def phi(n):
#*****************************************************************************80
#
## PHI computes the number of relatively prime predecessors of an integer.
#
# Definition:
#
# PHI(N) is the number of integers between 1 and N which are
# relatively prime to N. I and J are relatively prime if they
# have no common factors. The function PHI(N) is known as
# "Euler's totient function".
#
# By convention, 1 and N are relatively prime.
#
# First values:
#
# N PHI(N)
#
# 1 1
# 2 1
# 3 2
# 4 2
# 5 4
# 6 2
# 7 6
# 8 4
# 9 6
# 10 4
# 11 10
# 12 4
# 13 12
# 14 6
# 15 8
# 16 8
# 17 16
# 18 6
# 19 18
# 20 8
#
# Formula:
#
# PHI(U*V) = PHI(U) * PHI(V) if U and V are relatively prime.
#
# PHI(P^K) = P^(K-1) * ( P - 1 ) if P is prime.
#
# PHI(N) = N * Product ( P divides N ) ( 1 - 1 / P )
#
# N = Sum ( D divides N ) PHI(D).
#
# Licensing:
#
# This code is distributed under the GNU LGPL license.
#
# Modified:
#
# 25 February 2015
#
# Author:
#
# John Burkardt
#
# Parameters:
#
# Input, integer N, the value to be analyzed.
#
# Output, integer VALUE, the value of PHI(N). If N is less than
# or equal to 0, PHI will be returned as 0. If there is not enough
# room for full factoring of N, PHI will be returned as -1.
#
from i4_factor import i4_factor
if (n <= 0):
value = 0
return value
if (n == 1):
value = 1
return value
#
# Factor N.
#
nfactor, factor, power, nleft = i4_factor(n)
if (nleft != 1):
print ''
print 'PHI - Fatal error!'
print ' Not enough factorization space.'
value = 1
for i in range(0, nfactor):
value = value * factor[i]**(power[i] - 1) * (factor[i] - 1)
return value
def legendre_symbol ( q, p ):
