Elements of the ring \(\ZZ\) of integers#
Sage has highly optimized and extensive functionality for arithmetic with integers and the ring of integers.
EXAMPLES:
Add 2 integers:
sage: a = Integer(3) ; b = Integer(4)
sage: a + b == 7
True
Add an integer and a real number:
sage: a + 4.0
7.00000000000000
Add an integer and a rational number:
sage: a + Rational(2)/5
17/5
Add an integer and a complex number:
sage: b = ComplexField().0 + 1.5
sage: loads((a+b).dumps()) == a+b
True
sage: z = 32
sage: -z
-32
sage: z = 0; -z
0
sage: z = -0; -z
0
sage: z = -1; -z
1
Multiplication:
sage: a = Integer(3) ; b = Integer(4)
sage: a * b == 12
True
sage: loads((a * 4.0).dumps()) == a*b
True
sage: a * Rational(2)/5
6/5
sage: [2,3] * 4
[2, 3, 2, 3, 2, 3, 2, 3]
sage: 'sage'*Integer(3)
'sagesagesage'
COERCIONS:
Return version of this integer in the multi-precision floating real field R:
sage: n = 9390823
sage: RR = RealField(200)
sage: RR(n)
9.3908230000000000000000000000000000000000000000000000000000e6
AUTHORS:
William Stein (2005): initial version
Gonzalo Tornaria (2006-03-02): vastly improved python/GMP conversion; hashing
Didier Deshommes (2006-03-06): numerous examples and docstrings
William Stein (2006-03-31): changes to reflect GMP bug fixes
William Stein (2006-04-14): added GMP factorial method (since it’s now very fast).
David Harvey (2006-09-15): added nth_root, exact_log
David Harvey (2006-09-16): attempt to optimise Integer constructor
Rishikesh (2007-02-25): changed quo_rem so that the rem is positive
David Harvey, Martin Albrecht, Robert Bradshaw (2007-03-01): optimized Integer constructor and pool
Pablo De Napoli (2007-04-01): multiplicative_order should return +infinity for non zero numbers
Robert Bradshaw (2007-04-12): is_perfect_power, Jacobi symbol (with Kronecker extension). Convert some methods to use GMP directly rather than PARI, Integer(), PY_NEW(Integer)
David Roe (2007-03-21): sped up valuation and is_square, added val_unit, is_power, is_power_of and divide_knowing_divisible_by
Robert Bradshaw (2008-03-26): gamma function, multifactorials
Robert Bradshaw (2008-10-02): bounded squarefree part
David Loeffler (2011-01-15): fixed bug #10625 (inverse_mod should accept an ideal as argument)
Vincent Delecroix (2010-12-28): added unicode in Integer.__init__
David Roe (2012-03): deprecate
is_power()
in favour ofis_perfect_power()
(see trac ticket #12116)Vincent Delecroix (2017-05-03): faster integer-rational comparisons
Vincent Klein (2017-05-11): add __mpz__() to class Integer
Vincent Klein (2017-05-22): Integer constructor support gmpy2.mpz parameter
Samuel Lelièvre (2018-08-02): document that divisors are sorted (trac ticket #25983)
- sage.rings.integer.GCD_list(v)#
Return the greatest common divisor of a list of integers.
INPUT:
v
– list or tuple
Elements of \(v\) are converted to Sage integers. An empty list has GCD zero.
This function is used, for example, by
rings/arith.py
.EXAMPLES:
sage: from sage.rings.integer import GCD_list sage: w = GCD_list([3,9,30]); w 3 sage: type(w) <class 'sage.rings.integer.Integer'>
Check that the bug reported in trac ticket #3118 has been fixed:
sage: sage.rings.integer.GCD_list([2,2,3]) 1
The inputs are converted to Sage integers.
sage: w = GCD_list([int(3), int(9), '30']); w 3 sage: type(w) <class 'sage.rings.integer.Integer'>
Check that the GCD of the empty list is zero (trac ticket #17257):
sage: GCD_list([]) 0
- class sage.rings.integer.Integer#
Bases:
sage.structure.element.EuclideanDomainElement
The
Integer
class represents arbitrary precision integers. It derives from theElement
class, so integers can be used as ring elements anywhere in Sage.Integer() interprets strings that begin with
0o
as octal numbers, strings that begin with0x
as hexadecimal numbers and strings that begin with0b
as binary numbers.The class
Integer
is implemented in Cython, as a wrapper of the GMPmpz_t
integer type.EXAMPLES:
sage: Integer(123) 123 sage: Integer("123") 123
Sage Integers support PEP 3127 literals:
sage: Integer('0x12') 18 sage: Integer('-0o12') -10 sage: Integer('+0b101010') 42
Conversion from PARI:
sage: Integer(pari('-10380104371593008048799446356441519384')) -10380104371593008048799446356441519384 sage: Integer(pari('Pol([-3])')) -3
Conversion from gmpy2:
sage: from gmpy2 import mpz sage: Integer(mpz(3)) 3
- __pow__(left, right, modulus)#
Return
(left ^ right) % modulus
.EXAMPLES:
sage: 2^-6 1/64 sage: 2^6 64 sage: 2^0 1 sage: 2^-0 1 sage: (-1)^(1/3) (-1)^(1/3)
For consistency with Python and MPFR, 0^0 is defined to be 1 in Sage:
sage: 0^0 1
See also http://www.faqs.org/faqs/sci-math-faq/0to0/ and https://math.stackexchange.com/questions/11150/zero-to-the-zero-power-is-00-1.
The base need not be a Sage integer. If it is a Python type, the result is a Python type too:
sage: r = int(2) ^ 10; r; type(r) 1024 <... 'int'> sage: r = int(3) ^ -3; r; type(r) 0.037037037037037035 <... 'float'> sage: r = float(2.5) ^ 10; r; type(r) 9536.7431640625 <... 'float'>
We raise 2 to various interesting exponents:
sage: 2^x # symbolic x 2^x sage: 2^1.5 # real number 2.82842712474619 sage: 2^float(1.5) # python float abs tol 3e-16 2.8284271247461903 sage: 2^I # complex number 2^I sage: r = 2 ^ int(-3); r; type(r) 1/8 <class 'sage.rings.rational.Rational'> sage: f = 2^(sin(x)-cos(x)); f 2^(-cos(x) + sin(x)) sage: f(x=3) 2^(-cos(3) + sin(3))
A symbolic sum:
sage: x,y,z = var('x,y,z') sage: 2^(x+y+z) 2^(x + y + z) sage: 2^(1/2) sqrt(2) sage: 2^(-1/2) 1/2*sqrt(2)
- additive_order()#
Return the additive order of self.
EXAMPLES:
sage: ZZ(0).additive_order() 1 sage: ZZ(1).additive_order() +Infinity
- as_integer_ratio()#
Return the pair
(self.numerator(), self.denominator())
, which is(self, 1)
.EXAMPLES:
sage: x = -12 sage: x.as_integer_ratio() (-12, 1)
- balanced_digits(base=10, positive_shift=True)#
Return the list of balanced digits for
self
in the given base.The balanced base
b
usesb
digits centered around zero. Thus ifb
is odd, there is only one possibility, namely digits between-b//2
andb//2
(both included). For instance in base 9, one uses digits from -4 to 4. Ifb
is even, one has to choose between digits from-b//2
tob//2 - 1
or-b//2 + 1
tob//2
(base 10 for instance: either -5 to 4 or -4 to 5), and this is defined by the value ofpositive_shift
.INPUT:
base
– integer (default: 10); whenbase
is 2, only the nonnegative or the nonpositive integers can be represented bybalanced_digits
. Thus we say base must be greater than 2.positive_shift
– boolean (default: True); for even bases, the representation uses digits from-b//2 + 1
tob//2
if set to True, and from-b//2
tob//2 - 1
otherwise. This has no effect for odd bases.
