All Roads Lead To Rome Conjecture

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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
6 years, 6 months ago

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9 votes

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

We must show that the sequence n,f(n),f(f(n)),f(f(f(n))),n, f(n), f(f(n)), f(f(f(n))), \ldots has 66 in it.

Note that f(5)=6f(5) = 6, 6=66 = 6, and f(f(7))=f(8)=6f(f(7)) = f(8) = 6, Clearly, the conjecture holds for 5n75 \le n \le 7.


If n8n \ge 8 is composite, let n=pin = \prod p_i for some primes pip_i not necessarily distinct. Then f(n)=pif(n) = \sum p_i.

If f(n)<5f(n) < 5, then we must have the sum of two or more primes less than 55.

The only possibility is 2+2=42+2 = 4, but then n=22=4<8n = 2 \cdot 2 = 4 < 8, a contradiction. Therefore, f(n)5f(n) \ge 5.

Also, for each ii, we have pin2p_i \le \dfrac{n}{2}. So, lnpipiln(n/2)(n/2)\dfrac{\ln p_i}{p_i} \ge \dfrac{\ln (n/2)}{(n/2)}, and thus, lnn=lnpiln(n/2)(n/2)pi\ln n = \sum \ln p_i \ge \dfrac{\ln (n/2)}{(n/2)} \sum p_i.

Therefore, f(n)=pin2lnnln(n/2)=n2lnnlnnln2=n211ln2lnnn232=34nn2f(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 n8n \ge 8 is composite, we have 5f(n)n25 \le f(n) \le n-2.


If n8n \ge 8 is prime, then n+18n+1 \ge 8 is composite. Therefore, 5f(f(n))=f(n+1)n15 \le f(f(n)) = f(n+1) \le n-1.


Therefore, if n8n \ge 8, we have either 5f(n)n25 \le f(n) \le n-2 or 5f(f(n))n15 \le f(f(n)) \le n-1.

So, by principle of infinite descent, we have 5f(k)(n)75 \le f^{(k)}(n) \le 7 for some finite positive integer kk.

Then, since f(5)=6f(5) = 6, 6=66 = 6, and f(f(7))=f(8)=6f(f(7)) = f(8) = 6, the conjecture holds for all n8n \ge 8.

Therefore, the conjecture holds for all n5n \ge 5 as desired.

Jimmy Kariznov - 6 years, 6 months 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 - 6 years, 6 months ago

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Of course, it is much harder (if not impossible) to show that n,s(n),s(s(n)),n, s(n), s(s(n)), \ldots eventually has a term less than nn. This makes that problem (as well as the Collatz Conjecture) difficult to solve.

Jimmy Kariznov - 6 years, 6 months 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 23=82^{3} = 8?

Nishanth Hegde - 6 years, 6 months ago

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Yeah, 88 has three prime factors 22.

Tim Vermeulen - 6 years, 6 months ago

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