Number Theory

Ntheory Class Reference

class sympy.ntheory.generate.Sieve[source]

An infinite list of prime numbers, implemented as a dynamically growing sieve of Eratosthenes. When a lookup is requested involving an odd number that has not been sieved, the sieve is automatically extended up to that number.

Examples

>>> from sympy import sieve
>>> sieve._reset() # this line for doctest only
>>> 25 in sieve
False
>>> sieve._list
array('l', [2, 3, 5, 7, 11, 13, 17, 19, 23])
extend(n)[source]

Grow the sieve to cover all primes <= n (a real number).

Examples

>>> from sympy import sieve
>>> sieve._reset() # this line for doctest only
>>> sieve.extend(30)
>>> sieve[10] == 29
True
extend_to_no(i)[source]

Extend to include the ith prime number.

i must be an integer.

The list is extended by 50% if it is too short, so it is likely that it will be longer than requested.

Examples

>>> from sympy import sieve
>>> sieve._reset() # this line for doctest only
>>> sieve.extend_to_no(9)
>>> sieve._list
array('l', [2, 3, 5, 7, 11, 13, 17, 19, 23])
primerange(a, b)[source]

Generate all prime numbers in the range [a, b).

Examples

>>> from sympy import sieve
>>> print([i for i in sieve.primerange(7, 18)])
[7, 11, 13, 17]
search(n)[source]

Return the indices i, j of the primes that bound n.

If n is prime then i == j.

Although n can be an expression, if ceiling cannot convert it to an integer then an n error will be raised.

Examples

>>> from sympy import sieve
>>> sieve.search(25)
(9, 10)
>>> sieve.search(23)
(9, 9)

Ntheory Functions Reference

sympy.ntheory.generate.prime(nth)[source]

Return the nth prime, with the primes indexed as prime(1) = 2, prime(2) = 3, etc…. The nth prime is approximately n*log(n).

Logarithmic integral of x is a pretty nice approximation for number of primes <= x, i.e. li(x) ~ pi(x) In fact, for the numbers we are concerned about( x<1e11 ), li(x) - pi(x) < 50000

Also, li(x) > pi(x) can be safely assumed for the numbers which can be evaluated by this function.

Here, we find the least integer m such that li(m) > n using binary search. Now pi(m-1) < li(m-1) <= n,

We find pi(m - 1) using primepi function.

Starting from m, we have to find n - pi(m-1) more primes.

For the inputs this implementation can handle, we will have to test primality for at max about 10**5 numbers, to get our answer.

See also

sympy.ntheory.primetest.isprime
Test if n is prime
primerange
Generate all primes in a given range
primepi
Return the number of primes less than or equal to n

References

Examples

>>> from sympy import prime
>>> prime(10)
29
>>> prime(1)
2
>>> prime(100000)
1299709
sympy.ntheory.generate.primepi(n)[source]

Return the value of the prime counting function pi(n) = the number of prime numbers less than or equal to n.

Algorithm Description:

In sieve method, we remove all multiples of prime p except p itself.

Let phi(i,j) be the number of integers 2 <= k <= i which remain after sieving from primes less than or equal to j. Clearly, pi(n) = phi(n, sqrt(n))

If j is not a prime, phi(i,j) = phi(i, j - 1)

if j is a prime, We remove all numbers(except j) whose smallest prime factor is j.

Let x= j*a be such a number, where 2 <= a<= i / j Now, after sieving from primes <= j - 1, a must remain (because x, and hence a has no prime factor <= j - 1) Clearly, there are phi(i / j, j - 1) such a which remain on sieving from primes <= j - 1

Now, if a is a prime less than equal to j - 1, x= j*a has smallest prime factor = a, and has already been removed(by sieving from a). So, we don’t need to remove it again. (Note: there will be pi(j - 1) such x)

Thus, number of x, that will be removed are: phi(i / j, j - 1) - phi(j - 1, j - 1) (Note that pi(j - 1) = phi(j - 1, j - 1))

=> phi(i,j) = phi(i, j - 1) - phi(i / j, j - 1) + phi(j - 1, j - 1)

So,following recursion is used and implemented as dp:

phi(a, b) = phi(a, b - 1), if b is not a prime phi(a, b) = phi(a, b-1)-phi(a / b, b-1) + phi(b-1, b-1), if b is prime

Clearly a is always of the form floor(n / k), which can take at most 2*sqrt(n) values. Two arrays arr1,arr2 are maintained arr1[i] = phi(i, j), arr2[i] = phi(n // i, j)

Finally the answer is arr2[1]

See also

sympy.ntheory.primetest.isprime
Test if n is prime
primerange
Generate all primes in a given range
prime
Return the nth prime

Examples

>>> from sympy import primepi
>>> primepi(25)
9
sympy.ntheory.generate.nextprime(n, ith=1)[source]

Return the ith prime greater than n.

i must be an integer.

See also

prevprime
Return the largest prime smaller than n
primerange
Generate all primes in a given range

Notes

Potential primes are located at 6*j +/- 1. This property is used during searching.

>>> from sympy import nextprime
>>> [(i, nextprime(i)) for i in range(10, 15)]
[(10, 11), (11, 13), (12, 13), (13, 17), (14, 17)]
>>> nextprime(2, ith=2) # the 2nd prime after 2
5
sympy.ntheory.generate.prevprime(n)[source]

Return the largest prime smaller than n.

See also

nextprime
Return the ith prime greater than n
primerange
Generates all primes in a given range

Notes

Potential primes are located at 6*j +/- 1. This property is used during searching.

>>> from sympy import prevprime
>>> [(i, prevprime(i)) for i in range(10, 15)]
[(10, 7), (11, 7), (12, 11), (13, 11), (14, 13)]
sympy.ntheory.generate.primerange(a, b)[source]

Generate a list of all prime numbers in the range [a, b).

If the range exists in the default sieve, the values will be returned from there; otherwise values will be returned but will not modify the sieve.

See also

nextprime
Return the ith prime greater than n
prevprime
Return the largest prime smaller than n
randprime
Returns a random prime in a given range
primorial
Returns the product of primes based on condition
Sieve.primerange
return range from already computed primes or extend the sieve to contain the requested range.

Notes

Some famous conjectures about the occurrence of primes in a given range are [1]:

  • Twin primes: though often not, the following will give 2 primes
    an infinite number of times:
    primerange(6*n - 1, 6*n + 2)
  • Legendre’s: the following always yields at least one prime
    primerange(n**2, (n+1)**2+1)
  • Bertrand’s (proven): there is always a prime in the range
    primerange(n, 2*n)
  • Brocard’s: there are at least four primes in the range
    primerange(prime(n)**2, prime(n+1)**2)

The average gap between primes is log(n) [2]; the gap between primes can be arbitrarily large since sequences of composite numbers are arbitrarily large, e.g. the numbers in the sequence n! + 2, n! + 3 … n! + n are all composite.

References

  1. http://en.wikipedia.org/wiki/Prime_number
  2. http://primes.utm.edu/notes/gaps.html

Examples

>>> from sympy import primerange, sieve
>>> print([i for i in primerange(1, 30)])
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29]

The Sieve method, primerange, is generally faster but it will occupy more memory as the sieve stores values. The default instance of Sieve, named sieve, can be used:

>>> list(sieve.primerange(1, 30))
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
sympy.ntheory.generate.randprime(a, b)[source]

Return a random prime number in the range [a, b).

Bertrand’s postulate assures that randprime(a, 2*a) will always succeed for a > 1.

