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This something like Waring meets the primes. Take some prime p(n) and divide it by all the primes less than one-half p(n) to find the sum of the remainders that are themselves also primes. Is there some sum of these remainders for p(s1) that is greater than p(n) for all subsequent primes greater than p(n)? Is there some p(s2) such than all primes greater than p(s2) will have a sum of remainders twice p(s2)? Some p(s3) for three times the sum of all its prime remainders for all primes greater than p(s3)? This could be continued for some p(s4), [p(s5)... Do you know if anyone has investigated this?

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  • $\begingroup$ Could you please format your question a little bit, using different paragraphs, clear notations and MathJax? Currently very difficult to read. $\endgroup$ – Arnaud Mortier Mar 16 '18 at 17:28
  • $\begingroup$ just so I understand, for 31 (n=12?), the sum of remainders of division by primes less than 15.5 is 20. and 31>20 so s1>12 yes? $\endgroup$ – phdmba7of12 Mar 16 '18 at 18:04
  • $\begingroup$ for 37 (n=13?) one of the remainders is 4 so you leave that out of the sum, correct? $\endgroup$ – phdmba7of12 Mar 16 '18 at 18:10
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    $\begingroup$ For 31, 31mod2=1, 31mod3=1, 31mod5=1, 31mod7=3, 31mod11=9, 31mod13=5--->add only the primes 3 and 5 to get 8. $\endgroup$ – J. M. Bergot Mar 16 '18 at 18:27
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Here is a totally heuristic non-rigorous attempt.

For a prime $p$, there are about $n =\dfrac{p}{2 \ln p}$ primes less than $p/2$.

For each of these primes, say $q$, the probability of the remainder being prime is about $\dfrac{1}{\ln q}$.

Therefore the sum of these remainders is about $\sum_{q < p/2} \dfrac{q}{\ln q} \sim \left(\dfrac{p}{2\ln p}\right)^2 $.

This is larger than $p$ by a factor of $\dfrac{p}{(2\ln p)^2}$ which is greater than $1$, so the sum is usually much greater than $p$.

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  • $\begingroup$ You can test the truth of what you contend by finding p(s1), p(s2) and a few others just to see how long it takes. You'd have to go a bit beyond your first answer to insure than no exceptional prime does not meet the requirement of having the sum of its prime remainders LESS than what is required; then you'd need a higher value in p(s1) or p(s2). $\endgroup$ – J. M. Bergot Mar 16 '18 at 18:10
  • $\begingroup$ For 17 divide by 3 to get rem=2; divide by 5 to get rem=2; divide by 7 to get rem=3-->2+2+3=7 and less than 17. $\endgroup$ – J. M. Bergot Mar 16 '18 at 18:20
  • $\begingroup$ It seems that your readers enjoy Grand Theory but do not bother to do the simple thing of testing if it is valid by doing some numerical examples by way of experimentation. Whatever happened to empiricism? $\endgroup$ – J. M. Bergot Mar 17 '18 at 17:45

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