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a(0) = 1; a(n) is the smallest integer k > a(n-1) such that 2^(k-1) == 1 (mod a(n-1)*k).
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%I #22 Jun 06 2021 07:59:27

%S 1,3,5,13,37,73,109,181,541,1621,4861,9721,10531,17551,29251,87751,

%T 526501,3159001,5528251,11056501,44226001,49385701,98771401,172849951,

%U 345699901,352755001,564408001,634959001,793698751,793886887,4763321317,4822127753

%N a(0) = 1; a(n) is the smallest integer k > a(n-1) such that 2^(k-1) == 1 (mod a(n-1)*k).

%C For n > 0, a(n) is prime or pseudoprime (a Fermat pseudoprime to base 2).

%C It seems that for any odd initial term a(0), this recursion gives at most finitely many composite terms (which were not found in this sequence).

%C Conjecture: a(n) is prime for every n > 0, namely a(n) is the smallest odd prime p > a(n-1) such that 2^(p-1) == 1 (mod a(n-1)), with a(0) = 1.

%H Amiram Eldar, <a href="/A306826/b306826.txt">Table of n, a(n) for n = 0..1000</a>

%t A = {1}; While[Length[A] < 500, a = Last[A]; r = MultiplicativeOrder[2, a]; k = a + r; While[PowerMod[2, k - 1, k a] != 1, k = k + r]; AppendTo[A, k]]; Take[A, 75] (* Emmanuel Vantieghem, Apr 02 2019 *)

%Y Cf. A175257, A307244.

%K nonn

%O 0,2

%A _Thomas Ordowski_, Mar 12 2019

%E More terms from _Amiram Eldar_, Mar 12 2019