See also

primerange
Generate all primes in a given range

References

Examples

>>> from sympy import randprime, isprime
>>> randprime(1, 30) 
13
>>> isprime(randprime(1, 30))
True
sympy.ntheory.generate.primorial(n, nth=True)[source]

Returns the product of the first n primes (default) or the primes less than or equal to n (when nth=False).

>>> from sympy.ntheory.generate import primorial, randprime, primerange
>>> from sympy import factorint, Mul, primefactors, sqrt
>>> primorial(4) # the first 4 primes are 2, 3, 5, 7
210
>>> primorial(4, nth=False) # primes <= 4 are 2 and 3
6
>>> primorial(1)
2
>>> primorial(1, nth=False)
1
>>> primorial(sqrt(101), nth=False)
210

One can argue that the primes are infinite since if you take a set of primes and multiply them together (e.g. the primorial) and then add or subtract 1, the result cannot be divided by any of the original factors, hence either 1 or more new primes must divide this product of primes.

In this case, the number itself is a new prime:

>>> factorint(primorial(4) + 1)
{211: 1}

In this case two new primes are the factors:

>>> factorint(primorial(4) - 1)
{11: 1, 19: 1}

Here, some primes smaller and larger than the primes multiplied together are obtained:

>>> p = list(primerange(10, 20))
>>> sorted(set(primefactors(Mul(*p) + 1)).difference(set(p)))
[2, 5, 31, 149]

See also

primerange
Generate all primes in a given range
sympy.ntheory.generate.cycle_length(f, x0, nmax=None, values=False)[source]

For a given iterated sequence, return a generator that gives the length of the iterated cycle (lambda) and the length of terms before the cycle begins (mu); if values is True then the terms of the sequence will be returned instead. The sequence is started with value x0.

Note: more than the first lambda + mu terms may be returned and this is the cost of cycle detection with Brent’s method; there are, however, generally less terms calculated than would have been calculated if the proper ending point were determined, e.g. by using Floyd’s method.

>>> from sympy.ntheory.generate import cycle_length

This will yield successive values of i <– func(i):

>>> def iter(func, i):
...     while 1:
...         ii = func(i)
...         yield ii
...         i = ii
...

A function is defined:

>>> func = lambda i: (i**2 + 1) % 51

and given a seed of 4 and the mu and lambda terms calculated:

>>> next(cycle_length(func, 4))
(6, 2)

We can see what is meant by looking at the output:

>>> n = cycle_length(func, 4, values=True)
>>> list(ni for ni in n)
[17, 35, 2, 5, 26, 14, 44, 50, 2, 5, 26, 14]

There are 6 repeating values after the first 2.

If a sequence is suspected of being longer than you might wish, nmax can be used to exit early (and mu will be returned as None):

>>> next(cycle_length(func, 4, nmax = 4))
(4, None)
>>> [ni for ni in cycle_length(func, 4, nmax = 4, values=True)]
[17, 35, 2, 5]
Code modified from:
http://en.wikipedia.org/wiki/Cycle_detection.
sympy.ntheory.generate.composite(nth)[source]

Return the nth composite number, with the composite numbers indexed as composite(1) = 4, composite(2) = 6, etc….

See also

sympy.ntheory.primetest.isprime
Test if n is prime
primerange
Generate all primes in a given range
primepi
Return the number of primes less than or equal to n
prime
Return the nth prime
compositepi
Return the number of positive composite numbers less than or equal to n

Examples

>>> from sympy import composite
>>> composite(36)
52
>>> composite(1)
4
>>> composite(17737)
20000
sympy.ntheory.generate.compositepi(n)[source]

Return the number of positive composite numbers less than or equal to n. The first positive composite is 4, i.e. compositepi(4) = 1.

See also

sympy.ntheory.primetest.isprime
Test if n is prime
primerange
Generate all primes in a given range
prime
Return the nth prime
primepi
Return the number of primes less than or equal to n
composite
Return the nth composite number

Examples

>>> from sympy import compositepi
>>> compositepi(25)
15
>>> compositepi(1000)
831
sympy.ntheory.factor_.smoothness(n)[source]

Return the B-smooth and B-power smooth values of n.

The smoothness of n is the largest prime factor of n; the power- smoothness is the largest divisor raised to its multiplicity.

>>> from sympy.ntheory.factor_ import smoothness
>>> smoothness(2**7*3**2)
(3, 128)
>>> smoothness(2**4*13)
(13, 16)
>>> smoothness(2)
(2, 2)
sympy.ntheory.factor_.smoothness_p(n, m=-1, power=0, visual=None)[source]

Return a list of [m, (p, (M, sm(p + m), psm(p + m)))…] where:

  1. p**M is the base-p divisor of n
  2. sm(p + m) is the smoothness of p + m (m = -1 by default)
  3. psm(p + m) is the power smoothness of p + m

The list is sorted according to smoothness (default) or by power smoothness if power=1.

The smoothness of the numbers to the left (m = -1) or right (m = 1) of a factor govern the results that are obtained from the p +/- 1 type factoring methods.

>>> from sympy.ntheory.factor_ import smoothness_p, factorint
>>> smoothness_p(10431, m=1)
(1, [(3, (2, 2, 4)), (19, (1, 5, 5)), (61, (1, 31, 31))])
>>> smoothness_p(10431)
(-1, [(3, (2, 2, 2)), (19, (1, 3, 9)), (61, (1, 5, 5))])
>>> smoothness_p(10431, power=1)
(-1, [(3, (2, 2, 2)), (61, (1, 5, 5)), (19, (1, 3, 9))])

If visual=True then an annotated string will be returned:

>>> print(smoothness_p(21477639576571, visual=1))
p**i=4410317**1 has p-1 B=1787, B-pow=1787
p**i=4869863**1 has p-1 B=2434931, B-pow=2434931

This string can also be generated directly from a factorization dictionary and vice versa:

>>> factorint(17*9)
{3: 2, 17: 1}
>>> smoothness_p(_)
'p**i=3**2 has p-1 B=2, B-pow=2\np**i=17**1 has p-1 B=2, B-pow=16'
>>> smoothness_p(_)
{3: 2, 17: 1}

The table of the output logic is:


Visual
Input True False other
dict str tuple str
str str tuple dict
tuple str tuple str
n str tuple tuple
mul str tuple tuple

See also

factorint, smoothness

sympy.ntheory.factor_.trailing(n)[source]

Count the number of trailing zero digits in the binary representation of n, i.e. determine the largest power of 2 that divides n.

Examples

>>> from sympy import trailing
>>> trailing(128)
7
>>> trailing(63)
0
sympy.ntheory.factor_.multiplicity(p, n)[source]

Find the greatest integer m such that p**m divides n.

Examples

>>> from sympy.ntheory import multiplicity
>>> from sympy.core.numbers import Rational as R
>>> [multiplicity(5, n) for n in [8, 5, 25, 125, 250]]
[0, 1, 2, 3, 3]
>>> multiplicity(3, R(1, 9))
-2
sympy.ntheory.factor_.perfect_power(n, candidates=None, big=True, factor=True)[source]

Return (b, e) such that n == b**e if n is a perfect power; otherwise return False.

By default, the base is recursively decomposed and the exponents collected so the largest possible e is sought. If big=False then the smallest possible e (thus prime) will be chosen.

If candidates for exponents are given, they are assumed to be sorted and the first one that is larger than the computed maximum will signal failure for the routine.

