determination of even abundant numbers with one odd prime factor

determination of even abundant numbers with one odd prime factor

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

2

m

⁢

p

n

.
By the first

theorem

, for such a number to be abundant, we must have

(

1

-

2

-

m

-

1

)

⁢

(

1

-

p

-

n

-

1

)

1

2

⁢

(

1

-

p

-

1

)

>

2

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

2

m

⁢

p

n

.

Theorem 1

.

Let

p

>

2

be prime. Then, for all

n

≥

0

, we find that

2

m

⁢

p

n

is abundant for

m

sufficiently large.

Proof.

When

n

≥

0

, we have

1

-

p

-

n

-

1

>

1

-

p

-

1

.

Since

lim

m

→

∞

⁡

2

-

m

-

1

=

0

, it follows that

(

1

-

2

-

m

-

1

)

⁢

(

1

-

p

-

n

-

1

)

>

1

-

p

-

1

.

for

m

sufficiently large, hence

2

m

⁢

p

n

will be abundant for

m

sufficiently large.
∎

Theorem 2

.

Let

p

>

2

be prime. Then there exists

m

0

such that:

1.

If

m

<

m

0

, then

2

m

⁢

p

n

is not abundant for any

n

.

2.

if

m

≥

m

0

, then

2

m

⁢

p

n

is abundant for

n

sufficiently large.

Proof.

Set

m

0

to be the smallest

integer

such that

2

m

0

+

1

>

p

.
This inequality is

equivalent

to

1

-

2

-

m

0

-

1

>

1

-

p

-

1

.

On the one hand, if

m

<

m

0

, then it will be impossible to

satisfy

our criterion for any choice of

n

. On the other hand, if

m

≥

m

0

,
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

m

0

be the unique integer such that

2

m

0

+

1

>

p

>

2

m

0

.

Then

2

m

0

⁢

p

is either perfect or abundant and every abundant
number of the form

2

m

⁢

p

n

is a multiple of

2

m

0

⁢

p

.

Proof.

We begin with the equation

p

+

1

≤

2

m

0

+

1

.

Making some algebraic manipulations, we obtain the following:

p

≤

2

m

0

+

1

1

+

p

-

1

p

-

1

≥

2

-

m

0

-

1

⁢

(

1

+

p

-

1

)

p

-

1

-

2

-

m

0

-

1

⁢

(

1

+

p

-

1

)

≥

0

1

+

p

-

1

-

2

-

m

0

-

1

⁢

(

1

+

p

-

1

)

≥

1

(

1

-

2

-

m

0

-

1

)

⁢

(

1

+

p

-

1

)

≥

1

(

1

-

2

-

m

0

-

1

)

⁢

(

1

+

p

-

2

)

≥

1

-

p

-

1

According to our earlier inequality, this means that

2

m

0

⁢

p

is either perfect or abundant.

Suppose that

2

m

⁢

p

n

is abundant. Then, by the previous
result,

m

≤

m

0

; since powers of

2

are deficient,

n

>

0

,
so

2

m

0

p

∣

2

m

p

n

.
∎

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

2

m

0

⁢

p

is
perfect, we have listed

2

m

0

+

1

⁢

p

and

2

m

0

⁢

p

2

as
these numbers are abundant.

3

12

(

2

2

⋅

3

)

,

18

(

2

⋅

3

2

)

5

20

(

2

2

⋅

5

)

7

56

(

2

3

⋅

7

)

,

196

(

2

2

⋅

7

2

)

11

88

(

2

3

⋅

11

)

13

104

(

2

3

⋅

13

)

17

272

(

2

4

⋅

17

)

19

304

(

2

4

⋅

19

)

23

368

(

2

4

⋅

23

)

29

464

(

2

4

⋅

29

)

31

992

(

2

5

⋅

31

)

,

7688

(

2

5

⋅

31

2

)

37

1184

(

2

5

⋅

37

)

41

1312

(

2

5

⋅

41

)

43

1376

(

2

5

⋅

43

)

47

1504

(

2

5

⋅

47

)

53

1696

(

2

5

⋅

53

)

59

1888

(

2

5

⋅

59

)

61

1952

(

2

5

⋅

61

)

67

4208

(

2

6

⋅

67

)

71

4544

(

2

6

⋅

71

)

73

4672

(

2

6

⋅

73

)

79

5056

(

2

6

⋅

79

)

83

5312

(

2

6

⋅

83

)

89

5696

(

2

6

⋅

89

)

97

6208

(

2

6

⋅

97

)

101

6464

(

2

6

⋅

101

)

103

6592

(

2

6

⋅

103

)

107

6848

(

2

6

⋅

107

)

109

6976

(

2

6

⋅

109

)

113

7232

(

2

6

⋅

113

)

127

16256

(

2

7

⋅

127

)

,

1032256

(

2

6

⋅

127

2

)

131

16768

(

2

7

⋅

131

)

137

17536

(

2

7

⋅

137

)

139

17792

(

2

7

⋅

139

)

149

19072

(

2

7

⋅

149

)

151

19328

(

2

7

⋅

151

)

157

20096

(

2

7

⋅

157

)

163

20864

(

2

7

⋅

163

)

167

21376

(

2

7

⋅

167

)

173

22144

(

2

7

⋅

173

)

179

22912

(

2

7

⋅

179

)

181

23168

(

2

7

⋅

181

)

191

24448

(

2

7

⋅

191

)

193

24704

(

2

7

⋅

193

)

197

25216

(

2

7

⋅

197

)

199

25472

(

2

7

⋅

199

)

211

27008

(

2

7

⋅

211

)

223

28544

(

2

7

⋅

223

)

227

29056

(

2

7

⋅

227

)

229

29312

(

2

7

⋅

229

)

233

29824

(

2

7

⋅

233

)

239

30592

(

2

7

⋅

239

)

241

30848

(

2

7

⋅

241

)

251

32128

(

2

7

⋅

251

)

Title

determination of even abundant numbers with one odd prime factor

Canonical name

DeterminationOfEvenAbundantNumbersWithOneOddPrimeFactor

Date of creation

2013-03-22 16:47:51

Last modified on

2013-03-22 16:47:51

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

24

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 11A05

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