#*****************************************************************************80
#
## LEGENDRE_SYMBOL evaluates the Legendre symbol (Q/P).
#
# Definition:
#
# Let P be an odd prime. Q is a QUADRATIC RESIDUE modulo P
# if there is an integer R such that R^2 = Q ( mod P ).
# The Legendre symbol ( Q / P ) is defined to be:
#
# + 1 if Q ( mod P ) /= 0 and Q is a quadratic residue modulo P,
# - 1 if Q ( mod P ) /= 0 and Q is not a quadratic residue modulo P,
# 0 if Q ( mod P ) == 0.
#
# We can also define ( Q / P ) for P = 2 by:
#
# + 1 if Q ( mod P ) /= 0
# 0 if Q ( mod P ) == 0
#
# Example:
#
# (0/7) = 0
# (1/7) = + 1 ( 1^2 = 1 mod 7 )
# (2/7) = + 1 ( 3^2 = 2 mod 7 )
# (3/7) = - 1
# (4/7) = + 1 ( 2^2 = 4 mod 7 )
# (5/7) = - 1
# (6/7) = - 1
#
# Note:
#
# For any prime P, exactly half of the integers from 1 to P-1
# are quadratic residues.
#
# ( 0 / P ) = 0.
#
# ( Q / P ) = ( mod ( Q, P ) / P ).
#
# ( Q / P ) = ( Q1 / P ) * ( Q2 / P ) if Q = Q1 * Q2.
#
# If Q is prime, and P is prime and greater than 2, then:
#
# if ( Q == 1 ) then
#
# ( Q / P ) = 1
#
# else if ( Q == 2 ) then
#
# ( Q / P ) = + 1 if mod ( P, 8 ) = 1 or mod ( P, 8 ) = 7,
# ( Q / P ) = - 1 if mod ( P, 8 ) = 3 or mod ( P, 8 ) = 5.
#
# else
#
# ( Q / P ) = - ( P / Q ) if Q = 3 ( mod 4 ) and P = 3 ( mod 4 ),
# = ( P / Q ) otherwise.
#
# Licensing:
#
# This code is distributed under the GNU LGPL license.
#
# Modified:
#
# 16 February 2015
#
# Author:
#
# John Burkardt
#
# Reference:
#
# Charles Pinter,
# A Book of Abstract Algebra,
# McGraw Hill, 1982, pages 236-237.
#
# Daniel Zwillinger,
# CRC Standard Mathematical Tables and Formulae,
# 30th Edition,
# CRC Press, 1996, pages 86-87.
#
# Parameters:
#
# Input, integer Q, an integer whose Legendre symbol with
# respect to P is desired.
#
# Input, integer P, a prime number, greater than 1, with respect
# to which the Legendre symbol of Q is desired.
#
# Output, integer L, the Legendre symbol (Q/P).
# Ordinarily, L will be -1, 0 or 1.
# L = -2, P is less than or equal to 1.
# L = -3, P is not prime.
# L = -4, the internal stack of factors overflowed.
# L = -5, not enough factorization space.
#
import numpy as np
from i4_factor import i4_factor
from i4_is_prime import i4_is_prime
from sys import exit
l = 0
#
# P must be greater than 1.
#
if ( p <= 1 ):
print ''
print 'LEGENDRE_SYMBOL - Fatal error!'
print ' P must be greater than 1.'
l = -2
exit ( 'LEGENDRE_SYMBOL - Fatal error!' )
#
# P must be prime.
#
if ( not ( i4_is_prime ( p ) ) ):
print ''
print 'LEGENDRE_SYMBOL - Fatal error!'
print ' P is not prime.'
l = -3
exit ( 'LEGENDRE_SYMBOL - Fatal error!' )
#
# ( k*P / P ) = 0.
#
if ( ( q % p ) == 0 ):
l = 0
return l
#
# For the special case P = 2, (Q/P) = 1 for all odd numbers.
#
if ( p == 2 ):
l = 1
return l
#
# Make a copy of Q, and force it to be nonnegative.
#
qq = q
while ( qq < 0 ):
qq = qq + p
nstack = 0
pstack = np.zeros ( 100 )
qstack = np.zeros ( 100 )
pp = p
l = 1
while ( True ):
qq = ( qq % pp )
#
# Decompose QQ into factors of prime powers.
#
nfactor, factor, power, nleft = i4_factor ( qq )
if ( nleft != 1 ):
print ''
print 'LEGENDRE_SYMBOL - Fatal error!'
print ' Not enough factorization space.'
l = -5
exit ( 'LEGENDRE_SYMBOL - Fatal error!' )
#
# Each factor which is an odd power is added to the stack.
#
nmore = 0
for i in range ( 0, nfactor ):
if ( ( power[i] % 2 ) == 1 ):
nmore = nmore + 1
pstack[nstack] = pp
qstack[nstack] = factor[i]
nstack = nstack + 1
hop = False
if ( nmore != 0 ):
nstack = nstack - 1
qq = qstack[nstack]
#
# Check for a QQ of 1 or 2.
#
if ( qq == 1 ):
l = + 1 * l
elif ( qq == 2 and ( ( pp % 8 ) == 1 or ( pp % 8 ) == 7 ) ):
l = + 1 * l
elif ( qq == 2 and ( ( pp % 8 ) == 3 or ( pp % 8 ) == 5 ) ):
l = - 1 * l
else:
if ( ( pp % 4 ) == 3 and ( qq % 4 ) == 3 ):
l = - 1 * l
rr = pp
pp = qq
qq = rr
hop = True
#
# If the stack is empty, we're done.
#
if ( not hop ):
if ( nstack == 0 ):
break
#
# Otherwise, get the last P and Q from the stack, and process them.
#
nstack = nstack - 1
pp = pstack[nstack]
qq = qstack[nstack]
return l
def sigma ( n ):
