Problems & Puzzles: Puzzles

 Puzzle 893. Primes as a concatenation of a series of powers of 3, with 6m terms. This is a follow-up to the results obtained to Puzzle 892. For S(p, 6m) = p^0&p^1&...&p^(6m-1) Jan van Delden & Emmanuel Vantieghem proved that S(p, 6m) is always an integer divisible by 3, for any prime p>3 and for any m>0 value. What if p=3 and the quantity of concatenated terms is still 6m? May S(3,6m) = 3^0&3^1&...^&3^(6m-1) be a prime for some m value? As both of them stated, in such a case S(3,6m) is an odd integer never divided by 3, so it could produce a prime integer. The search was extended until 666 concatenated terms or 105987 digits by Jan van Delden and no probable prime was still found (See Q. Can you extend this search until a probable prime is detected?

Contribuion came from Jan van Delden

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Jan wrote on Set 21, 2017:

I extended the search until a sequence of length 1398,  S(3,1396) has 466608 digits. No primes were found.

I decided to stop since one Miller-Rabin test takes about 18 hours (single thread) for numbers this size.,,For k>666 I only tested k=6m.

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