EXAMPLES:
sage: 8.balanced_digits(3) [-1, 0, 1] sage: (-15).balanced_digits(5) [0, 2, -1] sage: 17.balanced_digits(6) [-1, 3] sage: 17.balanced_digits(6, positive_shift=False) [-1, -3, 1] sage: (-46).balanced_digits() [4, 5, -1] sage: (-46).balanced_digits(positive_shift=False) [4, -5] sage: (-23).balanced_digits(12) [1, -2] sage: (-23).balanced_digits(12, positive_shift=False) [1, -2] sage: 0.balanced_digits(7) [] sage: 14.balanced_digits(5.8) Traceback (most recent call last): ... ValueError: base must be an integer sage: 14.balanced_digits(2) Traceback (most recent call last): ... ValueError: base must be > 2
See also
- binary()#
Return the binary digits of
self
as a string.EXAMPLES:
sage: print(Integer(15).binary()) 1111 sage: print(Integer(16).binary()) 10000 sage: print(Integer(16938402384092843092843098243).binary()) 1101101011101100011110001110010010100111010001101010001111111000101000000000101111000010000011
- binomial(m, algorithm='gmp')#
Return the binomial coefficient “self choose m”.
INPUT:
m
– an integeralgorithm
–'gmp'
(default),'mpir'
(an alias forgmp
), or'pari'
;'gmp'
is faster for smallm
, and'pari'
tends to be faster for largem
OUTPUT:
integer
EXAMPLES:
sage: 10.binomial(2) 45 sage: 10.binomial(2, algorithm='pari') 45 sage: 10.binomial(-2) 0 sage: (-2).binomial(3) -4 sage: (-3).binomial(0) 1
The argument
m
or (self-m
) must fit into unsigned long:sage: (2**256).binomial(2**256) 1 sage: (2**256).binomial(2**256-1) 115792089237316195423570985008687907853269984665640564039457584007913129639936 sage: (2**256).binomial(2**128) Traceback (most recent call last): ... OverflowError: m must fit in an unsigned long
- bit_length()#
Return the number of bits required to represent this integer.
Identical to
int.bit_length()
.EXAMPLES:
sage: 500.bit_length() 9 sage: 5.bit_length() 3 sage: 0.bit_length() == len(0.bits()) == 0.ndigits(base=2) True sage: 12345.bit_length() == len(12345.binary()) True sage: 1023.bit_length() 10 sage: 1024.bit_length() 11
- bits()#
Return the bits in self as a list, least significant first. The result satisfies the identity
x == sum(b*2^e for e, b in enumerate(x.bits()))
Negative numbers will have negative “bits”. (So, strictly speaking, the entries of the returned list are not really members of \(\ZZ/2\ZZ\).)
This method just calls
digits()
withbase=2
.See also
bit_length()
, a faster way to computelen(x.bits())
binary()
, which returns a string in perhaps more familiar notation
EXAMPLES:
sage: 500.bits() [0, 0, 1, 0, 1, 1, 1, 1, 1] sage: 11.bits() [1, 1, 0, 1] sage: (-99).bits() [-1, -1, 0, 0, 0, -1, -1]
- ceil()#
Return the ceiling of self, which is self since self is an integer.
EXAMPLES:
sage: n = 6 sage: n.ceil() 6
- class_number(proof=True)#
Return the class number of the quadratic order with this discriminant.
INPUT:
self
– an integer congruent to \(0\) or \(1\mod4\) which is not a squareproof
(boolean, defaultTrue
) – ifFalse
then for negative discriminants a faster algorithm is used by the PARI library which is known to give incorrect results when the class group has many cyclic factors.
OUTPUT:
(integer) the class number of the quadratic order with this discriminant.
Note
This is not always equal to the number of classes of primitive binary quadratic forms of discriminant \(D\), which is equal to the narrow class number. The two notions are the same when \(D<0\), or \(D>0\) and the fundamental unit of the order has negative norm; otherwise the number of classes of forms is twice this class number.
EXAMPLES:
sage: (-163).class_number() 1 sage: (-104).class_number() 6 sage: [((4*n+1),(4*n+1).class_number()) for n in [21..29]] [(85, 2), (89, 1), (93, 1), (97, 1), (101, 1), (105, 2), (109, 1), (113, 1), (117, 1)]
- conjugate()#
Return the complex conjugate of this integer, which is the integer itself.
EXAMPLES:
sage: n = 205 sage: n.conjugate() 205
- coprime_integers(m)#
Return the non-negative integers \(< m\) that are coprime to this integer.
EXAMPLES:
sage: n = 8 sage: n.coprime_integers(8) [1, 3, 5, 7] sage: n.coprime_integers(11) [1, 3, 5, 7, 9] sage: n = 5; n.coprime_integers(10) [1, 2, 3, 4, 6, 7, 8, 9] sage: n.coprime_integers(5) [1, 2, 3, 4] sage: n = 99; n.coprime_integers(99) [1, 2, 4, 5, 7, 8, 10, 13, 14, 16, 17, 19, 20, 23, 25, 26, 28, 29, 31, 32, 34, 35, 37, 38, 40, 41, 43, 46, 47, 49, 50, 52, 53, 56, 58, 59, 61, 62, 64, 65, 67, 68, 70, 71, 73, 74, 76, 79, 80, 82, 83, 85, 86, 89, 91, 92, 94, 95, 97, 98]
AUTHORS:
Naqi Jaffery (2006-01-24): examples
David Roe (2017-10-02): Use sieving
Jeroen Demeyer (2018-06-25): allow returning zero (only relevant for 1.coprime_integers(n))
ALGORITHM:
Create an integer with \(m\) bits and set bits at every multiple of a prime \(p\) that divides this integer and is less than \(m\). Then return a list of integers corresponding to the unset bits.
- crt(y, m, n)#
Return the unique integer between \(0\) and \(mn\) that is congruent to the integer modulo \(m\) and to \(y\) modulo \(n\). We assume that \(m\) and \(n\) are coprime.
EXAMPLES:
sage: n = 17 sage: m = n.crt(5, 23, 11); m 247 sage: m%23 17 sage: m%11 5
- denominator()#
Return the denominator of this integer, which of course is always 1.
EXAMPLES:
sage: x = 5 sage: x.denominator() 1 sage: x = 0 sage: x.denominator() 1
- digits(base=10, digits=None, padto=0)#
Return a list of digits for
self
in the given base in little endian order.The returned value is unspecified if self is a negative number and the digits are given.
INPUT:
base
- integer (default: 10)digits
- optional indexable object as source for the digitspadto
- the minimal length of the returned list, sufficient number of zeros are added to make the list minimum that length (default: 0)
As a shorthand for
digits(2)
, you can usebits()
.Also see
ndigits()
.EXAMPLES:
sage: 17.digits() [7, 1] sage: 5.digits(base=2, digits=["zero","one"]) ['one', 'zero', 'one'] sage: 5.digits(3) [2, 1] sage: 0.digits(base=10) # 0 has 0 digits [] sage: 0.digits(base=2) # 0 has 0 digits [] sage: 10.digits(16,'0123456789abcdef') ['a'] sage: 0.digits(16,'0123456789abcdef') [] sage: 0.digits(16,'0123456789abcdef',padto=1) ['0'] sage: 123.digits(base=10,padto=5) [3, 2, 1, 0, 0] sage: 123.digits(base=2,padto=3) # padto is the minimal length [1, 1, 0, 1, 1, 1, 1] sage: 123.digits(base=2,padto=10,digits=(1,-1)) [-1, -1, 1, -1, -1, -1, -1, 1, 1, 1] sage: a=9939082340; a.digits(10) [0, 4, 3, 2, 8, 0, 9, 3, 9, 9] sage: a.digits(512) [100, 302, 26, 74] sage: (-12).digits(10) [-2, -1] sage: (-12).digits(2) [0, 0, -1, -1]
We support large bases.
sage: n=2^6000 sage: n.digits(2^3000) [0, 0, 1]
sage: base=3; n=25 sage: l=n.digits(base) sage: # the next relationship should hold for all n,base sage: sum(base^i*l[i] for i in range(len(l)))==n True sage: base=3; n=-30; l=n.digits(base); sum(base^i*l[i] for i in range(len(l)))==n True
The inverse of this method – constructing an integer from a list of digits and a base – can be done using the above method or by simply using
ZZ()
with a base:sage: x = 123; ZZ(x.digits(), 10) 123 sage: x == ZZ(x.digits(6), 6) True sage: x == ZZ(x.digits(25), 25) True
Using
sum()
andenumerate()
to do the same thing is slightly faster in many cases (andbalanced_sum()
may be faster yet). Of course it gives the same result:sage: base = 4 sage: sum(digit * base^i for i, digit in enumerate(x.digits(base))) == ZZ(x.digits(base), base) True
Note: In some cases it is faster to give a digits collection. This would be particularly true for computing the digits of a series of small numbers. In these cases, the code is careful to allocate as few python objects as reasonably possible.
sage: digits = list(range(15)) sage: l = [ZZ(i).digits(15,digits) for i in range(100)] sage: l[16] [1, 1]
This function is comparable to
str
for speed.sage: n=3^100000 sage: n.digits(base=10)[-1] # slightly slower than str 1 sage: n=10^10000 sage: n.digits(base=10)[-1] # slightly faster than str 1
AUTHORS:
Joel B. Mohler (2008-03-02): significantly rewrote this entire function
- divide_knowing_divisible_by(right)#
Returns the integer self / right when self is divisible by right.
