## 6.2.12

### OPN Research - Feb 06, 2012

Let N = {q^k}{n^2} be an odd perfect number with Euler
factor q^k.

Suppose sigma(n)/q < sigma(q)/n.
Then n*sigma(n) < q*sigma(q).

Since I(q) < 5/4 < I(n) (where I(x) is the abundancy index of x),
then sigma(q)/q < sigma(n)/n.  This implies that:

n*sigma(q) < q*sigma(n)

Therefore:
n*(sigma(q) + sigma(n)) < q*(sigma(q) + sigma(n))

Consequently:
n < q

Result #1:  If sigma(n)/q < sigma(q)/n, then n < q.

Note that, in this case:
sigma(n)/q^k <= sigma(n)/q < sigma(q)/n <= sigma(q^k)/n.

Corollary #1:  If sigma(n)/q^k < sigma(q^k)/n, then
n < q^k.

Proof of Corollary #1:
Suppose sigma(n)/q^k < sigma(q^k)/n.   Then

n*sigma(n) < {q^k}*sigma(q^k)

Since I(q^k) < 5/4 < I(n), then

sigma(q^k)/q^k < sigma(n)/n, which gives:

n*sigma(q^k) < {q^k}*sigma(n)

Therefore:

n*(sigma(q^k) + sigma(n)) < {q^k}*(sigma(q^k) + sigma(n))

Consequently:

n < q^k.

Now, suppose that sigma(q)/n < sigma(n)/q.  Then:

q*sigma(q) < n*sigma(n)

Again:

sigma(q)/q < sigma(n)/n, which gives:

n*sigma(q) < q*sigma(n)

Therefore:

(q + n)*sigma(q) < (q + n)*sigma(n)

Consequently:

sigma(q) < sigma(n)

Result #2:  If sigma(q)/n < sigma(n)/q,
then sigma(q) < sigma(n).

Corollary #2:  If sigma(q^k)/n < sigma(n)/q^k,
then sigma(q^k) < sigma(n).

Now, if sigma(q) < sigma(n) then:

q + 1 < sigma(n) < 2n

(q + 1)/2 < n