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)
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)
def _eval_rewrite_as_Sum(self, arg):
from diofant 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 _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)
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
def eval(cls, n, alpha, x):
# L_{n}^{0}(x) ---> L_{n}(x)
if alpha == S.Zero:
return laguerre(n, x)
if not n.is_Number:
# We can evaluate for some special values of x
if x == S.Zero:
return binomial(n + alpha, alpha)
elif x == S.Infinity and n.is_positive:
return S.NegativeOne**n * S.Infinity
elif x == S.NegativeInfinity and n.is_positive:
return S.Infinity
else:
# n is a given fixed integer, evaluate into polynomial
if n.is_negative:
raise ValueError(
"The index n must be nonnegative integer (got %r)" % n)
else:
return cls._eval_at_order(n, x, alpha)
def _eval_rewrite_as_binomial(self, n):
return binomial(2 * n, n) / (n + 1)
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 diofant.functions.combinatorial.numbers import nC
>>> from diofant.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
========
diofant.utilities.iterables.multiset_combinations
"""
from diofant.functions.combinatorial.factorials import binomial
from diofant.core.mul import prod
if isinstance(n, DIOFANT_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)