If self is not divisible by right, the return value is undefined, and may not even be close to self/right for multi-word integers.
EXAMPLES:
sage: a = 8; b = 4 sage: a.divide_knowing_divisible_by(b) 2 sage: (100000).divide_knowing_divisible_by(25) 4000 sage: (100000).divide_knowing_divisible_by(26) # close (random) 3846
However, often it’s way off.
sage: a = 2^70; a 1180591620717411303424 sage: a // 11 # floor divide 107326510974310118493 sage: a.divide_knowing_divisible_by(11) # way off and possibly random 43215361478743422388970455040
- divides(n)#
Return True if self divides n.
EXAMPLES:
sage: Z = IntegerRing() sage: Z(5).divides(Z(10)) True sage: Z(0).divides(Z(5)) False sage: Z(10).divides(Z(5)) False
- divisors(method=None)#
Return the list of all positive integer divisors of this integer, sorted in increasing order.
EXAMPLES:
sage: (-3).divisors() [1, 3] sage: 6.divisors() [1, 2, 3, 6] sage: 28.divisors() [1, 2, 4, 7, 14, 28] sage: (2^5).divisors() [1, 2, 4, 8, 16, 32] sage: 100.divisors() [1, 2, 4, 5, 10, 20, 25, 50, 100] sage: 1.divisors() [1] sage: 0.divisors() Traceback (most recent call last): ... ValueError: n must be nonzero sage: (2^3 * 3^2 * 17).divisors() [1, 2, 3, 4, 6, 8, 9, 12, 17, 18, 24, 34, 36, 51, 68, 72, 102, 136, 153, 204, 306, 408, 612, 1224] sage: a = odd_part(factorial(31)) sage: v = a.divisors() sage: len(v) 172800 sage: prod(e + 1 for p, e in factor(a)) 172800 sage: all(t.divides(a) for t in v) True
sage: n = 2^551 - 1 sage: L = n.divisors() sage: len(L) 256 sage: L[-1] == n True
Note
If one first computes all the divisors and then sorts it, the sorting step can easily dominate the runtime. Note, however, that (non-negative) multiplication on the left preserves relative order. One can leverage this fact to keep the list in order as one computes it using a process similar to that of the merge sort algorithm.
- euclidean_degree()#
Return the degree of this element as an element of an Euclidean domain.
If this is an element in the ring of integers, this is simply its absolute value.
EXAMPLES:
sage: ZZ(1).euclidean_degree() 1
- exact_log(m)#
Returns the largest integer \(k\) such that \(m^k \leq \text{self}\), i.e., the floor of \(\log_m(\text{self})\).
This is guaranteed to return the correct answer even when the usual log function doesn’t have sufficient precision.
INPUT:
m
- integer >= 2
AUTHORS:
David Harvey (2006-09-15)
- Joel B. Mohler (2009-04-08) – rewrote this to handle small cases
and/or easy cases up to 100x faster..
EXAMPLES:
sage: Integer(125).exact_log(5) 3 sage: Integer(124).exact_log(5) 2 sage: Integer(126).exact_log(5) 3 sage: Integer(3).exact_log(5) 0 sage: Integer(1).exact_log(5) 0 sage: Integer(178^1700).exact_log(178) 1700 sage: Integer(178^1700-1).exact_log(178) 1699 sage: Integer(178^1700+1).exact_log(178) 1700 sage: # we need to exercise the large base code path too sage: Integer(1780^1700-1).exact_log(1780) 1699 sage: # The following are very very fast. sage: # Note that for base m a perfect power of 2, we get the exact log by counting bits. sage: n=2983579823750185701375109835; m=32 sage: n.exact_log(m) 18 sage: # The next is a favorite of mine. The log2 approximate is exact and immediately provable. sage: n=90153710570912709517902579010793251709257901270941709247901209742124;m=213509721309572 sage: n.exact_log(m) 4
sage: x = 3^100000 sage: RR(log(RR(x), 3)) 100000.000000000 sage: RR(log(RR(x + 100000), 3)) 100000.000000000
sage: x.exact_log(3) 100000 sage: (x+1).exact_log(3) 100000 sage: (x-1).exact_log(3) 99999
sage: x.exact_log(2.5) Traceback (most recent call last): ... TypeError: Attempt to coerce non-integral RealNumber to Integer
- exp(prec=None)#
Returns the exponential function of self as a real number.
This function is provided only so that Sage integers may be treated in the same manner as real numbers when convenient.
INPUT:
prec
- integer (default: None): if None, returns symbolic, else to given bits of precision as in RealField
EXAMPLES:
sage: Integer(8).exp() e^8 sage: Integer(8).exp(prec=100) 2980.9579870417282747435920995 sage: exp(Integer(8)) e^8
For even fairly large numbers, this may not be useful.
sage: y=Integer(145^145) sage: y.exp() e^25024207011349079210459585279553675697932183658421565260323592409432707306554163224876110094014450895759296242775250476115682350821522931225499163750010280453185147546962559031653355159703678703793369785727108337766011928747055351280379806937944746847277089168867282654496776717056860661614337004721164703369140625 sage: y.exp(prec=53) # default RealField precision +infinity
- factor(algorithm='pari', proof=None, limit=None, int_=False, verbose=0)#
Return the prime factorization of this integer as a formal Factorization object.
INPUT:
algorithm
- string'pari'
- (default) use the PARI library'kash'
- use the KASH computer algebra system (requires kash)'magma'
- use the MAGMA computer algebra system (requires an installation of MAGMA)'qsieve'
- use Bill Hart’s quadratic sieve code; WARNING: this may not work as expected, see qsieve? for more information'ecm'
- use ECM-GMP, an implementation of Hendrik Lenstra’s elliptic curve method.
proof
- bool (default: True) whether or not to prove primality of each factor (only applicable for'pari'
and'ecm'
).limit
- int or None (default: None) if limit is given it must fit in a signed int, and the factorization is done using trial division and primes up to limit.
OUTPUT:
a Factorization object containing the prime factors and their multiplicities
EXAMPLES:
sage: n = 2^100 - 1; n.factor() 3 * 5^3 * 11 * 31 * 41 * 101 * 251 * 601 * 1801 * 4051 * 8101 * 268501
This factorization can be converted into a list of pairs \((p, e)\), where \(p\) is prime and \(e\) is a positive integer. Each pair can also be accessed directly by its index (ordered by increasing size of the prime):
sage: f = 60.factor() sage: list(f) [(2, 2), (3, 1), (5, 1)] sage: f[2] (5, 1)
Similarly, the factorization can be converted to a dictionary so the exponent can be extracted for each prime:
sage: f = (3^6).factor() sage: dict(f) {3: 6} sage: dict(f)[3] 6
We use proof=False, which doesn’t prove correctness of the primes that appear in the factorization:
sage: n = 920384092842390423848290348203948092384082349082 sage: n.factor(proof=False) 2 * 11 * 1531 * 4402903 * 10023679 * 619162955472170540533894518173 sage: n.factor(proof=True) 2 * 11 * 1531 * 4402903 * 10023679 * 619162955472170540533894518173
We factor using trial division only:
sage: n.factor(limit=1000) 2 * 11 * 41835640583745019265831379463815822381094652231
We factor using a quadratic sieve algorithm:
sage: p = next_prime(10^20) sage: q = next_prime(10^21) sage: n = p*q sage: n.factor(algorithm='qsieve') doctest:... RuntimeWarning: the factorization returned by qsieve may be incomplete (the factors may not be prime) or even wrong; see qsieve? for details 100000000000000000039 * 1000000000000000000117
We factor using the elliptic curve method:
sage: p = next_prime(10^15) sage: q = next_prime(10^21) sage: n = p*q sage: n.factor(algorithm='ecm') 1000000000000037 * 1000000000000000000117
- factorial()#
Return the factorial \(n! = 1 \cdot 2 \cdot 3 \cdots n\).
