A306833 a(1) = 3; a(n+1) is the smallest k > a(n) such that 2^(k-1) == 1 (mod a(n)).

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

Offset

1,1

Comments

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

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

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.

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

Links

Table of n, a(n) for n=1..15.

Prog

(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

Crossrefs

Cf. A002326, A036259, A036260.

Sequence in context: A144933 A133033 A145365 * A146433 A134672 A321024

Adjacent sequences: A306830 A306831 A306832 * A306834 A306835 A306836

Keyword

nonn,fini,full

Author

Thomas Ordowski, Mar 12 2019

Status

approved