If factor=True then simultaneous factorization of n is attempted since finding a factor indicates the only possible root for n. This is True by default since only a few small factors will be tested in the course of searching for the perfect power.

Examples

>>> from sympy import perfect_power
>>> perfect_power(16)
(2, 4)
>>> perfect_power(16, big = False)
(4, 2)
sympy.ntheory.factor_.pollard_rho(n, s=2, a=1, retries=5, seed=1234, max_steps=None, F=None)[source]

Use Pollard’s rho method to try to extract a nontrivial factor of n. The returned factor may be a composite number. If no factor is found, None is returned.

The algorithm generates pseudo-random values of x with a generator function, replacing x with F(x). If F is not supplied then the function x**2 + a is used. The first value supplied to F(x) is s. Upon failure (if retries is > 0) a new a and s will be supplied; the a will be ignored if F was supplied.

The sequence of numbers generated by such functions generally have a a lead-up to some number and then loop around back to that number and begin to repeat the sequence, e.g. 1, 2, 3, 4, 5, 3, 4, 5 – this leader and loop look a bit like the Greek letter rho, and thus the name, ‘rho’.

For a given function, very different leader-loop values can be obtained so it is a good idea to allow for retries:

>>> from sympy.ntheory.generate import cycle_length
>>> n = 16843009
>>> F = lambda x:(2048*pow(x, 2, n) + 32767) % n
>>> for s in range(5):
...     print('loop length = %4i; leader length = %3i' % next(cycle_length(F, s)))
...
loop length = 2489; leader length =  42
loop length =   78; leader length = 120
loop length = 1482; leader length =  99
loop length = 1482; leader length = 285
loop length = 1482; leader length = 100

Here is an explicit example where there is a two element leadup to a sequence of 3 numbers (11, 14, 4) that then repeat:

>>> x=2
>>> for i in range(9):
...     x=(x**2+12)%17
...     print(x)
...
16
13
11
14
4
11
14
4
11
>>> next(cycle_length(lambda x: (x**2+12)%17, 2))
(3, 2)
>>> list(cycle_length(lambda x: (x**2+12)%17, 2, values=True))
[16, 13, 11, 14, 4]

Instead of checking the differences of all generated values for a gcd with n, only the kth and 2*kth numbers are checked, e.g. 1st and 2nd, 2nd and 4th, 3rd and 6th until it has been detected that the loop has been traversed. Loops may be many thousands of steps long before rho finds a factor or reports failure. If max_steps is specified, the iteration is cancelled with a failure after the specified number of steps.

References

  • Richard Crandall & Carl Pomerance (2005), “Prime Numbers: A Computational Perspective”, Springer, 2nd edition, 229-231

Examples

>>> from sympy import pollard_rho
>>> n=16843009
>>> F=lambda x:(2048*pow(x,2,n) + 32767) % n
>>> pollard_rho(n, F=F)
257

Use the default setting with a bad value of a and no retries:

>>> pollard_rho(n, a=n-2, retries=0)

If retries is > 0 then perhaps the problem will correct itself when new values are generated for a:

>>> pollard_rho(n, a=n-2, retries=1)
257
sympy.ntheory.factor_.pollard_pm1(n, B=10, a=2, retries=0, seed=1234)[source]

Use Pollard’s p-1 method to try to extract a nontrivial factor of n. Either a divisor (perhaps composite) or None is returned.

The value of a is the base that is used in the test gcd(a**M - 1, n). The default is 2. If retries > 0 then if no factor is found after the first attempt, a new a will be generated randomly (using the seed) and the process repeated.

Note: the value of M is lcm(1..B) = reduce(ilcm, range(2, B + 1)).

A search is made for factors next to even numbers having a power smoothness less than B. Choosing a larger B increases the likelihood of finding a larger factor but takes longer. Whether a factor of n is found or not depends on a and the power smoothness of the even mumber just less than the factor p (hence the name p - 1).

Although some discussion of what constitutes a good a some descriptions are hard to interpret. At the modular.math site referenced below it is stated that if gcd(a**M - 1, n) = N then a**M % q**r is 1 for every prime power divisor of N. But consider the following:

>>> from sympy.ntheory.factor_ import smoothness_p, pollard_pm1
>>> n=257*1009
>>> smoothness_p(n)
(-1, [(257, (1, 2, 256)), (1009, (1, 7, 16))])

So we should (and can) find a root with B=16:

>>> pollard_pm1(n, B=16, a=3)
1009

If we attempt to increase B to 256 we find that it doesn’t work:

>>> pollard_pm1(n, B=256)
>>>

But if the value of a is changed we find that only multiples of 257 work, e.g.:

>>> pollard_pm1(n, B=256, a=257)
1009

Checking different a values shows that all the ones that didn’t work had a gcd value not equal to n but equal to one of the factors:

>>> from sympy.core.numbers import ilcm, igcd
>>> from sympy import factorint, Pow
>>> M = 1
>>> for i in range(2, 256):
...     M = ilcm(M, i)
...
>>> set([igcd(pow(a, M, n) - 1, n) for a in range(2, 256) if
...      igcd(pow(a, M, n) - 1, n) != n])
{1009}

But does aM % d for every divisor of n give 1?

>>> aM = pow(255, M, n)
>>> [(d, aM%Pow(*d.args)) for d in factorint(n, visual=True).args]
[(257**1, 1), (1009**1, 1)]

No, only one of them. So perhaps the principle is that a root will be found for a given value of B provided that:

  1. the power smoothness of the p - 1 value next to the root does not exceed B
  2. a**M % p != 1 for any of the divisors of n.

By trying more than one a it is possible that one of them will yield a factor.

References

Examples

With the default smoothness bound, this number can’t be cracked:

>>> from sympy.ntheory import pollard_pm1, primefactors
>>> pollard_pm1(21477639576571)

Increasing the smoothness bound helps:

>>> pollard_pm1(21477639576571, B=2000)
4410317

Looking at the smoothness of the factors of this number we find:

>>> from sympy.utilities import flatten
>>> from sympy.ntheory.factor_ import smoothness_p, factorint
>>> print(smoothness_p(21477639576571, visual=1))
p**i=4410317**1 has p-1 B=1787, B-pow=1787
p**i=4869863**1 has p-1 B=2434931, B-pow=2434931

The B and B-pow are the same for the p - 1 factorizations of the divisors because those factorizations had a very large prime factor:

>>> factorint(4410317 - 1)
{2: 2, 617: 1, 1787: 1}
>>> factorint(4869863-1)
{2: 1, 2434931: 1}

Note that until B reaches the B-pow value of 1787, the number is not cracked;

>>> pollard_pm1(21477639576571, B=1786)
>>> pollard_pm1(21477639576571, B=1787)
4410317

The B value has to do with the factors of the number next to the divisor, not the divisors themselves. A worst case scenario is that the number next to the factor p has a large prime divisisor or is a perfect power. If these conditions apply then the power-smoothness will be about p/2 or p. The more realistic is that there will be a large prime factor next to p requiring a B value on the order of p/2. Although primes may have been searched for up to this level, the p/2 is a factor of p - 1, something that we don’t know. The modular.math reference below states that 15% of numbers in the range of 10**15 to 15**15 + 10**4 are 10**6 power smooth so a B of 10**6 will fail 85% of the time in that range. From 10**8 to 10**8 + 10**3 the percentages are nearly reversed…but in that range the simple trial division is quite fast.

sympy.ntheory.factor_.factorint(n, limit=None, use_trial=True, use_rho=True, use_pm1=True, verbose=False, visual=None, multiple=False)[source]

Given a positive integer n, factorint(n) returns a dict containing the prime factors of n as keys and their respective multiplicities as values. For example:

>>> from sympy.ntheory import factorint
>>> factorint(2000)    # 2000 = (2**4) * (5**3)
{2: 4, 5: 3}
>>> factorint(65537)   # This number is prime
{65537: 1}

For input less than 2, factorint behaves as follows:

  • factorint(1) returns the empty factorization, {}
  • factorint(0) returns {0:1}
  • factorint(-n) adds -1:1 to the factors and then factors n

Partial Factorization:

If limit (> 3) is specified, the search is stopped after performing trial division up to (and including) the limit (or taking a corresponding number of rho/p-1 steps). This is useful if one has a large number and only is interested in finding small factors (if any). Note that setting a limit does not prevent larger factors from being found early; it simply means that the largest factor may be composite. Since checking for perfect power is relatively cheap, it is done regardless of the limit setting.