If the input does not fit in an
unsigned long int
anOverflowError
is raised.EXAMPLES:
sage: for n in srange(7): ....: print("{} {}".format(n, n.factorial())) 0 1 1 1 2 2 3 6 4 24 5 120 6 720
Large integers raise an
OverflowError
:sage: (2**64).factorial() Traceback (most recent call last): ... OverflowError: argument too large for factorial
And negative ones a
ValueError
:sage: (-1).factorial() Traceback (most recent call last): ... ValueError: factorial only defined for non-negative integers
- floor()#
Return the floor of self, which is just self since self is an integer.
EXAMPLES:
sage: n = 6 sage: n.floor() 6
- gamma()#
The gamma function on integers is the factorial function (shifted by one) on positive integers, and \(\pm \infty\) on non-positive integers.
EXAMPLES:
sage: gamma(5) 24 sage: gamma(0) Infinity sage: gamma(-1) Infinity sage: gamma(-2^150) Infinity
- global_height(prec=None)#
Returns the absolute logarithmic height of this rational integer.
INPUT:
prec
(int) – desired floating point precision (default: default RealField precision).
OUTPUT:
(real) The absolute logarithmic height of this rational integer.
ALGORITHM:
The height of the integer \(n\) is \(\log |n|\).
EXAMPLES:
sage: ZZ(5).global_height() 1.60943791243410 sage: ZZ(-2).global_height(prec=100) 0.69314718055994530941723212146 sage: exp(_) 2.0000000000000000000000000000
- hex()#
Return the hexadecimal digits of
self
in lower case.Note
‘0x’ is not prepended to the result like is done by the corresponding Python function on
int
orlong
. This is for efficiency sake–adding and stripping the string wastes time; since this function is used for conversions from integers to other C-library structures, it is important that it be fast.EXAMPLES:
sage: print(Integer(15).hex()) f sage: print(Integer(16).hex()) 10 sage: print(Integer(16938402384092843092843098243).hex()) 36bb1e3929d1a8fe2802f083
- imag()#
Returns the imaginary part of self, which is zero.
EXAMPLES:
sage: Integer(9).imag() 0
- inverse_mod(n)#
Returns the inverse of self modulo \(n\), if this inverse exists. Otherwise, raises a
ZeroDivisionError
exception.INPUT:
self
- Integern
- Integer, or ideal of integer ring
OUTPUT:
x
- Integer such that x*self = 1 (mod m), or raises ZeroDivisionError.
IMPLEMENTATION:
Call the mpz_invert GMP library function.
EXAMPLES:
sage: a = Integer(189) sage: a.inverse_mod(10000) 4709 sage: a.inverse_mod(-10000) 4709 sage: a.inverse_mod(1890) Traceback (most recent call last): ... ZeroDivisionError: inverse of Mod(189, 1890) does not exist sage: a = Integer(19)**100000 # long time sage: c = a.inverse_mod(a*a) # long time Traceback (most recent call last): ... ZeroDivisionError: inverse of Mod(..., ...) does not exist
We check that trac ticket #10625 is fixed:
sage: ZZ(2).inverse_mod(ZZ.ideal(3)) 2
We check that trac ticket #9955 is fixed:
sage: Rational(3) % Rational(-1) 0
- inverse_of_unit()#
Return inverse of self if self is a unit in the integers, i.e., self is -1 or 1. Otherwise, raise a ZeroDivisionError.
EXAMPLES:
sage: (1).inverse_of_unit() 1 sage: (-1).inverse_of_unit() -1 sage: 5.inverse_of_unit() Traceback (most recent call last): ... ArithmeticError: inverse does not exist sage: 0.inverse_of_unit() Traceback (most recent call last): ... ArithmeticError: inverse does not exist
- is_integer()#
Returns
True
as they are integersEXAMPLES:
sage: sqrt(4).is_integer() True
- is_integral()#
Return
True
since integers are integral, i.e., satisfy a monic polynomial with integer coefficients.EXAMPLES:
sage: Integer(3).is_integral() True
- is_irreducible()#
Returns
True
if self is irreducible, i.e. +/- primeEXAMPLES:
sage: z = 2^31 - 1 sage: z.is_irreducible() True sage: z = 2^31 sage: z.is_irreducible() False sage: z = 7 sage: z.is_irreducible() True sage: z = -7 sage: z.is_irreducible() True
- is_norm(K, element=False, proof=True)#
See
QQ(self).is_norm()
.EXAMPLES:
sage: K = NumberField(x^2 - 2, 'beta') sage: n = 4 sage: n.is_norm(K) True sage: 5.is_norm(K) False sage: 7.is_norm(QQ) True sage: n.is_norm(K, element=True) (True, -4*beta + 6) sage: n.is_norm(K, element=True)[1].norm() 4 sage: n = 5 sage: n.is_norm(K, element=True) (False, None) sage: n = 7 sage: n.is_norm(QQ, element=True) (True, 7)
- is_one()#
Returns
True
if the integer is \(1\), otherwiseFalse
.EXAMPLES:
sage: Integer(1).is_one() True sage: Integer(0).is_one() False
- is_perfect_power()#
Returns
True
ifself
is a perfect power, ie if there exist integers \(a\) and \(b\), \(b > 1\) withself
\(= a^b\).See also
perfect_power()
: Finds the minimal base for which this integer is a perfect power.is_power_of()
: If you know the base already this method is the fastest option.is_prime_power()
: Checks whether the base is prime.
EXAMPLES:
sage: Integer(-27).is_perfect_power() True sage: Integer(12).is_perfect_power() False sage: z = 8 sage: z.is_perfect_power() True sage: 144.is_perfect_power() True sage: 10.is_perfect_power() False sage: (-8).is_perfect_power() True sage: (-4).is_perfect_power() False
- is_power_of(n)#
Returns
True
if there is an integer b with \(\mathtt{self} = n^b\).See also
perfect_power()
: Finds the minimal base for which this integer is a perfect power.is_perfect_power()
: If you don’t know the base but just want to know if this integer is a perfect power, use this function.is_prime_power()
: Checks whether the base is prime.
EXAMPLES:
sage: Integer(64).is_power_of(4) True sage: Integer(64).is_power_of(16) False
Note
For large integers self, is_power_of() is faster than is_perfect_power(). The following examples gives some indication of how much faster.
sage: b = lcm(range(1,10000)) sage: b.exact_log(2) 14446 sage: t=cputime() sage: for a in range(2, 1000): k = b.is_perfect_power() sage: cputime(t) # random 0.53203299999999976 sage: t=cputime() sage: for a in range(2, 1000): k = b.is_power_of(2) sage: cputime(t) # random 0.0 sage: t=cputime() sage: for a in range(2, 1000): k = b.is_power_of(3) sage: cputime(t) # random 0.032002000000000308
sage: b = lcm(range(1, 1000)) sage: b.exact_log(2) 1437 sage: t=cputime() sage: for a in range(2, 10000): k = b.is_perfect_power() # note that we change the range from the example above sage: cputime(t) # random 0.17201100000000036 sage: t=cputime(); TWO=int(2) sage: for a in range(2, 10000): k = b.is_power_of(TWO) sage: cputime(t) # random 0.0040000000000000036 sage: t=cputime() sage: for a in range(2, 10000): k = b.is_power_of(3) sage: cputime(t) # random 0.040003000000000011 sage: t=cputime() sage: for a in range(2, 10000): k = b.is_power_of(a) sage: cputime(t) # random 0.02800199999999986
- is_prime(proof=None)#
Test whether
self
is prime.INPUT:
proof
– Boolean orNone
(default). If False, use a strong pseudo-primality test (seeis_pseudoprime()
). If True, use a provable primality test. If unset, use thedefault arithmetic proof flag
.
Note
Integer primes are by definition positive! This is different than Magma, but the same as in PARI. See also the
is_irreducible()
method.EXAMPLES:
sage: z = 2^31 - 1 sage: z.is_prime() True sage: z = 2^31 sage: z.is_prime() False sage: z = 7 sage: z.is_prime() True sage: z = -7 sage: z.is_prime() False sage: z.is_irreducible() True
sage: z = 10^80 + 129 sage: z.is_prime(proof=False) True sage: z.is_prime(proof=True) True
When starting Sage the arithmetic proof flag is True. We can change it to False as follows:
sage: proof.arithmetic() True sage: n = 10^100 + 267 sage: timeit("n.is_prime()") # not tested 5 loops, best of 3: 163 ms per loop sage: proof.arithmetic(False) sage: proof.arithmetic() False sage: timeit("n.is_prime()") # not tested 1000 loops, best of 3: 573 us per loop
ALGORITHM:
Calls the PARI
isprime
function.
