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Given a positive integer $n$, I've workout out a formula which involves the expression "sum of distinct primes dividing n" minus "number of distinct prime factors of n."

Are there any known inequalities for "sum of distinct primes dividing n" or for "number of distinct prime factors of n" that hold for all $n$?

I'm not interested in asymptotics.

  1. number of distinct prime factors: http://oeis.org/A001221
  2. sum of distinct prime factors: http://oeis.org/A008472
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  • $\begingroup$ What you ask in the question is not quite complete and understandable !! Please try to edit your question and make it more precise. You are just giving a function (or rather a number) and asking to find inequalities about that !! You see there are plenty of inequalities that can be created with a given number and that too without any computer aid and if you use a computer then you will get mesmerizing results and lots and lots of inequalities. So you edit this question or do something to make it more precise and exact. $\endgroup$ – adityaguharoy Jul 22 '17 at 6:07
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There are a number of obvious inequalities (thus "known" in the sense of "derivable by elementary methods") but they are usually too weak to be of interest. For prime numbers n, the sum of primes dividing them is n, for numbers of the form $p^aq^b$ the sum of their distinct prime factors is $p+q$, which is usually less than n and is less than n/2 when p is greater than 2 and q is greater than 5, but it is not clear how one goes from there or what one does with the inequalities derived. Similarly, the number of distinct prime factors of n is (for n not too small) less than log n , and is almost always less than (log n)/(log (log n)) (maybe there are 5 exceptions to this?), and again there are not so many uses for this. However, I am using these latter estimates in analysis of the resource usage of some number theory programs I am writing. Perhaps you could think of applications for such formulas and then search for those applications.

Gerhard "Make A Market For It" Paseman, 2017.07.21.

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  • $\begingroup$ It looks like several primorials and some of their factors n have more than log n/(log log n) distinct prime factors. More than 5 of them anyway. Gerhard "Still Believes Only Finitely Many" Paseman, 2017.07.21. $\endgroup$ – Gerhard Paseman Jul 22 '17 at 6:53
  • $\begingroup$ The post of GH from MO reminds me of oscillation results of Chebyshev's first function (log of primordial), so I now no longer believe there are finitely many exceptions. I suspect only finitely many of them are not primorials, however. Gerhard "Always Have A Fallback Position" Paseman, 2017.07.22. $\endgroup$ – Gerhard Paseman Jul 22 '17 at 16:18
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It is known that $$ \omega(n):=\sum_{p\mid n}1\leq(1 + o(1))\frac{\log n}{\log\log n},$$ and this is optimal (see e.g. Section 5.3 in Tenenbaum: Introduction to analytic and probabilistic number theory). Regarding the second quantity, we have the following lower bound that is optimal under $\omega(n)\to\infty$: $$ \sum_{p\mid n} p\geq \left(\frac{1}{2}+o(1)\right)\omega(n)^2\log\omega(n).$$ See also this somewhat related MO question.

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    $\begingroup$ Thanks for the first formula: it has adjusted my thinking. In what sense is the second formula true? I don't see how it applies to numbers n of the form twice a prime or three times a prime. Perhaps you are thinking of averaged values? Gerhard "Not Seeing The Miracle Occurring" Paseman, 2017.07.22. $\endgroup$ – Gerhard Paseman Jul 22 '17 at 16:45
  • $\begingroup$ @GerhardPaseman: The second formula is obviously nonsense as stated. I am on vacation and typed this without much thinking. Let me replace it with something that is true. $\endgroup$ – GH from MO Jul 22 '17 at 17:44

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