This number, for example, has two small factors and a huge semi-prime factor that cannot be reduced easily:

>>> from sympy.ntheory import isprime
>>> from sympy.core.compatibility import long
>>> a = 1407633717262338957430697921446883
>>> f = factorint(a, limit=10000)
>>> f == {991: 1, long(202916782076162456022877024859): 1, 7: 1}
True
>>> isprime(max(f))
False

This number has a small factor and a residual perfect power whose base is greater than the limit:

>>> factorint(3*101**7, limit=5)
{3: 1, 101: 7}

List of Factors:

If multiple is set to True then a list containing the prime factors including multiplicities is returned.

>>> factorint(24, multiple=True)
[2, 2, 2, 3]

Visual Factorization:

If visual is set to True, then it will return a visual factorization of the integer. For example:

>>> from sympy import pprint
>>> pprint(factorint(4200, visual=True))
 3  1  2  1
2 *3 *5 *7

Note that this is achieved by using the evaluate=False flag in Mul and Pow. If you do other manipulations with an expression where evaluate=False, it may evaluate. Therefore, you should use the visual option only for visualization, and use the normal dictionary returned by visual=False if you want to perform operations on the factors.

You can easily switch between the two forms by sending them back to factorint:

>>> from sympy import Mul, Pow
>>> regular = factorint(1764); regular
{2: 2, 3: 2, 7: 2}
>>> pprint(factorint(regular))
 2  2  2
2 *3 *7
>>> visual = factorint(1764, visual=True); pprint(visual)
 2  2  2
2 *3 *7
>>> print(factorint(visual))
{2: 2, 3: 2, 7: 2}

If you want to send a number to be factored in a partially factored form you can do so with a dictionary or unevaluated expression:

>>> factorint(factorint({4: 2, 12: 3})) # twice to toggle to dict form
{2: 10, 3: 3}
>>> factorint(Mul(4, 12, evaluate=False))
{2: 4, 3: 1}

The table of the output logic is:

   
Input True False other
dict mul dict mul
n mul dict dict
mul mul dict dict

Notes

Algorithm:

The function switches between multiple algorithms. Trial division quickly finds small factors (of the order 1-5 digits), and finds all large factors if given enough time. The Pollard rho and p-1 algorithms are used to find large factors ahead of time; they will often find factors of the order of 10 digits within a few seconds:

>>> factors = factorint(12345678910111213141516)
>>> for base, exp in sorted(factors.items()):
...     print('%s %s' % (base, exp))
...
2 2
2507191691 1
1231026625769 1

Any of these methods can optionally be disabled with the following boolean parameters:

  • use_trial: Toggle use of trial division
  • use_rho: Toggle use of Pollard’s rho method
  • use_pm1: Toggle use of Pollard’s p-1 method

factorint also periodically checks if the remaining part is a prime number or a perfect power, and in those cases stops.

For unevaluated factorial, it uses Legendre’s formula(theorem).

If verbose is set to True, detailed progress is printed.

sympy.ntheory.factor_.primefactors(n, limit=None, verbose=False)[source]

Return a sorted list of n’s prime factors, ignoring multiplicity and any composite factor that remains if the limit was set too low for complete factorization. Unlike factorint(), primefactors() does not return -1 or 0.

See also

divisors

Examples

>>> from sympy.ntheory import primefactors, factorint, isprime
>>> primefactors(6)
[2, 3]
>>> primefactors(-5)
[5]
>>> sorted(factorint(123456).items())
[(2, 6), (3, 1), (643, 1)]
>>> primefactors(123456)
[2, 3, 643]
>>> sorted(factorint(10000000001, limit=200).items())
[(101, 1), (99009901, 1)]
>>> isprime(99009901)
False
>>> primefactors(10000000001, limit=300)
[101]
sympy.ntheory.factor_.divisors(n, generator=False)[source]

Return all divisors of n sorted from 1..n by default. If generator is True an unordered generator is returned.

The number of divisors of n can be quite large if there are many prime factors (counting repeated factors). If only the number of factors is desired use divisor_count(n).

Examples

>>> from sympy import divisors, divisor_count
>>> divisors(24)
[1, 2, 3, 4, 6, 8, 12, 24]
>>> divisor_count(24)
8
>>> list(divisors(120, generator=True))
[1, 2, 4, 8, 3, 6, 12, 24, 5, 10, 20, 40, 15, 30, 60, 120]

This is a slightly modified version of Tim Peters referenced at: http://stackoverflow.com/questions/1010381/python-factorization

sympy.ntheory.factor_.divisor_count(n, modulus=1)[source]

Return the number of divisors of n. If modulus is not 1 then only those that are divisible by modulus are counted.

References

>>> from sympy import divisor_count
>>> divisor_count(6)
4
sympy.ntheory.factor_.udivisors(n, generator=False)[source]

Return all unitary divisors of n sorted from 1..n by default. If generator is True an unordered generator is returned.

The number of unitary divisors of n can be quite large if there are many prime factors. If only the number of unitary divisors is desired use udivisor_count(n).

References

Examples

>>> from sympy.ntheory.factor_ import udivisors, udivisor_count
>>> udivisors(15)
[1, 3, 5, 15]
>>> udivisor_count(15)
4
>>> sorted(udivisors(120, generator=True))
[1, 3, 5, 8, 15, 24, 40, 120]
sympy.ntheory.factor_.udivisor_count(n)[source]

Return the number of unitary divisors of n.

References

>>> from sympy.ntheory.factor_ import udivisor_count
>>> udivisor_count(120)
8
sympy.ntheory.factor_.antidivisors(n, generator=False)[source]

Return all antidivisors of n sorted from 1..n by default.

Antidivisors [R393] of n are numbers that do not divide n by the largest possible margin. If generator is True an unordered generator is returned.

References

[R393](1, 2) definition is described in http://oeis.org/A066272/a066272a.html

Examples

>>> from sympy.ntheory.factor_ import antidivisors
>>> antidivisors(24)
[7, 16]
>>> sorted(antidivisors(128, generator=True))
[3, 5, 15, 17, 51, 85]
sympy.ntheory.factor_.antidivisor_count(n)[source]

Return the number of antidivisors [R394] of n.

References

[R394](1, 2) formula from https://oeis.org/A066272

Examples

>>> from sympy.ntheory.factor_ import antidivisor_count
>>> antidivisor_count(13)
4
>>> antidivisor_count(27)
5
sympy.ntheory.factor_.totient(n)[source]

Calculate the Euler totient function phi(n)

totient(n) or \(\phi(n)\) is the number of positive integers \(\leq\) n that are relatively prime to n.