- is_prime_power(proof=None, get_data=False)#
Return
True
if this integer is a prime power, andFalse
otherwise.A prime power is a prime number raised to a positive power. Hence \(1\) is not a prime power.
For a method that uses a pseudoprimality test instead see
is_pseudoprime_power()
.INPUT:
proof
– Boolean orNone
(default). IfFalse
, use a strong pseudo-primality test (seeis_pseudoprime()
). IfTrue
, use a provable primality test. If unset, use the default arithmetic proof flag.get_data
– (defaultFalse
), ifTrue
return a pair(p,k)
such that this integer equalsp^k
withp
a prime andk
a positive integer or the pair(self,0)
otherwise.
See also
perfect_power()
: Finds the minimal base for which integer is a perfect power.is_perfect_power()
: Doesn’t test whether the base is prime.is_power_of()
: If you know the base already this method is the fastest option.is_pseudoprime_power()
: If the entry is very large.
EXAMPLES:
sage: 17.is_prime_power() True sage: 10.is_prime_power() False sage: 64.is_prime_power() True sage: (3^10000).is_prime_power() True sage: (10000).is_prime_power() False sage: (-3).is_prime_power() False sage: 0.is_prime_power() False sage: 1.is_prime_power() False sage: p = next_prime(10^20); p 100000000000000000039 sage: p.is_prime_power() True sage: (p^97).is_prime_power() True sage: (p+1).is_prime_power() False
With the
get_data
keyword set toTrue
:sage: (3^100).is_prime_power(get_data=True) (3, 100) sage: 12.is_prime_power(get_data=True) (12, 0) sage: (p^97).is_prime_power(get_data=True) (100000000000000000039, 97) sage: q = p.next_prime(); q 100000000000000000129 sage: (p*q).is_prime_power(get_data=True) (10000000000000000016800000000000000005031, 0)
The method works for large entries when \(proof=False\):
sage: proof.arithmetic(False) sage: ((10^500 + 961)^4).is_prime_power() True sage: proof.arithmetic(True)
We check that trac ticket #4777 is fixed:
sage: n = 150607571^14 sage: n.is_prime_power() True
- is_pseudoprime()#
Test whether
self
is a pseudoprime.This uses PARI’s Baillie-PSW probabilistic primality test. Currently, there are no known pseudoprimes for Baillie-PSW that are not actually prime. However it is conjectured that there are infinitely many.
See Wikipedia article Baillie-PSW_primality_test
EXAMPLES:
sage: z = 2^31 - 1 sage: z.is_pseudoprime() True sage: z = 2^31 sage: z.is_pseudoprime() False
- is_pseudoprime_power(get_data=False)#
Test if this number is a power of a pseudoprime number.
For large numbers, this method might be faster than
is_prime_power()
.INPUT:
get_data
– (defaultFalse
) ifTrue
return a pair \((p,k)\) such that this number equals \(p^k\) with \(p\) a pseudoprime and \(k\) a positive integer or the pair(self,0)
otherwise.
EXAMPLES:
sage: x = 10^200 + 357 sage: x.is_pseudoprime() True sage: (x^12).is_pseudoprime_power() True sage: (x^12).is_pseudoprime_power(get_data=True) (1000...000357, 12) sage: (997^100).is_pseudoprime_power() True sage: (998^100).is_pseudoprime_power() False sage: ((10^1000 + 453)^2).is_pseudoprime_power() True
- is_rational()#
Return
True
as an integer is a rational number.EXAMPLES:
sage: 5.is_rational() True
- is_square()#
Returns
True
if self is a perfect square.EXAMPLES:
sage: Integer(4).is_square() True sage: Integer(41).is_square() False
- is_squarefree()#
Returns True if this integer is not divisible by the square of any prime and False otherwise.
EXAMPLES:
sage: 100.is_squarefree() False sage: 102.is_squarefree() True sage: 0.is_squarefree() False
- is_unit()#
Returns
true
if this integer is a unit, i.e., 1 or \(-1\).EXAMPLES:
sage: for n in srange(-2,3): ....: print("{} {}".format(n, n.is_unit())) -2 False -1 True 0 False 1 True 2 False
- isqrt()#
Returns the integer floor of the square root of self, or raises an
ValueError
if self is negative.EXAMPLES:
sage: a = Integer(5) sage: a.isqrt() 2
sage: Integer(-102).isqrt() Traceback (most recent call last): ... ValueError: square root of negative integer not defined.
- jacobi(b)#
Calculate the Jacobi symbol \(\left(\frac{self}{b}\right)\).
EXAMPLES:
sage: z = -1 sage: z.jacobi(17) 1 sage: z.jacobi(19) -1 sage: z.jacobi(17*19) -1 sage: (2).jacobi(17) 1 sage: (3).jacobi(19) -1 sage: (6).jacobi(17*19) -1 sage: (6).jacobi(33) 0 sage: a = 3; b = 7 sage: a.jacobi(b) == -b.jacobi(a) True
- kronecker(b)#
Calculate the Kronecker symbol \(\left(\frac{self}{b}\right)\) with the Kronecker extension \((self/2)=(2/self)\) when \(self\) is odd, or \((self/2)=0\) when \(self\) is even.
EXAMPLES:
sage: z = 5 sage: z.kronecker(41) 1 sage: z.kronecker(43) -1 sage: z.kronecker(8) -1 sage: z.kronecker(15) 0 sage: a = 2; b = 5 sage: a.kronecker(b) == b.kronecker(a) True
- list()#
Return a list with this integer in it, to be compatible with the method for number fields.
EXAMPLES:
sage: m = 5 sage: m.list() [5]
- log(m=None, prec=None)#
Returns symbolic log by default, unless the logarithm is exact (for an integer argument). When precision is given, the RealField approximation to that bit precision is used.
This function is provided primarily so that Sage integers may be treated in the same manner as real numbers when convenient. Direct use of exact_log is probably best for arithmetic log computation.
INPUT:
m
- default: natural log base eprec
- integer (default: None): if None, returns symbolic, else to given bits of precision as in RealField
EXAMPLES:
sage: Integer(124).log(5) log(124)/log(5) sage: Integer(124).log(5,100) 2.9950093311241087454822446806 sage: Integer(125).log(5) 3 sage: Integer(125).log(5,prec=53) 3.00000000000000 sage: log(Integer(125)) 3*log(5)
For extremely large numbers, this works:
sage: x = 3^100000 sage: log(x,3) 100000
With the new Pynac symbolic backend, log(x) also works in a reasonable amount of time for this x:
sage: x = 3^100000 sage: log(x) log(1334971414230...5522000001)
But approximations are probably more useful in this case, and work to as high a precision as we desire:
sage: x.log(3,53) # default precision for RealField 100000.000000000 sage: (x+1).log(3,53) 100000.000000000 sage: (x+1).log(3,1000) 100000.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
We can use non-integer bases, with default e:
sage: x.log(2.5,prec=53) 119897.784671579
We also get logarithms of negative integers, via the symbolic ring, using the branch from \(-pi\) to \(pi\):
sage: log(-1) I*pi
The logarithm of zero is done likewise:
sage: log(0) -Infinity
Some rational bases yield integer logarithms (trac ticket #21517):
sage: ZZ(8).log(1/2) -3
Check that Python ints are accepted (trac ticket #21518):
sage: ZZ(8).log(int(2)) 3
- multifactorial(k)#
Compute the k-th factorial \(n!^{(k)}\) of self.
The multifactorial number \(n!^{(k)}\) is defined for non-negative integers \(n\) as follows. For \(k=1\) this is the standard factorial, and for \(k\) greater than \(1\) it is the product of every \(k\)-th terms down from \(n\) to \(1\). The recursive definition is used to extend this function to the negative integers \(n\).
This function uses direct call to GMP if \(k\) and \(n\) are non-negative and uses simple transformation for other cases.
