I am learning a paper about deficient-perfect numbers. I found a equation that I still can't understand it.
We start from definition: For a positive integer n, let σ(n) denote the sum of the positive divisors of n. Let d be a proper divisor of n. We call n a deficient-perfect number if σ(n)=2n−d.
For example, 10 is deficient perfect number with d=2. because $\sigma(10) =1+2+5+10=18$, we have $2n-d=2×10-2=18=\sigma(10) $. If n is an odd number, so we say n is an odd deficient perfect number.
Assume $n=3^{\alpha_1}×5^{\alpha_2}×p^{\alpha_3}$ is an odd-deficient perfect numbers with $p>5$ be a prime and $\alpha_i$ is even.
Let $f(\alpha_1,\alpha_2,\alpha_3)=(1-\frac{1}{3^{{\alpha_1}+1}})(1-\frac{1}{5^{{\alpha_2}+1}})(1-\frac{1}{p^{{\alpha_3}+1}})$
If I have sum of positive divisors of $n$ is
$$\sigma(n)=2\times{3^{\alpha_1}}{5^{\alpha_2}}{p^{\alpha_3}}-{3^{\beta_1}}{5^{\beta_2}}{p^{\beta_3}} \tag{2.1}$$ where $\beta_i<=\alpha_i$
Also, Let
$$g(\alpha_1,\alpha_2,\alpha_3)=\frac{2^4 \times(p-1)}{3\times5\times p}-\frac{2^3 \times(p-1)}{{3^{\alpha_1-\beta_1+1}}{5^{\alpha_2-\beta_2+1}}{p^{\alpha_3-\beta_3+1}}}$$
So, using Equation (2.1), we get $$f=g$$
I can't prove $f=g$. I was just told that function $g$ is defined in such a way using equation (2.1) such that $f=g$.
In the case above, I think this just use elementary techniques. I tried with some possible, like $f\sigma(n) =g\sigma(n) $, but I can't still found the way. This is my first time finding two function with different forms, but their can become equal using some way.
Is there anybody can help me? I really appreciate your answer.
For details, see on this image:
The full paper you can see here.