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determination of even abundant
numbers with one odd prime factor∗
rspuzio†
2013-03-21 22:13:53
In this entry, we will use the criterion of the parent entry to determine the
first few even abundant numbers. To keep things more managable, we shall take
advantage of the fact that a multiple of an abundant number is abundant and
only look for abundant numbers none of whose proper divisors are abundant.
Once we know these numbers, it becomes a rather easy matter to find the rest
of the abundant numbers by taking multiples.
To begin, we look at the criterion of the second thorem. Since 2/(2 − 1) = 2
and, for any p > 2, we have 1 < p/(p − 1) < 2, it follows that, for every prime
p, there will exist abundant numbers of the form 2m pn . By the first theorem,
for such a number to be abundant, we must have
(1 − 2−m−1 )(1 − p−n−1 )
>2
1
−1 )
2 (1 − p
or, after a little algebraic simplification,
(1 − 2−m−1 )(1 − p−n−1 ) > 1 − p−1 .
From this inequality, we can deduce a description of abundant numbers of the
form 2m pn .
Theorem 1. Let p > 2 be prime. Then, for all n ≥ 0, we find that 2m pn is
abundant for m sufficiently large.
Proof. When n ≥ 0, we have
1 − p−n−1 > 1 − p−1 .
Since limm→∞ 2−m−1 = 0, it follows that
(1 − 2−m−1 )(1 − p−n−1 ) > 1 − p−1 .
for m sufficiently large, hence 2m pn will be abundant for m sufficiently large.
∗ hDeterminationOfEvenAbundantNumbersWithOneOddPrimeFactori
created: h2013-0321i by: hrspuzioi version: h39031i Privacy setting: h1i hDefinitioni h11A05i
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1
Theorem 2. Let p > 2 be prime. Then there exists m0 such that:
1. If m < m0 , then 2m pn is not abundant for any n.
2. if m ≥ m0 , then 2m pn is abundant for n sufficiently large.
Proof. Set m0 to be the smallest integer such that 2m0 +1 > p. This inequality
is equivalent to
1 − 2−m0 −1 > 1 − p−1 .
On the one hand, if m < m0 , then it will be impossible to satisfy our criterion
for any choice of n. On the other hand, if m ≥ m0 , then, by ther same sort of
continuity argument employed previously, the criterion will be satisfied for m
sufficiently large.
Theorem 3. Let p > 2 be prime and let m0 be the unique integer such that
2m0 +1 > p > 2m0 .
Then 2m0 p is either perfect or abundant and every abundant number of the form
2m pn is a multiple of 2m0 p.
Proof. We begin with the equation
p + 1 ≤ 2m0 +1 .
Making some algebraic manipulations, we obtain the following:
2m0 + 1
1 + p−1
≥ 2−m0 −1 (1 + p−1 )
p≤
p−1
p−1 − 2−m0 −1 (1 + p−1 ) ≥ 0
1 + p−1 − 2−m0 −1 (1 + p−1 ) ≥ 1
(1 − 2−m0 −1 )(1 + p−1 ) ≥ 1
(1 − 2−m0 −1 )(1 + p−2 ) ≥ 1 − p−1
According to our earlier inequality, this means that 2m0 p is either perfect or
abundant.
Suppose that 2m pn is abundant. Then, by the previous result, m ≤ m0 ;
since powers of 2 are deficient, n > 0, so 2m0 p | 2m pn .
This result makes it rather easy to draw up a list of abundant numbers
with one odd prime factor none of whose proper factors are abundant starting
with a table of prime numbers, as has been done below. Note that, in the case
where 2m0 p is perfect, we have listed 2m0 +1 p and 2m0 p2 as these numbers are
abundant.
2
3
5
7
11
13
17
19
23
29
31
37
41
43
47
53
59
61
67
71
73
79
83
89
97
101
103
107
109
113
127
131
137
139
149
151
157
163
167
173
179
181
191
193
197
199
211
223
227
229
233
239
241
251
12 (22 · 3), 18 (2 · 32 )
20 (22 · 5)
56 (23 · 7), 196 (22 · 72 )
88 (23 · 11)
104 (23 · 13)
272 (24 · 17)
304 (24 · 19)
368 (24 · 23)
464 (24 · 29)
992 (25 · 31), 7688 (25 · 312 )
1184 (25 · 37)
1312 (25 · 41)
1376 (25 · 43)
1504 (25 · 47)
1696 (25 · 53)
1888 (25 · 59)
1952 (25 · 61)
4208 (26 · 67)
4544 (26 · 71)
4672 (26 · 73)
5056 (26 · 79)
5312 (26 · 83)
5696 (26 · 89)
6208 (26 · 97)
6464 (26 · 101)
6592 (26 · 103)
6848 (26 · 107)
6976 (26 · 109)
7232 (26 · 113)
16256 (27 · 127), 1032256 (26 · 1272 )
16768 (27 · 131)
17536 (27 · 137)
17792 (27 · 139)
19072 (27 · 149)
19328 (27 · 151)
20096 (27 · 157)
20864 (27 · 163)
21376 (27 · 167)
22144 (27 · 173)
22912 (27 · 179)
23168 (27 · 181)
24448 (27 · 191)
24704 (27 · 193)
25216 (27 · 197)
25472 (27 · 199)
27008 (27 · 211)
28544 (27 · 223)
29056 (27 · 227)
3
29312 (27 · 229)
7
29824 (2 · 233)
30592 (27 · 239)
30848 (27 · 241)
32128 (27 · 251)