EXAMPLES:
sage: 5.multifactorial(1) 120 sage: 5.multifactorial(2) 15 sage: 5.multifactorial(3) 10 sage: 23.multifactorial(2) 316234143225 sage: prod([1..23, step=2]) 316234143225 sage: (-29).multifactorial(7) 1/2640 sage: (-3).multifactorial(5) 1/2 sage: (-9).multifactorial(3) Traceback (most recent call last): ... ValueError: multifactorial undefined
When entries are too large an
OverflowError
is raised:sage: (2**64).multifactorial(2) Traceback (most recent call last): ... OverflowError: argument too large for multifactorial
- multiplicative_order()#
Return the multiplicative order of self.
EXAMPLES:
sage: ZZ(1).multiplicative_order() 1 sage: ZZ(-1).multiplicative_order() 2 sage: ZZ(0).multiplicative_order() +Infinity sage: ZZ(2).multiplicative_order() +Infinity
- nbits()#
Alias for
bit_length()
.
- ndigits(base=10)#
Return the number of digits of self expressed in the given base.
INPUT:
base
- integer (default: 10)
EXAMPLES:
sage: n = 52 sage: n.ndigits() 2 sage: n = -10003 sage: n.ndigits() 5 sage: n = 15 sage: n.ndigits(2) 4 sage: n = 1000**1000000+1 sage: n.ndigits() 3000001 sage: n = 1000**1000000-1 sage: n.ndigits() 3000000 sage: n = 10**10000000-10**9999990 sage: n.ndigits() 10000000
- next_prime(proof=None)#
Return the next prime after self.
This method calls the PARI
nextprime
function.INPUT:
proof
- bool or None (default: None, see proof.arithmetic or sage.structure.proof) Note that the global Sage default is proof=True
EXAMPLES:
sage: 100.next_prime() 101 sage: (10^50).next_prime() 100000000000000000000000000000000000000000000000151
Use
proof=False
, which is way faster since it does not need a primality proof:sage: b = (2^1024).next_prime(proof=False) sage: b - 2^1024 643
sage: Integer(0).next_prime() 2 sage: Integer(1001).next_prime() 1009
- next_prime_power(proof=None)#
Return the next prime power after self.
INPUT:
proof
- ifTrue
ensure that the returned value is the next prime power and if set toFalse
uses probabilistic methods (i.e. the result is not guaranteed). By default it uses global configuration variables to determine which alternative to use (seeproof.arithmetic
orsage.structure.proof
).
ALGORITHM:
The algorithm is naive. It computes the next power of 2 and go through the odd numbers calling
is_prime_power()
.EXAMPLES:
sage: (-1).next_prime_power() 2 sage: 2.next_prime_power() 3 sage: 103.next_prime_power() 107 sage: 107.next_prime_power() 109 sage: 2044.next_prime_power() 2048
- next_probable_prime()#
Return the next probable prime after self, as determined by PARI.
EXAMPLES:
sage: (-37).next_probable_prime() 2 sage: (100).next_probable_prime() 101 sage: (2^512).next_probable_prime() 13407807929942597099574024998205846127479365820592393377723561443721764030073546976801874298166903427690031858186486050853753882811946569946433649006084171 sage: 0.next_probable_prime() 2 sage: 126.next_probable_prime() 127 sage: 144168.next_probable_prime() 144169
- nth_root(n, truncate_mode=0)#
Returns the (possibly truncated) n’th root of self.
INPUT:
n
- integer >= 1 (must fit in C int type).truncate_mode
- boolean, whether to allow truncation if self is not an n’th power.
OUTPUT:
If truncate_mode is 0 (default), then returns the exact n’th root if self is an n’th power, or raises a ValueError if it is not.
If truncate_mode is 1, then if either n is odd or self is positive, returns a pair (root, exact_flag) where root is the truncated nth root (rounded towards zero) and exact_flag is a boolean indicating whether the root extraction was exact; otherwise raises a ValueError.
AUTHORS:
David Harvey (2006-09-15)
Interface changed by John Cremona (2009-04-04)
EXAMPLES:
sage: Integer(125).nth_root(3) 5 sage: Integer(124).nth_root(3) Traceback (most recent call last): ... ValueError: 124 is not a 3rd power sage: Integer(124).nth_root(3, truncate_mode=1) (4, False) sage: Integer(125).nth_root(3, truncate_mode=1) (5, True) sage: Integer(126).nth_root(3, truncate_mode=1) (5, False)
sage: Integer(-125).nth_root(3) -5 sage: Integer(-125).nth_root(3,truncate_mode=1) (-5, True) sage: Integer(-124).nth_root(3,truncate_mode=1) (-4, False) sage: Integer(-126).nth_root(3,truncate_mode=1) (-5, False)
sage: Integer(125).nth_root(2, True) (11, False) sage: Integer(125).nth_root(3, True) (5, True)
sage: Integer(125).nth_root(-5) Traceback (most recent call last): ... ValueError: n (=-5) must be positive
sage: Integer(-25).nth_root(2) Traceback (most recent call last): ... ValueError: cannot take even root of negative number
sage: a=9 sage: a.nth_root(3) Traceback (most recent call last): ... ValueError: 9 is not a 3rd power sage: a.nth_root(22) Traceback (most recent call last): ... ValueError: 9 is not a 22nd power sage: ZZ(2^20).nth_root(21) Traceback (most recent call last): ... ValueError: 1048576 is not a 21st power sage: ZZ(2^20).nth_root(21, truncate_mode=1) (1, False)
- numerator()#
Return the numerator of this integer.
EXAMPLES:
sage: x = 5 sage: x.numerator() 5
sage: x = 0 sage: x.numerator() 0
- oct()#
Return the digits of
self
in base 8.Note
‘0’ (or ‘0o’) is not prepended to the result like is done by the corresponding Python function on
int
orlong
. This is for efficiency sake–adding and stripping the string wastes time; since this function is used for conversions from integers to other C-library structures, it is important that it be fast.EXAMPLES:
sage: print(Integer(800).oct()) 1440 sage: print(Integer(8).oct()) 10 sage: print(Integer(-50).oct()) -62 sage: print(Integer(-899).oct()) -1603 sage: print(Integer(16938402384092843092843098243).oct()) 15535436162247215217705000570203
Behavior of Sage integers vs. Python integers:
sage: Integer(10).oct() '12' sage: oct(int(10)) '0o12' sage: Integer(-23).oct() '-27' sage: oct(int(-23)) '-0o27'
- odd_part()#
The odd part of the integer \(n\). This is \(n / 2^v\), where \(v = \mathrm{valuation}(n,2)\).
IMPLEMENTATION:
Currently returns 0 when self is 0. This behaviour is fairly arbitrary, and in Sage 4.6 this special case was not handled at all, eventually propagating a TypeError. The caller should not rely on the behaviour in case self is 0.
EXAMPLES:
sage: odd_part(5) 5 sage: odd_part(4) 1 sage: odd_part(factorial(31)) 122529844256906551386796875
- ord(p)#
Return the p-adic valuation of self.
INPUT:
p
- an integer at least 2.
EXAMPLES:
sage: n = 60 sage: n.valuation(2) 2 sage: n.valuation(3) 1 sage: n.valuation(7) 0 sage: n.valuation(1) Traceback (most recent call last): ... ValueError: You can only compute the valuation with respect to a integer larger than 1.
We do not require that p is a prime:
sage: (2^11).valuation(4) 5
- ordinal_str()#
Returns a string representation of the ordinal associated to self.
EXAMPLES:
sage: [ZZ(n).ordinal_str() for n in range(25)] ['0th', '1st', '2nd', '3rd', '4th', ... '10th', '11th', '12th', '13th', '14th', ... '20th', '21st', '22nd', '23rd', '24th'] sage: ZZ(1001).ordinal_str() '1001st' sage: ZZ(113).ordinal_str() '113th' sage: ZZ(112).ordinal_str() '112th' sage: ZZ(111).ordinal_str() '111th'
- p_primary_part(p)#
Return the p-primary part of
self
.INPUT:
p
– a prime integer.
OUTPUT: Largest power of
p
dividingself
.EXAMPLES:
sage: n = 40 sage: n.p_primary_part(2) 8 sage: n.p_primary_part(5) 5 sage: n.p_primary_part(7) 1 sage: n.p_primary_part(6) Traceback (most recent call last): ... ValueError: 6 is not a prime number
- perfect_power()#
Returns
(a, b)
, where this integer is \(a^b\) and \(b\) is maximal.If called on \(-1\), \(0\) or \(1\), \(b\) will be \(1\), since there is no maximal value of \(b\).
