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All Roads Lead To Rome Conjecture

Main post link -> http://www.piyushclipboards.blogspot.in/2013/07/all-roads-lead-to-rome-conjecture.html

Here is a very simple but interesting conjecture.

Take any natural number n>=5. If n is composite (a number having factors other than 1 and itself), add up all of its prime factors. If n is prime (a number having only two factors: 1 and the number itself), just add one to it. Repeat the process indefinitely. The conjecture states that no matter what number you start with, you shall always eventually reach the 'Perfect Number 6'.

Can anybody prove it??

Note by Piyush Gupta
4 years ago

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Let \(f(n)\) be the sum of the prime factors of \(n\) if \(n\) is composite and \(n+1\) if \(n\) is prime.

We must show that the sequence \(n, f(n), f(f(n)), f(f(f(n))), \ldots\) has \(6\) in it.

Note that \(f(5) = 6\), \(6 = 6\), and \(f(f(7)) = f(8) = 6\), Clearly, the conjecture holds for \(5 \le n \le 7\).


If \(n \ge 8\) is composite, let \(n = \prod p_i\) for some primes \(p_i\) not necessarily distinct. Then \(f(n) = \sum p_i\).

If \(f(n) < 5\), then we must have the sum of two or more primes less than \(5\).

The only possibility is \(2+2 = 4\), but then \(n = 2 \cdot 2 = 4 < 8\), a contradiction. Therefore, \(f(n) \ge 5\).

Also, for each \(i\), we have \(p_i \le \dfrac{n}{2}\). So, \(\dfrac{\ln p_i}{p_i} \ge \dfrac{\ln (n/2)}{(n/2)}\), and thus, \(\ln n = \sum \ln p_i \ge \dfrac{\ln (n/2)}{(n/2)} \sum p_i\).

Therefore, \(f(n) = \sum p_i \le \dfrac{n}{2} \dfrac{\ln n}{\ln(n/2)} = \dfrac{n}{2} \dfrac{\ln n}{\ln n - \ln 2} = \dfrac{n}{2} \dfrac{1}{1 - \tfrac{\ln 2}{\ln n}} \le \dfrac{n}{2} \cdot \dfrac{3}{2} = \dfrac{3}{4}n \le n-2\).

Hence, if \(n \ge 8\) is composite, we have \(5 \le f(n) \le n-2\).


If \(n \ge 8\) is prime, then \(n+1 \ge 8\) is composite. Therefore, \(5 \le f(f(n)) = f(n+1) \le n-1\).


Therefore, if \(n \ge 8\), we have either \(5 \le f(n) \le n-2\) or \(5 \le f(f(n)) \le n-1\).

So, by principle of infinite descent, we have \(5 \le f^{(k)}(n) \le 7\) for some finite positive integer \(k\).

Then, since \(f(5) = 6\), \(6 = 6\), and \(f(f(7)) = f(8) = 6\), the conjecture holds for all \(n \ge 8\).

Therefore, the conjecture holds for all \(n \ge 5\) as desired. Jimmy Kariznov · 4 years ago

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This reminds me of the mystery of the number 276. Find the sum of all of the proper divisors of 276, and then the sum of the proper divisors of that number, and repeat. Does this sequence ever settle down or become periodic? For perfect numbers, this sequence just repeats: s(6)=1+2+3=6...

If you can prove something about this, then it'll be worth publishing! Eric Edwards · 4 years ago

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@Eric Edwards Of course, it is much harder (if not impossible) to show that \(n, s(n), s(s(n)), \ldots\) eventually has a term less than \(n\). This makes that problem (as well as the Collatz Conjecture) difficult to solve. Jimmy Kariznov · 4 years ago

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I don't understand..i took 8, only prime factor is 2, i didn't end up in 6? am i supposed to add 2 three times since \(2^{3} = 8\)? Nishanth Hegde · 4 years ago

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@Nishanth Hegde Yeah, \(8\) has three prime factors \(2\). Tim Vermeulen · 4 years ago

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