See also

divisor_count

References

[R395]https://en.wikipedia.org/wiki/Euler%27s_totient_function
[R396]http://mathworld.wolfram.com/TotientFunction.html

Examples

>>> from sympy.ntheory import totient
>>> totient(1)
1
>>> totient(25)
20
sympy.ntheory.factor_.reduced_totient(n)[source]

Calculate the Carmichael reduced totient function lambda(n)

reduced_totient(n) or \(\lambda(n)\) is the smallest m > 0 such that \(k^m \equiv 1 \mod n\) for all k relatively prime to n.

See also

totient

References

[R397]https://en.wikipedia.org/wiki/Carmichael_function
[R398]http://mathworld.wolfram.com/CarmichaelFunction.html

Examples

>>> from sympy.ntheory import reduced_totient
>>> reduced_totient(1)
1
>>> reduced_totient(8)
2
>>> reduced_totient(30)
4
sympy.ntheory.factor_.divisor_sigma(n, k=1)[source]

Calculate the divisor function \(\sigma_k(n)\) for positive integer n

divisor_sigma(n, k) is equal to sum([x**k for x in divisors(n)])

If n’s prime factorization is:

\[n = \prod_{i=1}^\omega p_i^{m_i},\]

then

\[\sigma_k(n) = \prod_{i=1}^\omega (1+p_i^k+p_i^{2k}+\cdots + p_i^{m_ik}).\]
Parameters:

k : power of divisors in the sum

for k = 0, 1: divisor_sigma(n, 0) is equal to divisor_count(n) divisor_sigma(n, 1) is equal to sum(divisors(n))

Default for k is 1.

References

[R399]http://en.wikipedia.org/wiki/Divisor_function

Examples

>>> from sympy.ntheory import divisor_sigma
>>> divisor_sigma(18, 0)
6
>>> divisor_sigma(39, 1)
56
>>> divisor_sigma(12, 2)
210
>>> divisor_sigma(37)
38
sympy.ntheory.factor_.udivisor_sigma(n, k=1)[source]

Calculate the unitary divisor function \(\sigma_k^*(n)\) for positive integer n

udivisor_sigma(n, k) is equal to sum([x**k for x in udivisors(n)])

If n’s prime factorization is:

\[n = \prod_{i=1}^\omega p_i^{m_i},\]

then

\[\sigma_k^*(n) = \prod_{i=1}^\omega (1+ p_i^{m_ik}).\]
Parameters:

k : power of divisors in the sum

for k = 0, 1: udivisor_sigma(n, 0) is equal to udivisor_count(n) udivisor_sigma(n, 1) is equal to sum(udivisors(n))

Default for k is 1.

References

[R400]http://mathworld.wolfram.com/UnitaryDivisorFunction.html

Examples

>>> from sympy.ntheory.factor_ import udivisor_sigma
>>> udivisor_sigma(18, 0)
4
>>> udivisor_sigma(74, 1)
114
>>> udivisor_sigma(36, 3)
47450
>>> udivisor_sigma(111)
152
sympy.ntheory.factor_.core(n, t=2)[source]

Calculate core(n,t) = \(core_t(n)\) of a positive integer n

core_2(n) is equal to the squarefree part of n

If n’s prime factorization is:

\[n = \prod_{i=1}^\omega p_i^{m_i},\]

then

\[core_t(n) = \prod_{i=1}^\omega p_i^{m_i \mod t}.\]
Parameters:

t : core(n,t) calculates the t-th power free part of n

core(n, 2) is the squarefree part of n core(n, 3) is the cubefree part of n

Default for t is 2.

References

[R401]http://en.wikipedia.org/wiki/Square-free_integer#Squarefree_core

Examples

>>> from sympy.ntheory.factor_ import core
>>> core(24, 2)
6
>>> core(9424, 3)
1178
>>> core(379238)
379238
>>> core(15**11, 10)
15
sympy.ntheory.factor_.digits(n, b=10)[source]

Return a list of the digits of n in base b. The first element in the list is b (or -b if n is negative).

Examples

>>> from sympy.ntheory.factor_ import digits
>>> digits(35)
[10, 3, 5]
>>> digits(27, 2)
[2, 1, 1, 0, 1, 1]
>>> digits(65536, 256)
[256, 1, 0, 0]
>>> digits(-3958, 27)
[-27, 5, 11, 16]
sympy.ntheory.factor_.primenu(n)[source]

Calculate the number of distinct prime factors for a positive integer n.

If n’s prime factorization is:

\[n = \prod_{i=1}^k p_i^{m_i},\]

then primenu(n) or \(\nu(n)\) is:

\[\nu(n) = k.\]

See also

factorint

References

[R402]http://mathworld.wolfram.com/PrimeFactor.html

Examples

>>> from sympy.ntheory.factor_ import primenu
>>> primenu(1)
0
>>> primenu(30)
3
sympy.ntheory.factor_.primeomega(n)[source]

Calculate the number of prime factors counting multiplicities for a positive integer n.

If n’s prime factorization is:

\[n = \prod_{i=1}^k p_i^{m_i},\]

then primeomega(n) or \(\Omega(n)\) is:

\[\Omega(n) = \sum_{i=1}^k m_i.\]

See also

factorint

References

[R403]http://mathworld.wolfram.com/PrimeFactor.html

Examples

>>> from sympy.ntheory.factor_ import primeomega
>>> primeomega(1)
0
>>> primeomega(20)
3
sympy.ntheory.modular.symmetric_residue(a, m)[source]

Return the residual mod m such that it is within half of the modulus.

>>> from sympy.ntheory.modular import symmetric_residue
>>> symmetric_residue(1, 6)
1
>>> symmetric_residue(4, 6)
-2
sympy.ntheory.modular.crt(m, v, symmetric=False, check=True)[source]

Chinese Remainder Theorem.

The moduli in m are assumed to be pairwise coprime. The output is then an integer f, such that f = v_i mod m_i for each pair out of v and m. If symmetric is False a positive integer will be returned, else |f| will be less than or equal to the LCM of the moduli, and thus f may be negative.

If the moduli are not co-prime the correct result will be returned if/when the test of the result is found to be incorrect. This result will be None if there is no solution.

The keyword check can be set to False if it is known that the moduli are coprime.

As an example consider a set of residues U = [49, 76, 65] and a set of moduli M = [99, 97, 95]. Then we have:

>>> from sympy.ntheory.modular import crt, solve_congruence

>>> crt([99, 97, 95], [49, 76, 65])
(639985, 912285)

This is the correct result because:

>>> [639985 % m for m in [99, 97, 95]]
[49, 76, 65]

If the moduli are not co-prime, you may receive an incorrect result if you use check=False:

>>> crt([12, 6, 17], [3, 4, 2], check=False)
(954, 1224)
>>> [954 % m for m in [12, 6, 17]]
[6, 0, 2]
>>> crt([12, 6, 17], [3, 4, 2]) is None
True
>>> crt([3, 6], [2, 5])
(5, 6)

Note: the order of gf_crt’s arguments is reversed relative to crt, and that solve_congruence takes residue, modulus pairs.

Programmer’s note: rather than checking that all pairs of moduli share no GCD (an O(n**2) test) and rather than factoring all moduli and seeing that there is no factor in common, a check that the result gives the indicated residuals is performed – an O(n) operation.

