (i, n, m).count(None) != 2: raise ValueError('enter only 1 of i, n, or m') if i is not None: if isinstance(i, SYMPY_INTS): raise TypeError('items must be a list or dictionary') if not i: return S.Zero if type(i) is not dict: s = list(i) ms = multiset(s) elif type(i) is dict: all(ok(_) for _ in i.values()) ms = {k: v for k, v in i.items() if v} s = None if not ms: return S.Zero N = sum(ms.values()) counts = multiset(ms.values()) nkey = len(ms) elif n is not None: ok(n) if not n: return S.Zero return subfactorial(n) elif m is not None: if isinstance(m, dict): all(ok(i) and ok(j) for i, j in m.items()) counts = {k: v for k, v in m.items() if k*v} elif iterable(m) or isinstance(m, str): m = list(m) all(ok(i) for i in m) counts = multiset([i for i in m if i]) else: raise TypeError('expecting iterable') if not counts: return S.Zero N = sum(k*v for k, v in counts.items()) nkey = sum(counts.values()) s = None big = int(max(counts)) if big == 1: # no repetition return subfactorial(nkey) nval = len(counts) if big*2 > N: return S.Zero if big*2 == N: if nkey == 2 and nval == 1: return S.One # aaabbb if nkey - 1 == big: # one element repeated return factorial(big) # e.g. abc part of abcddd if N < 9 and brute is None or brute: # for all possibilities, this was found to be faster if s is None: s = [] i = 0 for m, v in counts.items(): for j in range(v): s.extend([i]*m) i += 1 return Integer(sum(1 for i in multiset_derangements(s))) from sympy.functions.elementary.exponential import exp return Integer(abs(integrate(exp(-x)*Mul(*[ laguerre(i, x)**m for i, m in counts.items()]), (x, 0, oo))))