#*****************************************************************************80
#
## SIGMA returns the value of SIGMA(N), the divisor sum.
#
# Definition:
#
# SIGMA(N) is the sum of the distinct divisors of N, including 1 and N.
#
# First values:
#
# N SIGMA(N)
#
# 1 1
# 2 3
# 3 4
# 4 7
# 5 6
# 6 12
# 7 8
# 8 15
# 9 13
# 10 18
# 11 12
# 12 28
# 13 14
# 14 24
# 15 24
# 16 31
# 17 18
# 18 39
# 19 20
# 20 42
#
# Formula:
#
# SIGMA(U*V) = SIGMA(U) * SIGMA(V) if U and V are relatively prime.
#
# SIGMA(P^K) = ( P^(K+1) - 1 ) / ( P - 1 ) if P is prime.
#
# Licensing:
#
# This code is distributed under the GNU LGPL license.
#
# Modified:
#
# 26 February 2015
#
# Author:
#
# John Burkardt
#
# Parameters:
#
# Input, integer N, the value to be analyzed.
#
# Output, integer VALUE, the value of SIGMA(N). If N is less than
# or equal to 0, VALUE will be returned as 0. If there is not
# enough room for factoring N, VALUE is returned as -1.
#
from i4_factor import i4_factor
maxfactor = 20
if ( n <= 0 ):
value = 0
return value
if ( n == 1 ):
value = 1
return value
#
# Factor N.
#
nfactor, factor, power, nleft = i4_factor ( n )
if ( nleft != 1 ):
print ''
print 'SIGMA - Fatal error!'
print ' Not enough factorization space.'
value = 1
for i in range ( 0, nfactor ):
value = ( value * ( factor[i] ** ( power[i] + 1 ) - 1 ) ) \
/ ( factor[i] - 1 )
return value
def legendre_symbol(q, p):
#*****************************************************************************80
#
## LEGENDRE_SYMBOL evaluates the Legendre symbol (Q/P).
#
# Definition:
#
# Let P be an odd prime. Q is a QUADRATIC RESIDUE modulo P
# if there is an integer R such that R^2 = Q ( mod P ).
# The Legendre symbol ( Q / P ) is defined to be:
#
# + 1 if Q ( mod P ) /= 0 and Q is a quadratic residue modulo P,
# - 1 if Q ( mod P ) /= 0 and Q is not a quadratic residue modulo P,
# 0 if Q ( mod P ) == 0.
#
# We can also define ( Q / P ) for P = 2 by:
#
# + 1 if Q ( mod P ) /= 0
# 0 if Q ( mod P ) == 0
#
# Example:
#
# (0/7) = 0
# (1/7) = + 1 ( 1^2 = 1 mod 7 )
# (2/7) = + 1 ( 3^2 = 2 mod 7 )
# (3/7) = - 1
# (4/7) = + 1 ( 2^2 = 4 mod 7 )
# (5/7) = - 1
# (6/7) = - 1
#
# Note:
#
# For any prime P, exactly half of the integers from 1 to P-1
# are quadratic residues.
#
# ( 0 / P ) = 0.
#
# ( Q / P ) = ( mod ( Q, P ) / P ).
#
# ( Q / P ) = ( Q1 / P ) * ( Q2 / P ) if Q = Q1 * Q2.
#
# If Q is prime, and P is prime and greater than 2, then:
#
# if ( Q == 1 ) then
#
# ( Q / P ) = 1
#
# else if ( Q == 2 ) then
#
# ( Q / P ) = + 1 if mod ( P, 8 ) = 1 or mod ( P, 8 ) = 7,
# ( Q / P ) = - 1 if mod ( P, 8 ) = 3 or mod ( P, 8 ) = 5.
#
# else
#
# ( Q / P ) = - ( P / Q ) if Q = 3 ( mod 4 ) and P = 3 ( mod 4 ),
# = ( P / Q ) otherwise.
#
# Licensing:
#
# This code is distributed under the GNU LGPL license.
#
# Modified:
#
# 16 February 2015
#
# Author:
#
# John Burkardt
#
# Reference:
#
# Charles Pinter,
# A Book of Abstract Algebra,
# McGraw Hill, 1982, pages 236-237.
#
# Daniel Zwillinger,
# CRC Standard Mathematical Tables and Formulae,
# 30th Edition,
# CRC Press, 1996, pages 86-87.
#
# Parameters:
#
# Input, integer Q, an integer whose Legendre symbol with
# respect to P is desired.
#
# Input, integer P, a prime number, greater than 1, with respect
# to which the Legendre symbol of Q is desired.
#
# Output, integer L, the Legendre symbol (Q/P).
# Ordinarily, L will be -1, 0 or 1.
# L = -2, P is less than or equal to 1.
# L = -3, P is not prime.
# L = -4, the internal stack of factors overflowed.
# L = -5, not enough factorization space.
#
import numpy as np
from i4_factor import i4_factor
from i4_is_prime import i4_is_prime
from sys import exit
l = 0
#
# P must be greater than 1.
#
if (p <= 1):
print ''
print 'LEGENDRE_SYMBOL - Fatal error!'
