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 2020-02-13, 12:52 #1 enzocreti   Mar 2018 72·11 Posts Constellations of multiples of 3 in this sequence 47, 51, 52, 59, 68, 70, 75, 79, 90, 94,95, 102, 111 This is a sequence of numbers that reduced mod 41 are powers. Are there infinitely many primes in this sequence? Last fiddled with by enzocreti on 2020-02-13 at 14:00
2020-02-14, 07:15   #2
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

19×541 Posts

Quote:
 Originally Posted by enzocreti 47, 51, 52, 59, 68, 70, 75, 79, 90, 94,95, 102, 111 This is a sequence of numbers that reduced mod 41 are powers. Are there infinitely many primes in this sequence?
Really? Whose power is 6? (i.e. .47 (mod 41))

2020-02-14, 12:20   #3
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

368210 Posts

Quote:
 Originally Posted by enzocreti 47, 51, 52, 59, 68, 70, 75, 79, 90, 94,95, 102, 111 This is a sequence of numbers that reduced mod 41 are powers. Are there infinitely many primes in this sequence?
Of course!!! By Dirichlet's theorem on arithmetic progressions, there are infinitely many primes = 4 mod 41, thus there are infinitely many such primes.

2020-02-14, 12:21   #4
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

2×7×263 Posts

Quote:
 Originally Posted by enzocreti 47, 51, 52, 59, 68, 70, 75, 79, 90, 94,95, 102, 111 This is a sequence of numbers that reduced mod 41 are powers. Are there infinitely many primes in this sequence?
Wait!!! 47%41 = 6, and 6 is not perfect power....

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