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 and, for any , we have , it follows that, for every prime , there will exist abundant numbers of the form . By the first theorem, for such a number to be abundant, we must have
or, after a little algebraic simplification,
From this inequality, we can deduce a description of abundant numbers of the form .
Let be prime. Then, for all , we find that is abundant for sufficiently large.
When , we have
Since , it follows that
for sufficiently large, hence will be abundant for sufficiently large. ∎
Let be prime. Then there exists such that:
If , then is not abundant for any .
if , then is abundant for sufficiently large.
On the one hand, if , then it will be impossible to satisfy our criterion for any choice of . On the other hand, if , then, by ther same sort of continuity argument employed previously, the criterion will be satisfied for sufficiently large. ∎
Let be prime and let be the unique integer such that
Then is either perfect or abundant and every abundant number of the form is a multiple of .
We begin with the equation
Making some algebraic manipulations, we obtain the following:
According to our earlier inequality, this means that is either perfect or abundant.
Suppose that is abundant. Then, by the previous result, ; since powers of are deficient, , so . ∎
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 is perfect, we have listed and as these numbers are abundant.
|Title||determination of even abundant numbers with one odd prime factor|
|Date of creation||2013-03-22 16:47:51|
|Last modified on||2013-03-22 16:47:51|
|Last modified by||rspuzio (6075)|