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Let p(n) be the nth prime. Find sum $p(n) + p(n+1)^2 + p(n+2)^3$ and examine the last digits for their distribution. Taking the 64 possible patterns for last digits in these sums, one get 14 for ending digits 1 and 9, and 12 each for ending digits 3,5,7. A few small experiments in primes close to 200,000 gives the order for most at 7, followed by 1,3 and then 9 and 5. Could one be able to predict the distribution once larger primes were considered? Do not forget the paper by Soundararajan and Oliver.
https://arxiv.org/pdf/1603.03720.pdf

J. M. Bergot
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  • Does "sum $p(n)+p(n+1)^2$ and $p(n+2)^3$" simply mean "$p(n)+p(n+1)^2+p(n+2)^3$? – Hagen von Eitzen Dec 06 '17 at 19:46
  • Anecdotal evidence using $10^6+2$ primes after $10^{14}\pi$: $220157\times 1, 179241\times 3, 188820\times 5, 196034\times 7, 215748\times 9$. With primes after $10^{20}e$, the counts are $220116, 180911, 188483, 194046, 216444$. The two stats look more similar to one another than to equideistribution – Hagen von Eitzen Dec 06 '17 at 19:55
  • https://www.google.ca/search?q=soundararajan+and+oliver+on+primes&rlz=1C1GGRV_enCA774&oq=soundararajan+and+oliver&aqs=chrome.0.69i59j69i57j0.11440j0j8&sourceid=chrome&ie=UTF-8 – J. M. Bergot Dec 06 '17 at 20:09
  • https://www.google.ca/search?q=soundararajan+and+oliver+on+primes&rlz=1C1GGRV_enCA774&oq=soundararajan+and+oliver&aqs=chrome.0.69i59j69i57j0.11440j0j8&sourceid=chrome&ie=UTF-8 – J. M. Bergot Dec 06 '17 at 20:10
  • I couldn't get the link to print. Google Soundararajan and Oliver to get the paper on primes. – J. M. Bergot Dec 06 '17 at 20:11
  • Yes as in 3 + 5^2 + 7^3 = 3 + 25 + 343 = 371. An experiment found that in 28 trials only five primes were found; three were from consecutive trials. – J. M. Bergot Dec 06 '17 at 20:12
  • The order then from most to least is ending digits 1,9,3,5,7 for both cases. I assume these were computed and that this order would not change even for high primes examined. Do you agree? Do you also agree that NO method exists to predict these results before any computations are done? – J. M. Bergot Dec 06 '17 at 20:23
  • Make the order from most to least the digits 1,9,7,5,3. – J. M. Bergot Dec 07 '17 at 18:15

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