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determination of abundant numbers
with specified prime factors∗
rspuzio†
2013-03-21 22:13:34
The formula for sums of factors may be used to find all abundant numbers with a specified set of prime factors or that no such numbers exist. To
accomplish this, we first do a little algebraic manipulation to our formula.
Qk
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Theorem 1. A number n whose factorization into prime numbers is i=1 pm
i
is abundant if and only if
k Y
1 − p−mi −1
i
> 2.
1 − p−1
i
i=1
Proof. By definition n is abundant, if the sum of the proper divisors of n is
greater than n. Using our formula, this is equivalent to the condition
k mi +1
Y
p
−1
i
pi − 1
i=1
>2
k
Y
i
pm
i .
i=1
Dividing the k-th term in the product on the left-hand side by the k-th term on
the right-hand side,
i
i +1
i −1
−1
pi−mi −1 pi − p−m
1 − p−m
1 pm
i
i
i
,
=
=
i
pm
pi − 1
pi − 1
p−1
1 − p−1
i
i
i
so the condition becomes
k Y
1 − p−mi −1
i
i=1
1 − p−1
i
>2
∗ hDeterminationOfAbundantNumbersWithSpecifiedPrimeFactorsi created: h2013-03-21i
by: hrspuzioi version: h39028i Privacy setting: h1i hTheoremi h11A05i
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Note that each of the terms in the product is bigger than 1. Furthemore,
the k-th term is bounded by
1
pi
−1 = p − 1 .
1 − pi
i
This means that it is only possible to have an abundant number whose prime
factors are {pi | 1 ≤ i ≤ k} if
k Y
pi
> 2.
pi − 1
i=1
As it turns out, the convers also holds, so we have a nice criterion for determining
when a set of prime numbers happens to be the set of prime divisors of an
abundant number.
Theorem 2. A finite set S of prime numbers is the set of prime divisors of an
abundant number if and only if
Y p > 2.
p−1
p∈S
Proof. As described above, if S is a set of prime factors of an abundant number,
then we may bound each term in the inequality of the previous theorem to obtain
the inequality in the current theorem. Assume, then that S is a finite set of
prime numbers which satisfies said inequality. Then, by continuity, there must
exist a real number > 0 such that
Y p
−x >2
p−1
p∈S
whenver 0 < x < . Since limk→∞ n−k = 0 when n > 1, we can, for every p ∈ S,
find an m(p) such that
pm(p)
< .
p−1
Hence,
Y
Y p
pm(p)
1 − p−m(p)−1
−
=
>2
p−1 p−1
1 − p−1
p∈S
so, by the previous theorem,
p∈S
Q
p∈S
pm(p) is an abundant number.
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