See also
is_perfect_power()
: testing whether an integer is a perfect power is usually faster than finding \(a\) and \(b\).is_prime_power()
: checks whether the base is prime.is_power_of()
: if you know the base already, this method is the fastest option.
EXAMPLES:
sage: 144.perfect_power() (12, 2) sage: 1.perfect_power() (1, 1) sage: 0.perfect_power() (0, 1) sage: (-1).perfect_power() (-1, 1) sage: (-8).perfect_power() (-2, 3) sage: (-4).perfect_power() (-4, 1) sage: (101^29).perfect_power() (101, 29) sage: (-243).perfect_power() (-3, 5) sage: (-64).perfect_power() (-4, 3)
- popcount()#
Return the number of 1 bits in the binary representation. If self<0, we return Infinity.
EXAMPLES:
sage: n = 123 sage: n.str(2) '1111011' sage: n.popcount() 6 sage: n = -17 sage: n.popcount() +Infinity
- powermod(exp, mod)#
Compute self**exp modulo mod.
EXAMPLES:
sage: z = 2 sage: z.powermod(31,31) 2 sage: z.powermod(0,31) 1 sage: z.powermod(-31,31) == 2^-31 % 31 True
As expected, the following is invalid:
sage: z.powermod(31,0) Traceback (most recent call last): ... ZeroDivisionError: cannot raise to a power modulo 0
- previous_prime(proof=None)#
Returns the previous prime before self.
This method calls the PARI
precprime
function.INPUT:
proof
- ifTrue
ensure that the returned value is the next prime power and if set toFalse
uses probabilistic methods (i.e. the result is not guaranteed). By default it uses global configuration variables to determine which alternative to use (seeproof.arithmetic
orsage.structure.proof
).
See also
EXAMPLES:
sage: 10.previous_prime() 7 sage: 7.previous_prime() 5 sage: 14376485.previous_prime() 14376463 sage: 2.previous_prime() Traceback (most recent call last): ... ValueError: no prime less than 2
An example using
proof=False
, which is way faster since it does not need a primality proof:sage: b = (2^1024).previous_prime(proof=False) sage: 2^1024 - b 105
- previous_prime_power(proof=None)#
Return the previous prime power before self.
INPUT:
proof
- ifTrue
ensure that the returned value is the next prime power and if set toFalse
uses probabilistic methods (i.e. the result is not guaranteed). By default it uses global configuration variables to determine which alternative to use (seeproof.arithmetic
orsage.structure.proof
).
ALGORITHM:
The algorithm is naive. It computes the previous power of 2 and go through the odd numbers calling the method
is_prime_power()
.EXAMPLES:
sage: 3.previous_prime_power() 2 sage: 103.previous_prime_power() 101 sage: 107.previous_prime_power() 103 sage: 2044.previous_prime_power() 2039 sage: 2.previous_prime_power() Traceback (most recent call last): ... ValueError: no prime power less than 2
- prime_divisors()#
Return the prime divisors of this integer, sorted in increasing order.
If this integer is negative, we do not include -1 among its prime divisors, since -1 is not a prime number.
EXAMPLES:
sage: a = 1; a.prime_divisors() [] sage: a = 100; a.prime_divisors() [2, 5] sage: a = -100; a.prime_divisors() [2, 5] sage: a = 2004; a.prime_divisors() [2, 3, 167]
- prime_factors()#
Return the prime divisors of this integer, sorted in increasing order.
If this integer is negative, we do not include -1 among its prime divisors, since -1 is not a prime number.
EXAMPLES:
sage: a = 1; a.prime_divisors() [] sage: a = 100; a.prime_divisors() [2, 5] sage: a = -100; a.prime_divisors() [2, 5] sage: a = 2004; a.prime_divisors() [2, 3, 167]
- prime_to_m_part(m)#
Returns the prime-to-m part of self, i.e., the largest divisor of
self
that is coprime tom
.INPUT:
m
- Integer
OUTPUT: Integer
EXAMPLES:
sage: 43434.prime_to_m_part(20) 21717 sage: 2048.prime_to_m_part(2) 1 sage: 2048.prime_to_m_part(3) 2048 sage: 0.prime_to_m_part(2) Traceback (most recent call last): ... ArithmeticError: self must be nonzero
- quo_rem(other)#
Returns the quotient and the remainder of self divided by other. Note that the remainder returned is always either zero or of the same sign as other.
INPUT:
other
- the divisor
OUTPUT:
q
- the quotient of self/otherr
- the remainder of self/other
EXAMPLES:
sage: z = Integer(231) sage: z.quo_rem(2) (115, 1) sage: z.quo_rem(-2) (-116, -1) sage: z.quo_rem(0) Traceback (most recent call last): ... ZeroDivisionError: Integer division by zero sage: a = ZZ.random_element(10**50) sage: b = ZZ.random_element(10**15) sage: q, r = a.quo_rem(b) sage: q*b + r == a True sage: 3.quo_rem(ZZ['x'].0) (0, 3)
- rational_reconstruction(m)#
Return the rational reconstruction of this integer modulo m, i.e., the unique (if it exists) rational number that reduces to self modulo m and whose numerator and denominator is bounded by sqrt(m/2).
INPUT:
self
– Integerm
– Integer
OUTPUT:
a
Rational
EXAMPLES:
sage: (3/7)%100 29 sage: (29).rational_reconstruction(100) 3/7
- real()#
Returns the real part of self, which is self.
EXAMPLES:
sage: Integer(-4).real() -4
- round(mode='away')#
Returns the nearest integer to
self
, which is self since self is an integer.EXAMPLES:
This example addresses trac ticket #23502:
sage: n = 6 sage: n.round() 6
- sign()#
Returns the sign of this integer, which is -1, 0, or 1 depending on whether this number is negative, zero, or positive respectively.
OUTPUT: Integer
EXAMPLES:
sage: 500.sign() 1 sage: 0.sign() 0 sage: (-10^43).sign() -1
- sqrt(prec=None, extend=True, all=False)#
The square root function.
INPUT:
prec
- integer (default: None): if None, return an exact square root; otherwise return a numerical square root, to the given bits of precision.extend
- bool (default: True); if True, return a square root in an extension ring, if necessary. Otherwise, raise a ValueError if the square is not in the base ring. Ignored if prec is not None.all
- bool (default: False); if True, return all square roots of self (a list of length 0, 1 or 2).
EXAMPLES:
sage: Integer(144).sqrt() 12 sage: sqrt(Integer(144)) 12 sage: Integer(102).sqrt() sqrt(102)
sage: n = 2 sage: n.sqrt(all=True) [sqrt(2), -sqrt(2)] sage: n.sqrt(prec=10) 1.4 sage: n.sqrt(prec=100) 1.4142135623730950488016887242 sage: n.sqrt(prec=100,all=True) [1.4142135623730950488016887242, -1.4142135623730950488016887242] sage: n.sqrt(extend=False) Traceback (most recent call last): ... ArithmeticError: square root of 2 is not an integer sage: (-1).sqrt(extend=False) Traceback (most recent call last): ... ArithmeticError: square root of -1 is not an integer sage: Integer(144).sqrt(all=True) [12, -12] sage: Integer(0).sqrt(all=True) [0]
- sqrtrem()#
Return (s, r) where s is the integer square root of self and r is the remainder such that \(\text{self} = s^2 + r\). Raises
ValueError
if self is negative.EXAMPLES:
sage: 25.sqrtrem() (5, 0) sage: 27.sqrtrem() (5, 2) sage: 0.sqrtrem() (0, 0)
sage: Integer(-102).sqrtrem() Traceback (most recent call last): ... ValueError: square root of negative integer not defined.
- squarefree_part(bound=-1)#
Return the square free part of \(x\) (=self), i.e., the unique integer \(z\) that \(x = z y^2\), with \(y^2\) a perfect square and \(z\) square-free.
Use
self.radical()
for the product of the primes that divide self.If self is 0, just returns 0.
