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a(n) is the smallest integer k such that the iterated multiplicative-order chain k -> ord_k(2) -> ord_{ord_k(2)}(2) -> ... reaches an even number in exactly n steps.
0

%I #32 Jul 05 2026 05:08:57

%S 2,3,7,47,1439,2879,214559,429119,858239,164255999

%N a(n) is the smallest integer k such that the iterated multiplicative-order chain k -> ord_k(2) -> ord_{ord_k(2)}(2) -> ... reaches an even number in exactly n steps.

%C The chain is formed by repeating the following: for any integer q, calculate ord_q(2) (the multiplicative order of 2 modulo q), then continue this process. The chain stops when it reaches an even number.

%e a(1): 3 -> 2.

%e a(2): 7 -> 3 -> 2.

%e a(3): 47 -> 23 -> 11 -> 10.

%e a(4): 1439 -> 719 -> 359 -> 179 -> 178.

%e a(5): 2879 -> 1439 -> 719 -> 359 -> 179 -> 178.

%e a(6): 214559 -> 107279 -> 53639 -> 2063 -> 1031 -> 515 -> 204.

%e a(7): 429119 -> 214559 -> 107279 -> 53639 -> 2063 -> 1031 -> 515 -> 204.

%e a(8): 858239 -> 429119 -> 214559 -> 107279 -> 53639 -> 2063 -> 1031 -> 515 -> 204.

%o (PARI) isok(k, n) = for (i=1, n-1, k = znorder(Mod(2, k)); if (!(k%2), return(0))); k = znorder(Mod(2, k)); if (k % 2, 0, 1);

%o a(n) = my(k=1); while(!isok(k, n), k+=2); k; \\ _Michel Marcus_, Jun 11 2026

%o (Python)

%o from sympy import n_order

%o def steps(k):

%o q, s = k, 0

%o while q % 2 == 1:

%o q = n_order(2, q)

%o s += 1

%o return s

%o def a(n):

%o k = 2

%o while steps(k) != n:

%o k += 1

%o return k

%o print([a(n) for n in range(6)])

%Y Cf. A005384, A059452, A002326.

%K nonn,hard,more,new

%O 0,1

%A _Nishant R. Gautam_, Jun 11 2026

%E a(9) from _Michel Marcus_, Jun 11 2026