print ' P must be greater than 1.'
l = -2
exit('LEGENDRE_SYMBOL - Fatal error!')
#
# P must be prime.
#
if (not (i4_is_prime(p))):
print ''
print 'LEGENDRE_SYMBOL - Fatal error!'
print ' P is not prime.'
l = -3
exit('LEGENDRE_SYMBOL - Fatal error!')
#
# ( k*P / P ) = 0.
#
if ((q % p) == 0):
l = 0
return l
#
# For the special case P = 2, (Q/P) = 1 for all odd numbers.
#
if (p == 2):
l = 1
return l
#
# Make a copy of Q, and force it to be nonnegative.
#
qq = q
while (qq < 0):
qq = qq + p
nstack = 0
pstack = np.zeros(100)
qstack = np.zeros(100)
pp = p
l = 1
while (True):
qq = (qq % pp)
#
# Decompose QQ into factors of prime powers.
#
nfactor, factor, power, nleft = i4_factor(qq)
if (nleft != 1):
print ''
print 'LEGENDRE_SYMBOL - Fatal error!'
print ' Not enough factorization space.'
l = -5
exit('LEGENDRE_SYMBOL - Fatal error!')
#
# Each factor which is an odd power is added to the stack.
#
nmore = 0
for i in range(0, nfactor):
if ((power[i] % 2) == 1):
nmore = nmore + 1
pstack[nstack] = pp
qstack[nstack] = factor[i]
nstack = nstack + 1
hop = False
if (nmore != 0):
nstack = nstack - 1
qq = qstack[nstack]
#
# Check for a QQ of 1 or 2.
#
if (qq == 1):
l = +1 * l
elif (qq == 2 and ((pp % 8) == 1 or (pp % 8) == 7)):
l = +1 * l
elif (qq == 2 and ((pp % 8) == 3 or (pp % 8) == 5)):
l = -1 * l
else:
if ((pp % 4) == 3 and (qq % 4) == 3):
l = -1 * l
rr = pp
pp = qq
qq = rr
hop = True
#
# If the stack is empty, we're done.
#
if (not hop):
if (nstack == 0):
break
#
# Otherwise, get the last P and Q from the stack, and process them.
#
nstack = nstack - 1
pp = pstack[nstack]
qq = qstack[nstack]
return l
def jacobi_symbol ( q, p ):
#*****************************************************************************80
#
## JACOBI_SYMBOL evaluates the Jacobi symbol (Q/P).
#
# Definition:
#
# If P is prime, then
#
# Jacobi Symbol (Q/P) = Legendre Symbol (Q/P)
#
# Else
#
# let P have the prime factorization
#
# P = Product ( 1 <= I <= N ) P(I)^E(I)
#
# Jacobi Symbol (Q/P) =
#
# Product ( 1 <= I <= N ) Legendre Symbol (Q/P(I))^E(I)
#
# Licensing:
#
# This code is distributed under the GNU LGPL license.
#
# Modified:
#
# 17 February 2015
#
# Author:
#
# John Burkardt
#
# Reference:
#
# Daniel Zwillinger,
# CRC Standard Mathematical Tables and Formulae,
# 30th Edition,
# CRC Press, 1996, pages 86-87.
#
# Parameters:
#
# Input, integer Q, an integer whose Jacobi symbol with
# respect to P is desired.
#
# Input, integer P, the number with respect to which the Jacobi
# symbol of Q is desired. P should be 2 or greater.
#
# Output, integer J, the Jacobi symbol (Q/P).
# Ordinarily, J will be -1, 0 or 1.
# -2, not enough factorization space.
# -3, an error during Legendre symbol calculation.
#
from i4_factor import i4_factor
from legendre_symbol import legendre_symbol
#
# P must be greater than 1.
#
if ( p <= 1 ):
print ''
print 'JACOBI_SYMBOL - Fatal error!'