EXAMPLES:
sage: squarefree_part(100) 1 sage: squarefree_part(12) 3 sage: squarefree_part(17*37*37) 17 sage: squarefree_part(-17*32) -34 sage: squarefree_part(1) 1 sage: squarefree_part(-1) -1 sage: squarefree_part(-2) -2 sage: squarefree_part(-4) -1
sage: a = 8 * 5^6 * 101^2 sage: a.squarefree_part(bound=2).factor() 2 * 5^6 * 101^2 sage: a.squarefree_part(bound=5).factor() 2 * 101^2 sage: a.squarefree_part(bound=1000) 2 sage: a.squarefree_part(bound=2**14) 2 sage: a = 7^3 * next_prime(2^100)^2 * next_prime(2^200) sage: a / a.squarefree_part(bound=1000) 49
- str(base=10)#
Return the string representation of
self
in the given base.EXAMPLES:
sage: Integer(2^10).str(2) '10000000000' sage: Integer(2^10).str(17) '394'
sage: two=Integer(2) sage: two.str(1) Traceback (most recent call last): ... ValueError: base (=1) must be between 2 and 36
sage: two.str(37) Traceback (most recent call last): ... ValueError: base (=37) must be between 2 and 36
sage: big = 10^5000000 sage: s = big.str() # long time (2s on sage.math, 2014) sage: len(s) # long time (depends on above defn of s) 5000001 sage: s[:10] # long time (depends on above defn of s) '1000000000'
- support()#
Return a sorted list of the primes dividing this integer.
OUTPUT: The sorted list of primes appearing in the factorization of this rational with positive exponent.
EXAMPLES:
sage: factorial(10).support() [2, 3, 5, 7] sage: (-999).support() [3, 37]
Trying to find the support of 0 gives an arithmetic error:
sage: 0.support() Traceback (most recent call last): ... ArithmeticError: Support of 0 not defined.
- test_bit(index)#
Return the bit at
index
.If the index is negative, returns 0.
Although internally a sign-magnitude representation is used for integers, this method pretends to use a two’s complement representation. This is illustrated with a negative integer below.
EXAMPLES:
sage: w = 6 sage: w.str(2) '110' sage: w.test_bit(2) 1 sage: w.test_bit(-1) 0 sage: x = -20 sage: x.str(2) '-10100' sage: x.test_bit(4) 0 sage: x.test_bit(5) 1 sage: x.test_bit(6) 1
- trailing_zero_bits()#
Return the number of trailing zero bits in self, i.e. the exponent of the largest power of 2 dividing self.
EXAMPLES:
sage: 11.trailing_zero_bits() 0 sage: (-11).trailing_zero_bits() 0 sage: (11<<5).trailing_zero_bits() 5 sage: (-11<<5).trailing_zero_bits() 5 sage: 0.trailing_zero_bits() 0
- trial_division(bound='LONG_MAX', start=2)#
Return smallest prime divisor of self up to bound, beginning checking at start, or abs(self) if no such divisor is found.
INPUT:
bound
– a positive integer that fits in a C signed longstart
– a positive integer that fits in a C signed long
OUTPUT:
a positive integer
EXAMPLES:
sage: n = next_prime(10^6)*next_prime(10^7); n.trial_division() 1000003 sage: (-n).trial_division() 1000003 sage: n.trial_division(bound=100) 10000049000057 sage: n.trial_division(bound=-10) Traceback (most recent call last): ... ValueError: bound must be positive sage: n.trial_division(bound=0) Traceback (most recent call last): ... ValueError: bound must be positive sage: ZZ(0).trial_division() Traceback (most recent call last): ... ValueError: self must be nonzero sage: n = next_prime(10^5) * next_prime(10^40); n.trial_division() 100003 sage: n.trial_division(bound=10^4) 1000030000000000000000000000000000000012100363 sage: (-n).trial_division(bound=10^4) 1000030000000000000000000000000000000012100363 sage: (-n).trial_division() 100003 sage: n = 2 * next_prime(10^40); n.trial_division() 2 sage: n = 3 * next_prime(10^40); n.trial_division() 3 sage: n = 5 * next_prime(10^40); n.trial_division() 5 sage: n = 2 * next_prime(10^4); n.trial_division() 2 sage: n = 3 * next_prime(10^4); n.trial_division() 3 sage: n = 5 * next_prime(10^4); n.trial_division() 5
You can specify a starting point:
sage: n = 3*5*101*103 sage: n.trial_division(start=50) 101
- trunc()#
Round this number to the nearest integer, which is self since self is an integer.
EXAMPLES:
sage: n = 6 sage: n.trunc() 6
- val_unit(p)#
Returns a pair: the p-adic valuation of self, and the p-adic unit of self.
INPUT:
p
- an integer at least 2.
OUTPUT:
v_p(self)
- the p-adic valuation ofself
u_p(self)
-self
/ \(p^{v_p(\mathrm{self})}\)
EXAMPLES:
sage: n = 60 sage: n.val_unit(2) (2, 15) sage: n.val_unit(3) (1, 20) sage: n.val_unit(7) (0, 60) sage: (2^11).val_unit(4) (5, 2) sage: 0.val_unit(2) (+Infinity, 1)
- valuation(p)#
Return the p-adic valuation of self.
INPUT:
p
- an integer at least 2.
EXAMPLES:
sage: n = 60 sage: n.valuation(2) 2 sage: n.valuation(3) 1 sage: n.valuation(7) 0 sage: n.valuation(1) Traceback (most recent call last): ... ValueError: You can only compute the valuation with respect to a integer larger than 1.
We do not require that p is a prime:
sage: (2^11).valuation(4) 5
- xgcd(n)#
Return the extended gcd of this element and
n
.INPUT:
n
– an integer
OUTPUT:
A triple
(g, s, t)
such thatg
is the non-negative gcd ofself
andn
, ands
andt
are cofactors satisfying the Bezout identity\[g = s \cdot \mathrm{self} + t \cdot n.\]Note
There is no guarantee that the cofactors will be minimal. If you need the cofactors to be minimal use
_xgcd()
. Also, using_xgcd()
directly might be faster in some cases, see trac ticket #13628.EXAMPLES:
sage: 6.xgcd(4) (2, 1, -1)
- class sage.rings.integer.IntegerWrapper#
Bases:
sage.rings.integer.Integer
Rationale for the
IntegerWrapper
class:With
Integers
, the allocation/deallocation function slots are hijacked with custom functions that stick already allocatedIntegers
(with initializedparent
andmpz_t
fields) into a pool on “deallocation” and then pull them out whenever a new one is needed. BecauseIntegers
are so common, this is actually a significant savings. However , this does cause issues with subclassing a Python class directly fromInteger
(but that’s ok for a Cython class).As a workaround, one can instead derive a class from the intermediate class
IntegerWrapper
, which sets statically its alloc/dealloc methods to the originalInteger
alloc/dealloc methods, before they are swapped manually for the custom ones.The constructor of
IntegerWrapper
further allows for specifying an alternative parent toIntegerRing()
.
- sage.rings.integer.free_integer_pool()#
- class sage.rings.integer.int_to_Z#
Bases:
sage.categories.morphism.Morphism
Morphism from Python ints to Sage integers.
EXAMPLES:
sage: f = ZZ.coerce_map_from(int) sage: type(f) <class 'sage.rings.integer.long_to_Z'> sage: f(5r) 5 sage: type(f(5r)) <class 'sage.rings.integer.Integer'> sage: 1 + 2r 3 sage: type(1 + 2r) <class 'sage.rings.integer.Integer'>
This is intended for internal use by the coercion system, to facilitate fast expressions mixing ints and more complex Python types. Note that (as with all morphisms) the input is forcably coerced to the domain
int
if it is not already of the correct type which may have undesirable results:sage: f.domain() Set of Python objects of class 'int' sage: f(1/3) 0 sage: f(1.7) 1 sage: f("10") 10
A pool is used for small integers:
sage: f(10) is f(10) True sage: f(-2) is f(-2) True
- sage.rings.integer.is_Integer(x)#
Return true if x is of the Sage integer type.
EXAMPLES:
sage: from sage.rings.integer import is_Integer sage: is_Integer(2) True sage: is_Integer(2/1) False sage: is_Integer(int(2)) False sage: is_Integer('5') False
- class sage.rings.integer.long_to_Z#
Bases:
sage.categories.morphism.Morphism
EXAMPLES:
sage: f = ZZ.coerce_map_from(int) sage: f Native morphism: From: Set of Python objects of class 'int' To: Integer Ring sage: f(1rL) 1
- sage.rings.integer.make_integer(s)#
Create a Sage integer from the base-32 Python string s. This is used in unpickling integers.
EXAMPLES:
sage: from sage.rings.integer import make_integer sage: make_integer('-29') -73 sage: make_integer(29) Traceback (most recent call last): ... TypeError: expected str...Integer found