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A307244 a(0) = 1; a(n) is the smallest integer k > a(n-1) such that 3^(k-1) == 1 (mod a(n-1)*k). 1

%I #19 Apr 10 2019 10:24:54

%S 1,2,5,13,19,37,73,97,193,241,601,751,2251,3001,4001,16001,96001,

%T 160001,1120001,4480001,13440001,20160001,23385601,29232001,36540001,

%U 38628001,115884001,231768001,579420001,1448550001,1931400001,2172825001,6518475001,22814662501,53234212501,425873700001,1703494800001

%N a(0) = 1; a(n) is the smallest integer k > a(n-1) such that 3^(k-1) == 1 (mod a(n-1)*k).

%C For n > 0, a(n) is prime or pseudoprime (a Fermat pseudoprime to base 3).

%C Conjecture: a(n) is prime for every n > 0, namely a(n) is the smallest prime p > a(n-1) different from 3 such that 3^(p-1) == 1 (mod a(n-1)), with a(0) = 1.

%C Generally: for a fixed integer base b > 1, a(n) is the smallest k > a(n-1) such that b^(k-1) == 1 (mod a(n-1)*k), with a(0) = 1. For n > 0, a(n) is prime or pseudoprime (a Fermat pseudoprime to base b). If for a base b, a(n) is a prime for every n > 0, then a(n) is the smallest prime p > a(n-1) that does not divide b such that b^(p-1) == 1 (mod a(n-1)), with a(0) = 1. For any integer base b > 1, a(n) is prime for almost all n. Seems that at most finitely many terms are composite.

%t A = {1}; While[Length[A] < 500, a = Last[A]; r = MultiplicativeOrder[3, a]; k = a + r; While[PowerMod[3, k - 1, k a] != 1, k = k + r]; AppendTo[A, k]]; Take[A, 75] (* _Emmanuel Vantieghem_, Mar 31 2019 *)

%Y Cf. A306826.

%K nonn

%O 0,2

%A _Thomas Ordowski_, Mar 30 2019

%E a(18)-a(29) from _Amiram Eldar_, Mar 30 2019

%E More terms from _Emmanuel Vantieghem_, Mar 31 2019

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Last modified May 3 06:44 EDT 2024. Contains 372206 sequences. (Running on oeis4.)