See also

solve_congruence

sympy.polys.galoistools.gf_crt
low level crt routine used by this routine
sympy.ntheory.modular.crt1(m)[source]

First part of Chinese Remainder Theorem, for multiple application.

Examples

>>> from sympy.ntheory.modular import crt1
>>> crt1([18, 42, 6])
(4536, [252, 108, 756], [0, 2, 0])
sympy.ntheory.modular.crt2(m, v, mm, e, s, symmetric=False)[source]

Second part of Chinese Remainder Theorem, for multiple application.

Examples

>>> from sympy.ntheory.modular import crt1, crt2
>>> mm, e, s = crt1([18, 42, 6])
>>> crt2([18, 42, 6], [0, 0, 0], mm, e, s)
(0, 4536)
sympy.ntheory.modular.solve_congruence(*remainder_modulus_pairs, **hint)[source]

Compute the integer n that has the residual ai when it is divided by mi where the ai and mi are given as pairs to this function: ((a1, m1), (a2, m2), …). If there is no solution, return None. Otherwise return n and its modulus.

The mi values need not be co-prime. If it is known that the moduli are not co-prime then the hint check can be set to False (default=True) and the check for a quicker solution via crt() (valid when the moduli are co-prime) will be skipped.

If the hint symmetric is True (default is False), the value of n will be within 1/2 of the modulus, possibly negative.

See also

crt
high level routine implementing the Chinese Remainder Theorem

Examples

>>> from sympy.ntheory.modular import solve_congruence

What number is 2 mod 3, 3 mod 5 and 2 mod 7?

>>> solve_congruence((2, 3), (3, 5), (2, 7))
(23, 105)
>>> [23 % m for m in [3, 5, 7]]
[2, 3, 2]

If you prefer to work with all remainder in one list and all moduli in another, send the arguments like this:

>>> solve_congruence(*zip((2, 3, 2), (3, 5, 7)))
(23, 105)

The moduli need not be co-prime; in this case there may or may not be a solution:

>>> solve_congruence((2, 3), (4, 6)) is None
True
>>> solve_congruence((2, 3), (5, 6))
(5, 6)

The symmetric flag will make the result be within 1/2 of the modulus:

>>> solve_congruence((2, 3), (5, 6), symmetric=True)
(-1, 6)
sympy.ntheory.multinomial.binomial_coefficients(n)[source]

Return a dictionary containing pairs \({(k1,k2) : C_kn}\) where \(C_kn\) are binomial coefficients and \(n=k1+k2\). Examples ========

>>> from sympy.ntheory import binomial_coefficients
>>> binomial_coefficients(9)
{(0, 9): 1, (1, 8): 9, (2, 7): 36, (3, 6): 84,
 (4, 5): 126, (5, 4): 126, (6, 3): 84, (7, 2): 36, (8, 1): 9, (9, 0): 1}
sympy.ntheory.multinomial.binomial_coefficients_list(n)[source]

Return a list of binomial coefficients as rows of the Pascal’s triangle.

Examples

>>> from sympy.ntheory import binomial_coefficients_list
>>> binomial_coefficients_list(9)
[1, 9, 36, 84, 126, 126, 84, 36, 9, 1]
sympy.ntheory.multinomial.multinomial_coefficients(m, n)[source]

Return a dictionary containing pairs {(k1,k2,..,km) : C_kn} where C_kn are multinomial coefficients such that n=k1+k2+..+km.

For example:

>>> from sympy.ntheory import multinomial_coefficients
>>> multinomial_coefficients(2, 5) # indirect doctest
{(0, 5): 1, (1, 4): 5, (2, 3): 10, (3, 2): 10, (4, 1): 5, (5, 0): 1}

The algorithm is based on the following result:

\[\binom{n}{k_1, \ldots, k_m} = \frac{k_1 + 1}{n - k_1} \sum_{i=2}^m \binom{n}{k_1 + 1, \ldots, k_i - 1, \ldots}\]

Code contributed to Sage by Yann Laigle-Chapuy, copied with permission of the author.

sympy.ntheory.multinomial.multinomial_coefficients_iterator(m, n, _tuple=<class 'tuple'>)[source]

multinomial coefficient iterator

This routine has been optimized for \(m\) large with respect to \(n\) by taking advantage of the fact that when the monomial tuples \(t\) are stripped of zeros, their coefficient is the same as that of the monomial tuples from multinomial_coefficients(n, n). Therefore, the latter coefficients are precomputed to save memory and time.

>>> from sympy.ntheory.multinomial import multinomial_coefficients
>>> m53, m33 = multinomial_coefficients(5,3), multinomial_coefficients(3,3)
>>> m53[(0,0,0,1,2)] == m53[(0,0,1,0,2)] == m53[(1,0,2,0,0)] == m33[(0,1,2)]
True

Examples

>>> from sympy.ntheory.multinomial import multinomial_coefficients_iterator
>>> it = multinomial_coefficients_iterator(20,3)
>>> next(it)
((3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), 1)
sympy.ntheory.partitions_.npartitions(n, verbose=False)[source]

Calculate the partition function P(n), i.e. the number of ways that n can be written as a sum of positive integers.

P(n) is computed using the Hardy-Ramanujan-Rademacher formula [R404].

The correctness of this implementation has been tested through 10**10.

References

[R404](1, 2) http://mathworld.wolfram.com/PartitionFunctionP.html

Examples

>>> from sympy.ntheory import npartitions
>>> npartitions(25)
1958
sympy.ntheory.primetest.mr(n, bases)[source]

Perform a Miller-Rabin strong pseudoprime test on n using a given list of bases/witnesses.

References

  • Richard Crandall & Carl Pomerance (2005), “Prime Numbers: A Computational Perspective”, Springer, 2nd edition, 135-138

A list of thresholds and the bases they require are here: http://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_test#Deterministic_variants_of_the_test

Examples

>>> from sympy.ntheory.primetest import mr
>>> mr(1373651, [2, 3])
False
>>> mr(479001599, [31, 73])
True
sympy.ntheory.primetest.isprime(n)[source]

Test if n is a prime number (True) or not (False). For n < 2^64 the answer is definitive; larger n values have a small probability of actually being pseudoprimes.

Negative numbers (e.g. -2) are not considered prime.

The first step is looking for trivial factors, which if found enables a quick return. Next, if the sieve is large enough, use bisection search on the sieve. For small numbers, a set of deterministic Miller-Rabin tests are performed with bases that are known to have no counterexamples in their range. Finally if the number is larger than 2^64, a strong BPSW test is performed. While this is a probable prime test and we believe counterexamples exist, there are no known counterexamples.

See also

sympy.ntheory.generate.primerange
Generates all primes in a given range
sympy.ntheory.generate.primepi
Return the number of primes less than or equal to n
sympy.ntheory.generate.prime
Return the nth prime

References

Examples

>>> from sympy.ntheory import isprime
>>> isprime(13)
True
>>> isprime(15)
False
sympy.ntheory.residue_ntheory.n_order(a, n)[source]

Returns the order of a modulo n.

The order of a modulo n is the smallest integer k such that a**k leaves a remainder of 1 with n.

