1Math::NumSeq::Abundant(U3s)er Contributed Perl DocumentatMiaotnh::NumSeq::Abundant(3)
2
6 Math::NumSeq::Abundant -- abundant numbers, greater than sum of
7 divisors
8
10 use Math::NumSeq::Abundant;
11 my $seq = Math::NumSeq::Abundant->new;
12 my ($i, $value) = $seq->next;
13
15 The abundant numbers, being integers greater than the sum of their
16 divisors,
17 12, 18, 20, 24, 30, 36, ...
19 starting i=1
20 For example 12 is abundant because its divisors 1,2,3,4,6 add up to 16
22 which is > 12.
23 This is often expressed as 2*n>sigma(n) where sigma(n) is the sum of
25 divisors of n including n itself.
26 Deficient
28 Option "abundant_type => "deficient"" is those integers n with n < sum
29 divisors,
30 abundant_type => "deficient"
32 1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13,
33 This is the opposite of abundant, except the few perfect numbers n ==
35 sum divisors are excluded (see "Perfect Numbers" below).
36 Primitive Abundant
38 Option "abundant_type => "primitive"" gives abundant numbers which are
39 not a multiple of some smaller abundant,
40 abundant_type => "primitive"
42 12, 18, 20, 30, 42, 56, 66, 70, 78, ...
43 If an integer n is abundant then so are all multiples 2*n, 3*n, 4*n,
45 etc. The "primitive" abundants are those which are not such a
46 multiple.
47 Option "abundant_type => "non-primitive"" gives abundant numbers which
49 are not primitive, ie. which have a divisor which is also abundant.
50 abundant_type => "non-primitive"
52 24, 36, 40, 48, 54, 60, 72, 80, 84, ...
53 The abundant are all either primitive or non-primitive.
55 Perfect Numbers
57 Numbers with n == sum divisors are the perfect numbers 6, 28, 496,
58 8128, 33550336, etc. There's nothing here for them currently. They're
59 quite sparse, with Euler proving the even ones are always
60 n=2^(k-1)*(2^k-1) for prime 2^k-1 (those being the Mersenne primes).
61 The existence of any odd perfect numbers is a famous unsolved problem.
62 If there are any odd perfect numbers then they're very big.
63
65 See "FUNCTIONS" in Math::NumSeq for behaviour common to all sequence
66 classes.
67 "$seq = Math::NumSeq::Abundant->new ()"
69 "$seq = Math::NumSeq::Abundant->new (abundant_type => $str)"
70 Create and return a new sequence object. "abundant_type" (a
71 string) can be
72 "abundant" n > sum divisors (the default)
74 "deficient" n < sum divisors
75 "primitive" abundant and not a multiple of an abundant
76 "non-primitive" abundant and also a multiple of an abundant
77 "$bool = $seq->pred($value)"
79 Return true if $value is abundant, deficient or primitive abundant
80 per $seq.
81 This check requires factorizing $value and in the current code a
83 hard limit of 2**32 is placed on values to be checked, in the
84 interests of not going into a near-infinite loop.
85
87 Predicate
88 For prime factorization n=p^a * q^b * ... the divisors are all of
89 divisor = p^A * q^B * ... for A=0 to a, B=0 to b, etc
91 This includes n itself with A=a,B=b,etc. The sum is formed by grouping
93 each with factor p^i, etc, resulting in a product,
94 sigma = (1 + p + p^2 + ... + p^a)
96 * (1 + q + q^2 + ... + q^a)
97 * ...
98 sigma = (p^(a+1)-1)/(p-1) * (q^(b+1)-1)/(q-1) * ...
100 So from the prime factorization of n the sigma is formed and compared
102 against n,
103 sigma > 2*n abundant
105 sigma < 2*n deficient
106 Predicate -- Primitive
108 For primitive abundant we want to know also that no divisor of n is
109 abundant.
110 For divisors of n it suffices to consider n reduced by a single prime,
112 so n/p. If taking out some non-prime such as n/(p*q) gives an abundant
113 then so is n/p because it's a multiple of n/(p*q). To testing an n/p
114 for abundance,
115 sigma(n/p) > 2*n/p means have an abundant divisor
117 sigma(n/p) can be calculated from sigma(n) by dividing out the p^a term
119 described above and replacing it with the term for p^(a-1).
120 oldterm = (p^(a+1) - 1)/(p-1)
122 newterm = (p^a - 1)/(p-1)
123 sigma(n) * newterm / oldterm > n/p
125 sigma(n) * p*newterm / oldterm > n
126 p*newterm/oldterm simplifies to
128 sigma(n) * (1 - 1/oldterm) > n means an abundant divisor
130 The left side is a maximum when the factor (1 - 1/oldterm) reduces
132 sigma(n) by the least, and that's when oldterm is the biggest. So to
133 test for primitive abundance note the largest term in the sigma(n)
134 calculation above.
135 if sigma(n) > 2*n
137 then n is abundant
138 if sigma(n) * (1-1/maxterm) > 2*n
140 then have an abundant divisor and so n is not primitive abundant
141
143 Math::NumSeq
144
146 <http://user42.tuxfamily.org/math-numseq/index.html>
147
149 Copyright 2010, 2011, 2012, 2013, 2014, 2016, 2019, 2020 Kevin Ryde
150 Math-NumSeq is free software; you can redistribute it and/or modify it
152 under the terms of the GNU General Public License as published by the
153 Free Software Foundation; either version 3, or (at your option) any
154 later version.
155 Math-NumSeq is distributed in the hope that it will be useful, but
157 WITHOUT ANY WARRANTY; without even the implied warranty of
158 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
159 General Public License for more details.
160 You should have received a copy of the GNU General Public License along
162 with Math-NumSeq. If not, see <http://www.gnu.org/licenses/>.
163perl v5.34.1 2022-06-06 Math::NumSeq::Abundant(3)