print ' P must be greater than 1.'
j = -2
return l
#
# Decompose P into factors of prime powers.
#
nfactor, factor, power, nleft = i4_factor ( p )
if ( nleft != 1 ):
print ''
print 'JACOBI_SYMBOL - Fatal error!'
print ' Not enough factorization space.'
j = -2
return j
#
# Force Q to be nonnegative.
#
qq = q
while ( qq < 0 ):
qq = qq + p
#
# For each prime factor, compute the Legendre symbol, and
# multiply the Jacobi symbol by the appropriate factor.
#
j = 1
for i in range ( 0, nfactor ):
pp = factor[i]
l = legendre_symbol ( qq, pp )
if ( l < -1 ):
print ''
print 'JACOBI_SYMBOL - Fatal error!'
print ' Error during Legendre symbol calculation.'
j = -3
j = j * l ** power[i]
return j
def jacobi_symbol(q, p):
#*****************************************************************************80
#
## JACOBI_SYMBOL evaluates the Jacobi symbol (Q/P).
#
# Definition:
#
# If P is prime, then
#
# Jacobi Symbol (Q/P) = Legendre Symbol (Q/P)
#
# Else
#
# let P have the prime factorization
#
# P = Product ( 1 <= I <= N ) P(I)^E(I)
#
# Jacobi Symbol (Q/P) =
#
# Product ( 1 <= I <= N ) Legendre Symbol (Q/P(I))^E(I)
#
# Licensing:
#
# This code is distributed under the GNU LGPL license.
#
# Modified:
#
# 17 February 2015
#
# Author:
#
# John Burkardt
#
# Reference:
#
# Daniel Zwillinger,
# CRC Standard Mathematical Tables and Formulae,
# 30th Edition,
# CRC Press, 1996, pages 86-87.
#
# Parameters:
#
# Input, integer Q, an integer whose Jacobi symbol with
# respect to P is desired.
#
# Input, integer P, the number with respect to which the Jacobi
# symbol of Q is desired. P should be 2 or greater.
#
# Output, integer J, the Jacobi symbol (Q/P).
# Ordinarily, J will be -1, 0 or 1.
# -2, not enough factorization space.
# -3, an error during Legendre symbol calculation.
#
from i4_factor import i4_factor
from legendre_symbol import legendre_symbol
#
# P must be greater than 1.
#
if (p <= 1):
print('')
print('JACOBI_SYMBOL - Fatal error!')
print(' P must be greater than 1.')
j = -2
return l
#
# Decompose P into factors of prime powers.
#
nfactor, factor, power, nleft = i4_factor(p)
if (nleft != 1):
print('')
print('JACOBI_SYMBOL - Fatal error!')
print(' Not enough factorization space.')
j = -2
return j
#
# Force Q to be nonnegative.
#
qq = q
while (qq < 0):
qq = qq + p
#
# For each prime factor, compute the Legendre symbol, and
# multiply the Jacobi symbol by the appropriate factor.
#
j = 1
for i in range(0, nfactor):
pp = factor[i]
l = legendre_symbol(qq, pp)
if (l < -1):
print('')
print('JACOBI_SYMBOL - Fatal error!')
print(' Error during Legendre symbol calculation.')
j = -3
j = j * l**power[i]
return j
def tau(n):