Examples

>>> from sympy.ntheory import n_order
>>> n_order(3, 7)
6
>>> n_order(4, 7)
3
sympy.ntheory.residue_ntheory.is_primitive_root(a, p)[source]

Returns True if a is a primitive root of p

a is said to be the primitive root of p if gcd(a, p) == 1 and totient(p) is the smallest positive number s.t.

a**totient(p) cong 1 mod(p)

Examples

>>> from sympy.ntheory import is_primitive_root, n_order, totient
>>> is_primitive_root(3, 10)
True
>>> is_primitive_root(9, 10)
False
>>> n_order(3, 10) == totient(10)
True
>>> n_order(9, 10) == totient(10)
False
sympy.ntheory.residue_ntheory.primitive_root(p)[source]

Returns the smallest primitive root or None

Parameters:p : positive integer

References

[R405]
  1. Stein “Elementary Number Theory” (2011), page 44
[R406]
  1. Hackman “Elementary Number Theory” (2009), Chapter C

Examples

>>> from sympy.ntheory.residue_ntheory import primitive_root
>>> primitive_root(19)
2
sympy.ntheory.residue_ntheory.sqrt_mod(a, p, all_roots=False)[source]

Find a root of x**2 = a mod p

Parameters:

a : integer

p : positive integer

all_roots : if True the list of roots is returned or None

Notes

If there is no root it is returned None; else the returned root is less or equal to p // 2; in general is not the smallest one. It is returned p // 2 only if it is the only root.

Use all_roots only when it is expected that all the roots fit in memory; otherwise use sqrt_mod_iter.

Examples

>>> from sympy.ntheory import sqrt_mod
>>> sqrt_mod(11, 43)
21
>>> sqrt_mod(17, 32, True)
[7, 9, 23, 25]
sympy.ntheory.residue_ntheory.quadratic_residues(p)[source]

Returns the list of quadratic residues.

Examples

>>> from sympy.ntheory.residue_ntheory import quadratic_residues
>>> quadratic_residues(7)
[0, 1, 2, 4]
sympy.ntheory.residue_ntheory.nthroot_mod(a, n, p, all_roots=False)[source]

Find the solutions to x**n = a mod p

Parameters:

a : integer

n : positive integer

p : positive integer

all_roots : if False returns the smallest root, else the list of roots

Examples

>>> from sympy.ntheory.residue_ntheory import nthroot_mod
>>> nthroot_mod(11, 4, 19)
8
>>> nthroot_mod(11, 4, 19, True)
[8, 11]
>>> nthroot_mod(68, 3, 109)
23
sympy.ntheory.residue_ntheory.is_nthpow_residue(a, n, m)[source]

Returns True if x**n == a (mod m) has solutions.

References

[R407]
  1. Hackman “Elementary Number Theory” (2009), page 76
sympy.ntheory.residue_ntheory.is_quad_residue(a, p)[source]

Returns True if a (mod p) is in the set of squares mod p, i.e a % p in set([i**2 % p for i in range(p)]). If p is an odd prime, an iterative method is used to make the determination:

>>> from sympy.ntheory import is_quad_residue
>>> sorted(set([i**2 % 7 for i in range(7)]))
[0, 1, 2, 4]
>>> [j for j in range(7) if is_quad_residue(j, 7)]
[0, 1, 2, 4]
sympy.ntheory.residue_ntheory.legendre_symbol(a, p)[source]

Returns the Legendre symbol \((a / p)\).

For an integer a and an odd prime p, the Legendre symbol is defined as

\[\begin{split}\genfrac(){}{}{a}{p} = \begin{cases} 0 & \text{if } p \text{ divides } a\\ 1 & \text{if } a \text{ is a quadratic residue modulo } p\\ -1 & \text{if } a \text{ is a quadratic nonresidue modulo } p \end{cases}\end{split}\]
Parameters:

a : integer

p : odd prime

Examples

>>> from sympy.ntheory import legendre_symbol
>>> [legendre_symbol(i, 7) for i in range(7)]
[0, 1, 1, -1, 1, -1, -1]
>>> sorted(set([i**2 % 7 for i in range(7)]))
[0, 1, 2, 4]
sympy.ntheory.residue_ntheory.jacobi_symbol(m, n)[source]

Returns the Jacobi symbol \((m / n)\).

For any integer m and any positive odd integer n the Jacobi symbol is defined as the product of the Legendre symbols corresponding to the prime factors of n:

\[\genfrac(){}{}{m}{n} = \genfrac(){}{}{m}{p^{1}}^{\alpha_1} \genfrac(){}{}{m}{p^{2}}^{\alpha_2} ... \genfrac(){}{}{m}{p^{k}}^{\alpha_k} \text{ where } n = p_1^{\alpha_1} p_2^{\alpha_2} ... p_k^{\alpha_k}\]

Like the Legendre symbol, if the Jacobi symbol \(\genfrac(){}{}{m}{n} = -1\) then m is a quadratic nonresidue modulo n.

But, unlike the Legendre symbol, if the Jacobi symbol \(\genfrac(){}{}{m}{n} = 1\) then m may or may not be a quadratic residue modulo n.

Parameters:

m : integer

n : odd positive integer

Examples

>>> from sympy.ntheory import jacobi_symbol, legendre_symbol
>>> from sympy import Mul, S
>>> jacobi_symbol(45, 77)
-1
>>> jacobi_symbol(60, 121)
1

The relationship between the jacobi_symbol and legendre_symbol can be demonstrated as follows:

>>> L = legendre_symbol
>>> S(45).factors()
{3: 2, 5: 1}
>>> jacobi_symbol(7, 45) == L(7, 3)**2 * L(7, 5)**1
True
sympy.ntheory.residue_ntheory.discrete_log(n, a, b, order=None, prime_order=None)[source]

Compute the discrete logarithm of a to the base b modulo n.

This is a recursive function to reduce the discrete logarithm problem in cyclic groups of composite order to the problem in cyclic groups of prime order.

It employs different algorithms depending on the problem (subgroup order size, prime order or not):

  • Trial multiplication
  • Baby-step giant-step
  • Pollard’s Rho
  • Pohlig-Hellman

References

[R408]http://mathworld.wolfram.com/DiscreteLogarithm.html
[R409]“Handbook of applied cryptography”, Menezes, A. J., Van, O. P. C., & Vanstone, S. A. (1997).

Examples

>>> from sympy.ntheory import discrete_log
>>> discrete_log(41, 15, 7)
3
sympy.ntheory.continued_fraction.continued_fraction_convergents(cf)[source]

Return an iterator over the convergents of a continued fraction (cf).

The parameter should be an iterable returning successive partial quotients of the continued fraction, such as might be returned by continued_fraction_iterator. In computing the convergents, the continued fraction need not be strictly in canonical form (all integers, all but the first positive). Rational and negative elements may be present in the expansion.

Examples

>>> from sympy.core import Rational, pi
>>> from sympy import S
>>> from sympy.ntheory.continued_fraction import             continued_fraction_convergents, continued_fraction_iterator
>>> list(continued_fraction_convergents([0, 2, 1, 2]))
[0, 1/2, 1/3, 3/8]
>>> list(continued_fraction_convergents([1, S('1/2'), -7, S('1/4')]))
[1, 3, 19/5, 7]
>>> it = continued_fraction_convergents(continued_fraction_iterator(pi))
>>> for n in range(7):
...     print(next(it))
3
22/7
333/106
355/113
103993/33102
104348/33215
208341/66317
sympy.ntheory.continued_fraction.continued_fraction_iterator(x)[source]

Return continued fraction expansion of x as iterator.

