A308964 Least k > 0 such that A014222(k) == A014222(k+1) (mod n).

1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 3, 1, 2, 2, 2, 2, 3, 2, 3, 2, 2, 3, 4, 1, 3, 2, 2, 2, 3, 2, 3, 2, 3, 3, 2, 2, 3, 3, 2, 2, 2, 2, 3, 3, 2, 4, 5, 2, 3, 3, 3, 2, 3, 2, 3, 2, 3, 3, 4, 2, 3, 3, 2, 3, 2, 3, 4, 3, 4, 2, 3, 2, 2, 3, 3, 3, 3, 2, 3, 2, 3, 2, 3, 2, 3, 3, 3

Offset: 1

Author

Jinyuan Wang, Aug 30 2019

Keywords

• nonn

Comments

a(n) is the least positive integer k such that the sequence {A014222(i) mod n} for i >= k is constant. Proof: if g(i) = A014222(i), then n divides 3^g(i-1)*(3^(g(i)-g(i-1)) - 1) when g(i) == g(i+1) (mod n). g(j)-g(j-1) divides g(j+1)-g(j) for any j > 0, therefore, 3^(g(i)-g(i-1)) - 1 divides 3^(g(i+1)-g(i)) - 1. Then n divides 3^g(i)*(3^(g(i+1)-g(i)) - 1) = g(i+2) - g(i+1), that is, g(i) == g(i+1) == g(i+2) (mod n). Since a(n) is the least positive integer k such that g(k) == g(k+1) (mod n), the sequence {A014222(i) mod n} for i >= k is constant.

A014222(k+1) - A014222(k) is the largest n such that a(n) = k.

Examples

			3, 27, 7625597484987, ... mod 5 equal 3, 2, 2, ..., so A014222(k) mod 5 = 2 for all k >= 2, hence a(5) = 2.
		

Crossrefs

Cf. A014222, A308972, A322418.

Programs

PARI

a(n) = {my(c=0, k=1, x=0, d=n); while(k==1, z=x++; y=0; b=1; while(z>0, while(y++
				

Formula

a(n) <= A003434(n).

a(n) <= a(A000010(n)) + 1.