#*****************************************************************************80
#
## TAU returns the value of TAU(N), the number of distinct divisors of N.
#
# Discussion:
#
# TAU(N) is the number of divisors of N, including 1 and N.
#
# First values:
#
# N TAU(N)
#
# 1 1
# 2 2
# 3 2
# 4 3
# 5 2
# 6 4
# 7 2
# 8 4
# 9 3
# 10 4
# 11 2
# 12 6
# 13 2
# 14 4
# 15 4
# 16 5
# 17 2
# 18 6
# 19 2
# 20 6
#
# Formula:
#
# If the prime factorization of N is
#
# N = P1^E1 * P2^E2 * ... * PM^EM,
#
# then
#
# TAU(N) = ( E1 + 1 ) * ( E2 + 1 ) * ... * ( EM + 1 ).
#
# Licensing:
#
# This code is distributed under the GNU LGPL license.
#
# Modified:
#
# 25 February 2015
#
# Author:
#
# John Burkardt
#
# Parameters:
#
# Input, integer N, the value to be analyzed. N must be 1 or
# greater.
#
# Output, integer TAUN, the value of TAU(N). But if N is 0 or
# less, TAUN is returned as 0, a nonsense value. If there is
# not enough room for factoring, TAUN is returned as -1.
#
from i4_factor import i4_factor
if (n <= 0):
value = 0
return value
if (n == 1):
value = 1
return value
#
# Factor N.
#
nfactor, factor, power, nleft = i4_factor(n)
if (nleft != 1):
print ''
print 'TAU - Fatal error!'
print ' Not enough factorization space.'
value = 1
for i in range(0, nfactor):
value = value * (power[i] + 1)
return value
def tau ( n ):
#*****************************************************************************80
#
## TAU returns the value of TAU(N), the number of distinct divisors of N.
#
# Discussion:
#
# TAU(N) is the number of divisors of N, including 1 and N.
#
# First values:
#
# N TAU(N)
#
# 1 1
# 2 2
# 3 2
# 4 3
# 5 2
# 6 4
# 7 2
# 8 4
# 9 3
# 10 4
# 11 2
# 12 6
# 13 2
# 14 4
# 15 4
# 16 5
# 17 2
# 18 6
# 19 2
# 20 6
#
# Formula:
#
# If the prime factorization of N is
#
# N = P1^E1 * P2^E2 * ... * PM^EM,
#
# then
#
# TAU(N) = ( E1 + 1 ) * ( E2 + 1 ) * ... * ( EM + 1 ).
#
# Licensing:
#
# This code is distributed under the GNU LGPL license.
#
# Modified:
#
# 25 February 2015
#
# Author:
#
# John Burkardt
#
# Parameters:
#
# Input, integer N, the value to be analyzed. N must be 1 or
# greater.
#
# Output, integer TAUN, the value of TAU(N). But if N is 0 or
# less, TAUN is returned as 0, a nonsense value. If there is
# not enough room for factoring, TAUN is returned as -1.
#
from i4_factor import i4_factor
if ( n <= 0 ):
value = 0
return value
if ( n == 1 ):
value = 1
return value
#
# Factor N.
#
nfactor, factor, power, nleft = i4_factor ( n )
if ( nleft != 1 ):
print ''
print 'TAU - Fatal error!'
print ' Not enough factorization space.'
value = 1
for i in range ( 0, nfactor ):
value = value * ( power[i] + 1 )
return value
def moebius ( n ):
#*****************************************************************************80
#
## MOEBIUS returns the value of MU(N), the Moebius function of N.
#
# Definition:
#
# MU(N) is defined as follows:
#
# MU(N) = 1 if N = 1;
# 0 if N is divisible by the square of a prime;
# (-1)^K, if N is the product of K distinct primes.
#
# First values:
#
# N MU(N)
#
# 1 1
# 2 -1
# 3 -1
# 4 0
# 5 -1
# 6 1
# 7 -1
# 8 0
# 9 0
# 10 1
# 11 -1
# 12 0
# 13 -1
# 14 1
# 15 1
# 16 0
# 17 -1
# 18 0
# 19 -1
# 20 0
#
# As special cases, MU(N) is -1 if N is a prime, and MU(N) is 0
# if N is a square, cube, etc.
#
# The Moebius function is related to Euler's totient function:
#
# PHI(N) = Sum ( D divides N ) MU(D) * ( N / D ).
#
# Licensing:
#
# This code is distributed under the GNU LGPL license.
#
# Modified:
#
# 23 February 2015
#
# Author:
#
# John Burkardt
#
# Parameters:
#
# Input, integer N, the value to be analyzed.
#
# Output, integer VALUE, the value of MU(N).
# If N is less than or equal to 0, MU will be returned as -2.
# If there was not enough internal space for factoring, MU
# is returned as -3.
#
from i4_factor import i4_factor
if ( n <= 0 ):
value = -2
return value
if ( n == 1 ):
value = 1
return value
#
# Factor N.
#
nfactor, factor, power, nleft = i4_factor ( n )
if ( nleft != 1 ):
print ''
print 'MOEBIUS - Fatal error!'