References

[R410]http://en.wikipedia.org/wiki/Continued_fraction

Examples

>>> from sympy.core import Rational, pi
>>> from sympy.ntheory.continued_fraction import continued_fraction_iterator
>>> list(continued_fraction_iterator(Rational(3, 8)))
[0, 2, 1, 2]
>>> list(continued_fraction_iterator(Rational(-3, 8)))
[-1, 1, 1, 1, 2]
>>> for i, v in enumerate(continued_fraction_iterator(pi)):
...     if i > 7:
...         break
...     print(v)
3
7
15
1
292
1
1
1
sympy.ntheory.continued_fraction.continued_fraction_periodic(p, q, d=0)[source]

Find the periodic continued fraction expansion of a quadratic irrational.

Compute the continued fraction expansion of a rational or a quadratic irrational number, i.e. \(\frac{p + \sqrt{d}}{q}\), where \(p\), \(q\) and \(d \ge 0\) are integers.

Returns the continued fraction representation (canonical form) as a list of integers, optionally ending (for quadratic irrationals) with repeating block as the last term of this list.

Parameters:

p : int

the rational part of the number’s numerator

q : int

the denominator of the number

d : int, optional

the irrational part (discriminator) of the number’s numerator

References

[R411]http://en.wikipedia.org/wiki/Periodic_continued_fraction
[R412]K. Rosen. Elementary Number theory and its applications. Addison-Wesley, 3 Sub edition, pages 379-381, January 1992.

Examples

>>> from sympy.ntheory.continued_fraction import continued_fraction_periodic
>>> continued_fraction_periodic(3, 2, 7)
[2, [1, 4, 1, 1]]

Golden ratio has the simplest continued fraction expansion:

>>> continued_fraction_periodic(1, 2, 5)
[[1]]

If the discriminator is zero or a perfect square then the number will be a rational number:

>>> continued_fraction_periodic(4, 3, 0)
[1, 3]
>>> continued_fraction_periodic(4, 3, 49)
[3, 1, 2]
sympy.ntheory.continued_fraction.continued_fraction_reduce(cf)[source]

Reduce a continued fraction to a rational or quadratic irrational.

Compute the rational or quadratic irrational number from its terminating or periodic continued fraction expansion. The continued fraction expansion (cf) should be supplied as a terminating iterator supplying the terms of the expansion. For terminating continued fractions, this is equivalent to list(continued_fraction_convergents(cf))[-1], only a little more efficient. If the expansion has a repeating part, a list of the repeating terms should be returned as the last element from the iterator. This is the format returned by continued_fraction_periodic.

For quadratic irrationals, returns the largest solution found, which is generally the one sought, if the fraction is in canonical form (all terms positive except possibly the first).

Examples

>>> from sympy.ntheory.continued_fraction import continued_fraction_reduce
>>> continued_fraction_reduce([1, 2, 3, 4, 5])
225/157
>>> continued_fraction_reduce([-2, 1, 9, 7, 1, 2])
-256/233
>>> continued_fraction_reduce([2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8]).n(10)
2.718281835
>>> continued_fraction_reduce([1, 4, 2, [3, 1]])
(sqrt(21) + 287)/238
>>> continued_fraction_reduce([[1]])
1/2 + sqrt(5)/2
>>> from sympy.ntheory.continued_fraction import continued_fraction_periodic
>>> continued_fraction_reduce(continued_fraction_periodic(8, 5, 13))
(sqrt(13) + 8)/5
class sympy.ntheory.mobius[source]

Möbius function maps natural number to {-1, 0, 1}

It is defined as follows:
  1. \(1\) if \(n = 1\).
  2. \(0\) if \(n\) has a squared prime factor.
  3. \((-1)^k\) if \(n\) is a square-free positive integer with \(k\) number of prime factors.

It is an important multiplicative function in number theory and combinatorics. It has applications in mathematical series, algebraic number theory and also physics (Fermion operator has very concrete realization with Möbius Function model).

Parameters:n : positive integer

References

[R413]http://en.wikipedia.org/wiki/M%C3%B6bius_function
[R414]Thomas Koshy “Elementary Number Theory with Applications”

Examples

>>> from sympy.ntheory import mobius
>>> mobius(13*7)
1
>>> mobius(1)
1
>>> mobius(13*7*5)
-1
>>> mobius(13**2)
0
sympy.ntheory.egyptian_fraction.egyptian_fraction(r, algorithm='Greedy')[source]

Return the list of denominators of an Egyptian fraction expansion [R415] of the said rational \(r\).

Parameters:

r : Rational

a positive rational number.

algorithm : { “Greedy”, “Graham Jewett”, “Takenouchi”, “Golomb” }, optional

Denotes the algorithm to be used (the default is “Greedy”).

Notes

Currently the following algorithms are supported:

  1. Greedy Algorithm

    Also called the Fibonacci-Sylvester algorithm [R416]. At each step, extract the largest unit fraction less than the target and replace the target with the remainder.

    It has some distinct properties:

    1. Given \(p/q\) in lowest terms, generates an expansion of maximum length \(p\). Even as the numerators get large, the number of terms is seldom more than a handful.
    2. Uses minimal memory.
    3. The terms can blow up (standard examples of this are 5/121 and 31/311). The denominator is at most squared at each step (doubly-exponential growth) and typically exhibits singly-exponential growth.
  2. Graham Jewett Algorithm

    The algorithm suggested by the result of Graham and Jewett. Note that this has a tendency to blow up: the length of the resulting expansion is always 2**(x/gcd(x, y)) - 1. See [R417].

  3. Takenouchi Algorithm

    The algorithm suggested by Takenouchi (1921). Differs from the Graham-Jewett algorithm only in the handling of duplicates. See [R417].

  4. Golomb’s Algorithm

    A method given by Golumb (1962), using modular arithmetic and inverses. It yields the same results as a method using continued fractions proposed by Bleicher (1972). See [R418].

If the given rational is greater than or equal to 1, a greedy algorithm of summing the harmonic sequence 1/1 + 1/2 + 1/3 + … is used, taking all the unit fractions of this sequence until adding one more would be greater than the given number. This list of denominators is prefixed to the result from the requested algorithm used on the remainder. For example, if r is 8/3, using the Greedy algorithm, we get [1, 2, 3, 4, 5, 6, 7, 14, 420], where the beginning of the sequence, [1, 2, 3, 4, 5, 6, 7] is part of the harmonic sequence summing to 363/140, leaving a remainder of 31/420, which yields [14, 420] by the Greedy algorithm. The result of egyptian_fraction(Rational(8, 3), “Golomb”) is [1, 2, 3, 4, 5, 6, 7, 14, 574, 2788, 6460, 11590, 33062, 113820], and so on.

References

[R415](1, 2) http://en.wikipedia.org/wiki/Egyptian_fraction
[R416](1, 2) https://en.wikipedia.org/wiki/Greedy_algorithm_for_Egyptian_fractions
[R417](1, 2, 3) http://www.ics.uci.edu/~eppstein/numth/egypt/conflict.html
[R418](1, 2) http://ami.ektf.hu/uploads/papers/finalpdf/AMI_42_from129to134.pdf

Examples

>>> from sympy import Rational
>>> from sympy.ntheory.egyptian_fraction import egyptian_fraction
>>> egyptian_fraction(Rational(3, 7))
[3, 11, 231]
>>> egyptian_fraction(Rational(3, 7), "Graham Jewett")
[7, 8, 9, 56, 57, 72, 3192]
>>> egyptian_fraction(Rational(3, 7), "Takenouchi")
[4, 7, 28]
>>> egyptian_fraction(Rational(3, 7), "Golomb")
[3, 15, 35]
>>> egyptian_fraction(Rational(11, 5), "Golomb")
[1, 2, 3, 4, 9, 234, 1118, 2580]