# Determining when $3 \cdot 5^a \cdot 7$ is abundant.

I would like to determine the values of $a$ for which $3 \cdot 5^a \cdot 7$ is abundant.

My work so far: $$\sigma(3 \cdot 5^a \cdot 7) > 2 \cdot 3 \cdot 5^a \cdot 7 = 42 \cdot 5^a \Leftrightarrow$$ $$\sigma(3) \cdot \sigma (5^a) \cdot \sigma (7) > 42 \cdot5^a \Leftrightarrow$$

$$(4) \cdot \left ( \sum_{k = 0}^a 5^k\right ) \cdot (8) > 42 \cdot 5^a$$ ... And since the sum contains $5^a$ in it, I thought about trying to cancel that from both sides, but am stuck.

Can I get a nudge in the right direction? (Also, if there is a theoretic result that I should be using, feel free to mention it!)

Added: Using Will Jagy's hint, I now have $$8 \cdot (5^{a + 1} - 1) = 40 \cdot 5^a - 8 > 42 \cdot 5^a$$ which appears to have no solution.

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And I wonder why there are so few questions with "abundant" in them on MSE? – The Chaz 2.0 Feb 29 '12 at 3:09
Indeed, abundant posts are not abundant. – Gone Feb 29 '12 at 3:14
Correct, no solution. Why look at this pattern? – Will Jagy Feb 29 '12 at 3:20
@Will: cramster – The Chaz 2.0 Feb 29 '12 at 3:30

For all $a\ge 0$, $\sigma(p^a)=(p^{a+1}-1)/(p-1)$, so $$\frac{\sigma(p^a)}{p^a}=\frac{p^{a+1}-1}{p^{a+1}-p^{a}}\le \frac{p^{a+1}}{p^{a+1}-p^a}=\frac{p}{p-1}.$$ Therefore, $$\frac{\sigma(3\cdot 5^a\cdot 7)}{3\cdot 5^a\cdot 7}\le \frac{3^2-1}{3^2-3}\cdot \frac{5}{5-1}\cdot \frac{7^2-1}{7^2-7} = \frac{40}{21}<2,$$ so no number of the form $3\cdot 5^a\cdot 7$ can be abundant.

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$$\sigma(p^a) = \frac{p^{a+1}-1}{p-1}$$

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 I have incorporated this into my most recent edit. – The Chaz 2.0 Feb 29 '12 at 3:19

There is a function, about 100 years old, that gives abundant numbers, indeed what are called colossally abundant numbers. I call $f(\delta)$ the corresponding number for $1 \geq \delta > 0.$

I calculate $$f(1) = 1, \; f(1/2) = 2, \; f(1/4) = 6, \; f(1/6) = 12, \; f(1/10) = 60, \; f(1/12) = 120,$$ then $$f(1/14) = 360, \; f(1/17) = 2520, \; f(1/25) = 5040, \; f(1/31) = 55440, \; f(1/39) = 720720,$$ and so on as $\delta$ decreases.

Given some $\delta > 0,$ the correct exponent for some prime $p$ is $$\left\lfloor \frac{\log (p^{1 + \delta} - 1) - \log(p^\delta - 1)}{\log p} \right\rfloor \; - \; 1.$$
This is Theorem 10 on page 455 of Alaoglu and Erdos. For a fixed $\delta,$ the exponents either stay the same or decrease for increasing $p,$ and eventually the exponent 0 is reached. In particular, if $$\frac{\log \left(1 + \frac{1}{p} \right)}{\log p} \; < \; \delta,$$ the prime $p$ is not a factor of the number.

If you want the first (largest) $\delta$ for which a favorite prime $p$ gets assigned exponent $k,$ let $$\delta = \frac{\log(p^{k+1} - 1) - \log(p^{k+1} - p)}{\log p}$$

See OEIS and WOOKIE

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 This will take me a little longer to process! – The Chaz 2.0 Feb 29 '12 at 4:10 @TheChaz, see if you can program a function that takes $1 \geq \delta > 0$ and gives you $f(\delta)$ and the prime factorization of $f(\delta),$ then experiment with that. If you can get it to find $\sigma(f(\delta))$ and, say, $\sigma(f(\delta)) / (f(\delta)),$ that would be educational. – Will Jagy Feb 29 '12 at 4:27