print ' Incomplete factorization.'
value = -3
value = 1
for i in range ( 0, nfactor ):
value = - value
if ( 1 < power[i] ):
value = 0
return value
return value
def omega ( n ) :
#*****************************************************************************80
#
## OMEGA returns OMEGA(N), the number of distinct prime divisors of N.
#
# First values:
#
# N OMEGA(N)
#
# 1 1
# 2 1
# 3 1
# 4 1
# 5 1
# 6 2
# 7 1
# 8 1
# 9 1
# 10 2
# 11 1
# 12 2
# 13 1
# 14 2
# 15 2
# 16 1
# 17 1
# 18 2
# 19 1
# 20 2
#
# Formula:
#
# If N = 1, then
#
# OMEGA(N) = 1
#
# else if the prime factorization of N is
#
# N = P1^E1 * P2^E2 * ... * PM^EM,
#
# then
#
# OMEGA(N) = M
#
# Licensing:
#
# This code is distributed under the GNU LGPL license.
#
# Modified:
#
# 24 February 2015
#
# Author:
#
# John Burkardt
#
# Parameters:
#
# Input, integer N, the value to be analyzed. N must be 1 or
# greater.
#
# Output, integer VALUE, the value of OMEGA(N). But if N is 0 or
# less, NDIV is returned as 0, a nonsense value. If there is
# not enough room for factoring, NDIV is returned as -1.
#
from i4_factor import i4_factor
if ( n <= 0 ):
value = 0
return value
if ( n == 1 ):
value = 1
return value
#
# Factor N.
#
nfactor, factor, power, nleft = i4_factor ( n )
if ( nleft != 1 ):
print ''
print 'OMEGA - Fatal error!'
print ' Not enough factorization space.'
value = nfactor
return value
def phi ( n ):
#*****************************************************************************80
#
## PHI computes the number of relatively prime predecessors of an integer.
#
# Definition:
#
# PHI(N) is the number of integers between 1 and N which are
# relatively prime to N. I and J are relatively prime if they
# have no common factors. The function PHI(N) is known as
# "Euler's totient function".
#
# By convention, 1 and N are relatively prime.
#
# First values:
#
# N PHI(N)
#
# 1 1
# 2 1
# 3 2
# 4 2
# 5 4
# 6 2
# 7 6
# 8 4
# 9 6
# 10 4
# 11 10
# 12 4
# 13 12
# 14 6
# 15 8
# 16 8
# 17 16
# 18 6
# 19 18
# 20 8
#
# Formula:
#
# PHI(U*V) = PHI(U) * PHI(V) if U and V are relatively prime.
#
# PHI(P^K) = P^(K-1) * ( P - 1 ) if P is prime.
#
# PHI(N) = N * Product ( P divides N ) ( 1 - 1 / P )
#
# N = Sum ( D divides N ) PHI(D).
#
# Licensing:
#
# This code is distributed under the GNU LGPL license.
#
# Modified:
#
# 25 February 2015
#
# Author:
#
# John Burkardt
#
# Parameters:
#
# Input, integer N, the value to be analyzed.
#
# Output, integer VALUE, the value of PHI(N). If N is less than
# or equal to 0, PHI will be returned as 0. If there is not enough
# room for full factoring of N, PHI will be returned as -1.
#
from i4_factor import i4_factor
if ( n <= 0 ):
value = 0
return value
if ( n == 1 ):
value = 1
return value
#
# Factor N.
#
nfactor, factor, power, nleft = i4_factor ( n )
if ( nleft != 1 ):
print ''
print 'PHI - Fatal error!'
print ' Not enough factorization space.'
value = 1
for i in range ( 0, nfactor ):
value = value * factor[i] ** ( power[i] - 1 ) * ( factor[i] - 1 )
return value