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a(1) = 3; a(n+1) is the smallest k > a(n) such that 2^(k-1) == 1 (mod a(n)).
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%I #14 Apr 10 2019 10:23:35

%S 3,5,9,13,25,41,61,121,221,241,265,313,469,529,760

%N a(1) = 3; a(n+1) is the smallest k > a(n) such that 2^(k-1) == 1 (mod a(n)).

%C This sequence is finite, the last term a(15) = 760 is even.

%C Conjecture: for any initial term a(1), this recursion gives a finite sequence (ends with an even term).

%C Theorem: for odd a(n), a(n+1) is even if and only if ord_{a(n)}(2) is odd and (a(n) mod ord_{a(n)}(2)) is odd.

%C The set of penultimate terms of the sequences is {A036259} \ {A036260}.

%o (PARI) lista(nn) = {a = 3; print1(a, ", "); for (n=2, nn, k = a+1; while (Mod(2, a)^(k-1) != 1, k++); a = k; print1(a, ", "); if (!(a%2), break););} \\ _Michel Marcus_, Mar 24 2019

%Y Cf. A002326, A036259, A036260.

%K nonn,fini,full

%O 1,1

%A _Thomas Ordowski_, Mar 12 2019