Determine whether n is prime. Return 2 if n is definitely prime, return 1 if n is probably prime (without being certain), or return 0 if n is definitely non-prime.
This function performs some trial divisions, then reps Miller-Rabin probabilistic primality tests. A higher reps value will reduce the chances of a non-prime being identified as “probably prime”. A composite number will be identified as a prime with a probability of less than 4^(-reps). Reasonable values of reps are between 15 and 50.
Set rop to the next prime greater than op.
This function uses a probabilistic algorithm to identify primes. For practical purposes it’s adequate, the chance of a composite passing will be extremely small.
Set rop to the greatest common divisor of op1 and op2. The result is always positive even if one or both input operands are negative. Except if both inputs are zero; then this function defines gcd(0,0) = 0.
Compute the greatest common divisor of op1 and op2. If
rop is not
NULL, store the result there.
If the result is small enough to fit in an
unsigned long int, it is
returned. If the result does not fit, 0 is returned, and the result is equal
to the argument op1. Note that the result will always fit if op2
Set g to the greatest common divisor of a and b, and in addition set s and t to coefficients satisfying a*s + b*t = g. The value in g is always positive, even if one or both of a and b are negative (or zero if both inputs are zero). The values in s and t are chosen such that normally, abs(s) < abs(b) / (2 g) and abs(t) < abs(a) / (2 g), and these relations define s and t uniquely. There are a few exceptional cases:
If abs(a) = abs(b), then s = 0, t = sgn(b).
Otherwise, s = sgn(a) if b = 0 or abs(b) = 2 g, and t = sgn(b) if a = 0 or abs(a) = 2 g.
In all cases, s = 0 if and only if g = abs(b), i.e., if b divides a or a = b = 0.
If t is
NULL then that value is not computed.
Set rop to the least common multiple of op1 and op2. rop is always positive, irrespective of the signs of op1 and op2. rop will be zero if either op1 or op2 is zero.
Compute the inverse of op1 modulo op2 and put the result in rop. If the inverse exists, the return value is non-zero and rop will satisfy 0 <= rop < abs(op2) (with rop = 0 possible only when abs(op2) = 1, i.e., in the somewhat degenerate zero ring). If an inverse doesn’t exist the return value is zero and rop is undefined. The behaviour of this function is undefined when op2 is zero.
Calculate the Jacobi symbol (a/b). This is defined only for b odd.
Calculate the Legendre symbol (a/p). This is defined only for p an odd positive prime, and for such p it’s identical to the Jacobi symbol.
Calculate the Jacobi symbol (a/b) with the Kronecker extension (a/2)=(2/a) when a odd, or (a/2)=0 when a even.
When b is odd the Jacobi symbol and Kronecker symbol are
mpz_kronecker_ui etc can be used for mixed
precision Jacobi symbols too.
For more information see Henri Cohen section 1.4.2 (see References),
or any number theory textbook. See also the example program
demos/qcn.c which uses
Remove all occurrences of the factor f from op and store the result in rop. The return value is how many such occurrences were removed.
Set rop to the factorial of n:
mpz_fac_ui computes the plain factorial n!,
mpz_2fac_ui computes the double-factorial n!!, and
Set rop to the primorial of n, i.e. the product of all positive prime numbers <=n.
Compute the binomial coefficient n over
k and store the result in rop. Negative values of n are
mpz_bin_ui, using the identity
bin(-n,k) = (-1)^k * bin(n+k-1,k), see Knuth volume 1 section 1.2.6
mpz_fib_ui sets fn to to F[n], the n’th Fibonacci
mpz_fib2_ui sets fn to F[n], and fnsub1 to
These functions are designed for calculating isolated Fibonacci numbers. When
a sequence of values is wanted it’s best to start with
iterate the defining F[n+1]=F[n]+F[n-1] or
mpz_lucnum_ui sets ln to to L[n], the n’th Lucas
mpz_lucnum2_ui sets ln to L[n], and lnsub1
These functions are designed for calculating isolated Lucas numbers. When a
sequence of values is wanted it’s best to start with
iterate the defining L[n+1]=L[n]+L[n-1] or
The Fibonacci numbers and Lucas numbers are related sequences, so it’s never
necessary to call both
formulas for going from Fibonacci to Lucas can be found in Lucas Numbers Algorithm, the reverse is straightforward too.